· contents 1 review of limits4 1.1 the geometry of limits. . . . . . . . . . . . . . . . . . . ....

Supplemental Notes (Incomplete)

Applied Calculus

Anthony D. VanHoyDepartment of Mathematics

Humber College Insitute of Technology and Advanced Learning

(Last revised: November 26, 2015)

Copyright c© 2013 by Anthony D. VanHoyAll Rights Reserved.

Contents

1 Review of Limits 41.1 The Geometry of Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Limits of Specific Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 More on Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 Limits Involving Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2 Introduction to Differentiation 142.1 Definition of the Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . 142.2 Rules for Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.3 Differentiation of Composite Functions . . . . . . . . . . . . . . . . . . . . 192.4 Differentiation of Product and Quotient Functions . . . . . . . . . . . . . 212.5 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.6 Higher-Order Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3 Differentiation of the Transcendental Functions 293.1 Differentiation of Sine and Cosine Functions . . . . . . . . . . . . . . . . . 293.2 Differentiation of the Tangent, Cotangent, Secant, and Cosecant Functions 333.3 Differentiation of Exponential and Logarithmic Functions . . . . . . . . . 34

4 Introduction to Integration 394.1 The Indefinite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.2 Basic Integration Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424.3 Constant of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464.4 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.5 Area Under the Curve of a Function . . . . . . . . . . . . . . . . . . . . . 50

5 Integration Methods 535.1 Integrals of Exponential and Logarithmic Functions . . . . . . . . . . . . 535.2 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555.3 Average and Root Mean Square Values . . . . . . . . . . . . . . . . . . . . 575.4 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615.5 Integrating Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . 655.6 Integration by Trigonometric Substitution . . . . . . . . . . . . . . . . . . 70

6 Differential Equations 786.1 Notation and Classification . . . . . . . . . . . . . . . . . . . . . . . . . . 786.2 Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 816.3 Applications of Differential Equations . . . . . . . . . . . . . . . . . . . . 84

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6.4 Second-Order Homogeneous Differential Equations (right side zero) . . . . 886.5 Second Order Nonhomogeneous Differential Equations . . . . . . . . . . . 936.6 RLC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

7 The Laplace Transform 1007.1 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1007.2 The Laplace Transform of a Function . . . . . . . . . . . . . . . . . . . . . 1027.3 Inverse Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 1077.4 Boundary Value Problems and the Laplace Transform . . . . . . . . . . . 109

8 Infinite Series 1128.1 Introduction to Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . 112

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1 Review of Limits

To begin we work through a quick review of the concept of limit. This material isstrictly for review purposes and is minimal, as the typical student in this course hasalready taken CALC103. Together with functions, limits are the foundation of calculusand we can do virtually no calculus without them. In this course we do not rigorouslyprove any of our concepts of limits as this belongs in an advanced calculus or analysiscourse at university or graduate school of mathematics.

1.1 The Geometry of Limits

Consider the following graph of a function:

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

(5, 4)

x

f(x)

Figure 1.1: Graph of the function f(x)

Let us ask the question: What happens to y = f(x) as x gets close to 0 (from either theleft hand side or the right hand side of 0?

We can answer this question by simply looking at the picture above and observing whathappens to the y values as the x values approach 0. We can see that the graph y valueapproaches 2. In fact the function is defined to be 2 at x = 0. Note that this last pointis completely irrelevant. When discussing limits we are not interested in what happensat a particular value, but what happens near a value.

We now formally restate the question and remind the student how we write the answerto this question in mathematical notation:

Example 1.1.1. Find limx→0

f(x).

4

Solution: The limx→0

f(x) = 2 because as x approaches 0 from both the left and the

right hand side, y gets close to 2. J

Next, let us consider another problems:


f(x) (refer to Figure 1.1).

Solution: Again we ask the similar question: What happens to y = f(x) as x getsclose to 5? The hole in the graph may confuse some students, however we are remindedthat we are not interested in what happens at x = 5 only nearby. The open hole meansthat the function is not even defined at x = 5! None-the-less we can look closely andsee that the open hole is labeled (5, 4). This tells us that y is getting close to 4 as xapproaches 5 from both the left and the right hand side.

Thus limx→5

f(x) = 4. J

Thirdly, we look at a more curious problem:


f(x) (refer to Figure 1.1).

Solution: Here the question is: What happens to y = f(x) as x gets close to 2? Inthis case the gap in the graph may confuse some students, however we are reminded thatwe are not interested in what happens at x = 2 only nearby.

Also note that the limit from the left is different from the limit from the right. We cansee that as we approach x = 2 from the left hand side of 2, y approaches 0. However,if we let x approach 2 from the right hand side of 2, y approaches 1. So it would seemthat the limit from the left is 0 and the limit from the right is 1. We actually have anotation for this in mathematics:

limx→2−

f(x) = 0

and

limx→2+

f(x) = 1

So it appears that our question has two different answers. Generally, in mathematicsour structures cannot have multiple simultaneous values. So in this case we say thatlimx→2

f(x) is undefined or does not exist. J

Property 1.1.1: In general, if limx→a−

f(x) = L and limx→a+

f(x) = L, then limx→a

f(x) = L.

and

Property 1.1.2: If limx→a−

f(x) 6= limx→a+

f(x), then limx→a

f(x) does not exist.

We have one final question we will ask about Figure 1.1:

Example 1.1.4. Find limx→−∞

f(x).

5

Solution: In this problem the −∞ symbol refers to large negative numbers. In otherwords, what happens to y = f(x) as x becomes large negative?

We can see that as the graph of f(x) goes off the left hand side of the graph it tracksalong and gets closer to the dashed asymptote at y = 3. Thus lim

x→−∞f(x) = 3. J

This concludes our discussion of the geometry of limits.

1.2 Limits of Specific Functions

In the next section we take a look at several examples of very specific functions and askquestions about their limits. In some cases we may not have a graph or picture to lookat and considerable work may be required to find some of these limits.

In example 1.1.1 we found the limit of f(x) as x approached 0. The answer was 2, whichhappened to be the same as f(0). In other words, the value of the limit was the sameas the value of the function evaluated at x = 0. This happened because there was no”weird” behavior of f ”near” x = 0. The words ”weird” and ”near” are used looselyhere. We will give this phenomenon a formal name shortly, but let us look at some moreexamples of where we may take advantage of such lack of ”weirdness” to see if we canfigure it out.

Example 1.2.1. If g(x) = x3 − 1, find limx→1

g(x).

Solution: Although we could, here we will not use a graph to help us. We will create atable of values to see what happens to y = g(x) as x approaches 1. We need to evaluateg(x) closer and closer to x = 1 from both the left and right hand sides.

x . 1 y = g(x) x & 1 y = g(x)

0.8 -0.488 1.2 0.728

0.9 -0.271 1.1 0.331

0.95 -0.142625 1.05 0.157625

0.999 -0.002997 1.001 0.003003

Table 1.1: Evaluation of g(x) Near x = 1

We can see from the left two columns of Table 1.1, as we let x get close to 1 from lessthan 1 (x . 1), y = g(x) ≈ 0. As we let x approach 1 from the right hand side (x & 1),y = g(x) ≈ 0 as well. Thus we conclude that lim

x→1g(x) = 0. Students should verify the

values in the table and further consider checking values of x even closer to 1. J

By now many students will see that had we just computed g(1) we would have gottenour limit without all the extra work of constructing Table 1.1. Note that g(1) = 0.The point of Examples 1.1.1 and 1.2.1 is that if there is no ”weird” behavior ”nearby”then we can find limits by simply substituting the value directly into the function. The

6

next several definitions and properties will allow us to more precisely understand anddiscuss this ”weirdness’ that can occur in functions.

Definition 1.2.1: A function y = f(x) is continuous at an interior point c of itsdomain if

limx→c

f(x) = f(c),

y = f(x) is continuous at a left endpoint a of its domain if

limx→a+

f(x) = f(a),

y = f(x) is continuous at a right endpoint a of its domain if

limx→a−

f(x) = f(a),

y = f(x) is continuous if it is continuous at every point of its domain.If y = f(x) is is not continuous at a point c, we say that f is discontinuous at c andcall c a point of discontinuity. /

These final two properties are key to finding limits of functions easily.

Property 1.2.1: A function y = f(x) is continuous if and only if limx→a

f(x) = f(a)

Property 1.2.2: Polynomial functions are continuous.

In summary, these properties and definitions tell us that we can find the limit of anypolynomial (or any continuous function) by simply evaluating the function at the x valuewe are interested in approaching. Further, the precise mathematical term for what weare calling ”weird” is a point of discontinuity.

Example 1.2.2. If g(x) = x3 − 1, find limx→−3

g(x).

Solution: Note here that g(x) is a polynomial function. Thus, according to Properties1.2.1 and 1.2.2, we may directly substitute x = −3 into g(x) to find the limit.

limx→−3

g(x) = g(−3)

= (−3)3 − 1

= −28

Thus our limit is -28. J

Example 1.2.3. If h(x) = 1− ex+2, find limx→0

h(x).

Solution: In this problem we observe that h(x) is not a polynomial function, it isexponential (with e ≈ 2.72). Exponential functions are continuous as long as theirdomain is all R. In this case we know that the domain of h is all R because for any real

7

number x there will be a corresponding range value h(x), thus h is continuous. So byProperty 1.2.1, we may directly substitute x = 0 into h(x) to find the limit.

limx→0

h(x) = h(0)

= 1− e0+2

= 1− e2

≈ −6.38906 . . .

Thus our limit is 1− e2. J

One should keep in mind that it is not always possible to find limits quickly because ofthe possibility of discontinuities. Sometimes our only option is to estimate limits andeven then there are situations where estimating limits with tables can fail. Fortunatelyfor us those situations do not often occur in the types of applications we will need thisfor.

1.3 More on Limits

In this final section on limits we will take a look at some useful properties of limits andlook at some fairly common situations where there is an algebraic ”trick” to finding acomplicated limit quickly. We will call upon the material from this section from time totime to help us prove something or work problems.

Property 1.3.1: Let limx→c

f(x) = L1 and limx→c

g(x) = L1 (where L1 and L2 are real

numbers,) then:

1. Sum Rule: limx→c

[f(x) + g(x)] = limx→c

f(x) + limx→c

g(x) = L1 + L2

2. Difference Rule: limx→c

[f(x)− g(x)] = limx→c

f(x)− limx→c

g(x) = L1 − L2

3. Product Rule: limx→c

[f(x) · g(x)] = limx→c

f(x) · limx→c

g(x) = L1 · L2

4. Constant Multiple Rule: limx→c

[k · f(x)] = k limx→c

f(x) = k · L1

5. Quotient Rule: limx→c

f(x)

g(x)=

limx→c

f(x)

limx→c

g(x)=L1

L2, if L2 6= 0

Secondly, we take a look at an example of where this is useful:


x3 + 4x2 − 3

x2 + 5.

Solution: In this problem note that we are dealing with a rational (fractional) function.Students should be aware that rational functions have domain restrictions such that thedenominator cannot be zero. However, in this example the denominator cannot be zerobecause x2 + 5 is always positive, thus there are no issues going forward.

8

We begin by using Property 1.3.1 (5) from above:

limx→1

x3 + 4x2 − 3

x2 + 5=

limx→1

(x3 + 4x2 − 3)

limx→1

(x2 + 5)

=13 + 4(1)2 − 3

12 + 5

=2

6

=1

3

Thus limx→1

x3 + 4x2 − 3

x2 + 5=

1

3. J

In the previous chapter we said that polynomial functions are continuous and that wecan find the limits of functions that are continuous by substitution. We now formallystate what we previously used.

Theorem 1.3.1: If f(x) = anxn + an−1x

n−1 + · · ·+ a1x+ a0 is a polynomial function,and c is any real number, then

limx→c

f(x) = f(c) = ancn + an−1c

n−1 + · · ·+ a1c+ a0

/

The next theorem could have been used to solve the previous example.

Theorem 1.3.2: If f(x) and g(x) are polynomials, and c is any real number, then

limx→c

f(x)

g(x)=f(c)

g(c)provided g(c) 6= 0.

/

In our next example we will use algebra to help us find the limit of a function at adiscontinuity. Note that none of the properties or theorems from above can help use findthe limit in the next example.


x2 − 6x+ 5

x− 1.

Solution: Here we have the limit of a rational function. As we know rational functionsare notorious for having discontinuities. This function has a discontinuity at x = 1because it is not defined at x = 1. Thus we need to be a bit more creative or clever hereto find the limit. We could construct a table of values however there is a trick that canbe used here using elementary algebra.

9

Note that the numerator factors:

limx→1

x2 − 6x+ 5

x− 1= lim

x→1

(x− 5)(x− 1)

x− 1

= limx→1

(x− 5)

= 1− 5

= −4

The trick is to factor and attempt to reduce the fraction so that the expression in thedenominator (that depends on x) will cancel. If the factor in the denominator cancelsthen this trick will work and you have your limit. If factorization does not cause thedenominator to disappear then one has to try something else.

So limx→1

x2 − 6x+ 5

x− 1= −4. J

In the next example we look at another algebraic ”trick” that uses a process calledrationalizing the numerator.


x− 1√x− 1

.

Solution: Here we have the limit of another rational function. As we know rationalfunctions are notorious for having discontinuities. This function has a discontinuity atx = 1 again because it is not defined at x = 1. We could construct a table of valueshowever the trick here is to rationalize the numerator.

Note that:

limx→1

x− 1√x− 1

= limx→1

[x− 1√x− 1

·√x+ 1√x+ 1

]= lim

x→1

[(x− 1)(

√x+ 1)

x− 1

]= lim

x→1(√x+ 1)

= (√

1 + 1)

= 2

So limx→1

x− 1√x− 1

= 2. J

There are many other algebraic ”tricks” to finding limits at discontinuities. All thatis required to understand them is a little research on the Internet or a textbook. Thisconcludes the section.

1.4 Limits Involving Infinity

In this section we look at limits were x or the domain elements of a function go to ±∞.We first need to discuss a more precise meaning of these types of limits with a simple

10

example.

Example 1.4.1. Find limx→∞

1

x.

Solution: We begin by describing what is meant by x→∞. This simply means as xgets large positive. Likewise if x→ −∞, then this means x becomes large negative. Wecan construct a table of values to see what happens to 1

x as x gets large positive:

x 1/x

1 1

2 0.5

5 0.2

10 0.1

50 0.02

100 0.01

10000 0.0001

Table 1.2: 1x as x gets large

We can see here that 1x is getting close to 0. Thus lim

x→∞

1

x= 0. J

A similar argument can be made for the same function as x→ −∞.


4.

Solution: We begin by describing what is meant by x→∞. This simply means as xgets large positive. Likewise if x→ −∞, then this means x becomes large negative. Wecan construct a table of values to see what happens to 4 as x gets large positive (tableis not really necessary):

x f(x)

-1 4

-2 4

-5 4

-10 4

-50 4

-100 4

-10000 4

Table 1.3: f(x) = 4 as x gets large

We can see here that 4 is not changing. Thus limx→∞

4 = 4. Students should have already

been aware of the fact that limx→c

k = k for any c (including ±∞) and any k. J

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A similar set of properties exist for domain values going to infinity as for limits where xgoes to a real number.

Property 1.4.1: Let limx→±∞

f(x) = L1 and limx→±∞

g(x) = L1 (where L1 and L2 are real

numbers,) then:

1. Sum Rule: limx→±∞

[f(x) + g(x)] = limx→±∞

f(x) + limx→±∞

g(x) = L1 + L2

2. Difference Rule: limx→±∞

[f(x)− g(x)] = limx→±∞

f(x)− limx→±∞

g(x) = L1 − L2

3. Product Rule: limx→±∞

[f(x) · g(x)] = limx→±∞

f(x) · limx→±∞

g(x) = L1 · L2

4. Constant Multiple Rule: limx→±∞

[k · f(x)] = k limx→±∞

f(x) = k · L1

5. Quotient Rule: limx→±∞

f(x)

g(x)=

limx→±∞

f(x)

limx→±∞

g(x)=L1

L2, if L2 6= 0

The idea Example 1.4.1 can be expanded for for any function of the form f(x) =a

xrfor

some real number a and natural number n.

Theorem 1.4.1: For any fixed real number a and r ≥ 1, limx→±∞

a

xr= 0. /

This theorem is easily proven with the properties above.

Finding Limits Involving Infinity

There are more algebraic ”tricks” to finding limits involving infinity. Such as when wefind the limit of a rational function as x→ ±∞. As it turns out we can divide all termsof the rational expression by the independent variable raised to the highest power in thedenominator then use the facts above to evaluate each limit one at a time.


5x2 + 8x− 3

3x2 + 2.

Solution: Here we have the limit of another rational function as x gets large. Wecannot simply substitute −∞ in for x because it is not a real number. Perhaps weshould look for a way to use the above idea.

12

Note that:

limx→−∞

5x2 + 8x− 3

3x2 + 2= lim

x→−∞

5 + 8/x− 3/x2

3 + 2/x2

=lim

x→−∞5 + lim

x→−∞8/x− lim

x→−∞3/x2

limx→−∞

3 + limx→−∞

2/x2

=5 + 0− 0

3 + 0

=5

3

So limx→−∞

5x2 + 8x− 3

3x2 + 2=

5

3. J

This concludes our discussion of limits involving infinity and our review of limits in thiscourse. Students are encouraged to do their own research to better understand limitsand their application as this will better prepare students for this course.

13

2 Introduction to Differentiation

After developing a clear understand of functions and limits one is typically ready tomove on to the two major tiers of calculus, the derivative and integral. Generally thederivative is covered first, but not necessarily always. We will stay with tradition in thiscourse.

In the first section we give a quick reminder of the definition of the derivative of afunction. The student should not forget this definition as it useful in generating much ofthe theory in calculus. In later sections we will review rules of differentiation and workthrough applications.

2.1 Definition of the Derivative

Students in this course should previously have had experience with finding the derivative.The process of finding the derivative is called differentiation.

Definition 2.1.1: The derivative of a function f(x) is the functiondf

dxor f ′(x) whose

value at x is

df

dx= lim

h→0

f(x+ h)− f(x)

h(if the limit exists.)

/

If the limit in the definition above exists, then we say that f has a derivative (isdifferentiable) at x. If f has a derivative at every point of its domain, we call fdifferentiable.. If f is differentiable, we call its graph a differentiable curve.

As you may recall there are many names for the derivative. When f ′(x) exists for aparticular x, it is called the slope of the curve y = f(x) at x. The line that passesthrough the point P (x, f(x)) with slope f ′(x) is the tangent to the curve at P .

There are many common notations for the derivative of a function y = f(x) besides thetwo above. Some common derivative notation and how we speak it can be found on thenext table.

14

Notation In English

dfdx ”d f d x”

ddx(f) ”d dx of f”

dydx ”d y d x”

f ′(x) ”f prime of x”

y′ ”y prime”

y′(x) ”y prime of x”

Dx(f) ”D x of f”

Table 2.1: Common Derivative Notation

It is important to remember that each of these notations have exactly the same meaning.

For the remainder of this section we look at how to use the definition to find the deriva-tive.

Example 2.1.1. Find the derivative of f(x) = 3x+ 2.

Solution: To find the derivative we must compute the limit from Definition 2.1.1:

df

dx= lim

h→0

f(x+ h)− f(x)

h

= limh→0

3(x+ h) + 2− (3x+ 2)

h

= limh→0

3x+ 3h+ 2− 3x− 2

h

= limh→0

3h

h

= limh→0

3

= 3

So the derivative of f(x) is 3. J

Example 2.1.2. Find the derivative of g(x) = −4x2.

Solution: Again, to find the derivative we must compute the limit from Definition

15

2.1.1 using g instead of f :

dg

dx= lim

h→0

g(x+ h)− g(x)

h

= limh→0

−4(x+ h)2 + 4x2

h

= limh→0

−4(x2 + 2xh+ h2) + 4x2

h

= limh→0

−4x2 − 8xh− h2 + 4x2

h

= limh→0

−8xh+ h2

h

= limh→0

h(−8x+ h)

h

= limh→0

(−8x+ h)

= −8x

So the derivative of g(x) is −8x. J

Unfortunately, as we deal with increasingly more complicated functions the definitionbecomes impractical for finding the derivative. Consider the function h(x) = 5x200 −3x150 + 30x50, which is very tedious to differentiate with the definition. In the nextsection we look at shortcut methods, one of which will make h easy to differentiate.

2.2 Rules for Differentiation

Here we look at a few basic rules of differentiation.

Differentiation of a Power Function

We begin with likely the most important differentiation rule.

Definition 2.2.1: A Power Function is any function of the form

f(x) = xn,

with any real number n 6= 0. /

As you may have seen in a previous course the derivative of a power function is givenquickly with the power rule.

Proposition 2.2.1: If n is a fixed positive integer and f(x) = xn, then

df

dx= nxn−1.

16

Proof. In order to prove this proposition we need the Binomial Theorem from elementaryalgebra:

(x+ h)n = xn + nxn−1h+n(n− 1)

2!xn−1h2 + . . .+ hn

The proof uses the definition of the derivative and the above formula with a = x andb = h:

df

dx= lim

h→0

f(x+ h)− f(x)

h

= limh→0

(x+ h)n − xn

h

= limh→0

(xn + nxn−1h+ n(n−1)

2! xn−1h2 + . . .+ hn)− xn

h

= limh→0

nxn−1h+ n(n−1)2! xn−1h2 + . . .+ hn

h

The important thing to realize at this point is that each of the terms in the numeratorhave a multiple of h. Thus we can factor out h.

= limh→0

h(nxn−1 + n(n−1)

2! xn−1h+ . . .+ hn−1)

h

= limh→0

(nxn−1 +

n(n− 1)

2!xn−1h+ . . .+ hn−1

)

At this point we are dealing with a polynomial in terms of the variable x because nis constant. Thus we can let h go to zero. Notice all of the terms except for the firstbecome zero.

= nxn−1 +n(n− 1)

2!xn−1(0) + . . .+ (0)n−1

= nxn−1

Thusdf

dx= nxn−1, the desired result.

The Power Rule can be further generalized for any function of the form f(x) = axn

where a is any real number. Limit Laws 4 helps us here:

df

dx=

d

dx(axn)

= ad

dx(xn)

= a(nxn−1

)= anxn−1

17

We illustrate this and the Power Rule in the next example.

Example 2.2.1. Find the derivative of y = −7x5.

Solution: Using the power rule we have

dy

dx= −7(5)x5−1

= −35x4

So the derivative of y is −35x4. J

We can also extend the Power Rule to sums and differences of Power Functions.

Theorem 2.2.1: If f and g are both differentiable, then

d

dx[f(x)± g(x)] =

d

dxf(x)± d

dxg(x)

/

We omit the proof of this theorem, as it is quite simple.

Example 2.2.2. Write the equation of the line that is tangent to f(x) = 1− 2x− 3x2

and passes through (−2,−7).

Solution: We need to things to write the equation of the line: the slope and a point.Conveniently we have the point, we just need to find the slope of the tangent line (thederivative) of y at x = −2 (the x value of the point).

df

dx= −2− 6x

We now institute some notation. The following implies that we are to evaluate thederivative at x = −2.

df

dx

∣∣∣x=−2

= −2− 6(−2)

= 10 The slope of our line

Next we use the slope and the point to write the equation of our line. Here we useslope-intercept form with m as the slope and b as the y-intercept.

y = mx+ b

−7 = 10(−2) + b

13 = b

Thus our line has slope 10 and y-intercept 13, giving us the equation of our line asy = 10x+ 13. J

18

Finally, you can see a graph of the previous problem.

−3 −2 −1 1

−10

−8

−6

−4

−2

2

(−2,−7)

f(x)

y = 10x+ 13

x

y

Figure 2.1: Graph of f(x) and its Tangent Line

This concludes our section on the basic rules of differentiation.

2.3 Differentiation of Composite Functions

In this section we take a look at a very powerful differentiation rule, the Chain Rule.In concert with the Power Rule, the Chain Rule will allow us to differentiate a largecollection of Real Functions.

Definition 2.3.1: Let y(x) and u(x) be real valued functions where the domain of yis the range of u, then any function of the form

y(u(x))

is called a composite function. /

Example 2.3.1. The following functions are composite:

1. f1(x) = (x + 4)3 is composite where f1(x) = y(u(x)) with y(u) = u3 and u(x) =x+ 4.

2. f2(x) =√x2 − 1 is composite where f2(x) = y(u(x)) with y(u) =

√u and u(x) =

x2 − 1.

3. f3(x) = esinx is composite where f3(x) = y(u(x)) with y(u) = eu and u(x) = sinx.

To simplify our notation we will simply refer to y(u) as y and u(x) as u from this pointforward.

Naturally, we would like to know how to differentiate such functions since many functionscan be expressed as composite. The Chain Rule will enable us to do so.

19

Proposition 2.3.1: (Chain Rule) Let f , y, and u be differentiable functions withdomain(y) = range(u) such that f(x) = y(u(x)), then

dy

dx=dy

du· dudx

Proof. We actual detailed proof is somewhat more complicated, however, for our pur-poses a more compact version will be enough.

dy

dx=dy

dx· 1

=dy

dx· dudu

=dy

du· dudx

This proof depends on each of the differentials (dy, du, and dx) being small, but notzero. It should also be noted that many of the conclusions in calculus depend on theidea that differentials are small but not zero.

The main application of the Chain Rule is to find the decomposition of a function (if itcan be expressed as composite) then use the chain rule to find its derivative.

Example 2.3.2. Find the derivative of f(x) = (x+ 4)3 without expanding.

Solution: We first need to find the decomposition of f = y(u). Here y(u) = u3 andu(x) = x+ 4. Next, we apply the Chain Rule:

dy

dx=dy

du· dudx

= 3u2 · 1= 3(x+ 4)2

Thusdy

dx= 3(x+ 4)2. Also, it is important not to forget to remove the dummy variable

u. Many students may forget and leave u in their answer. This is a common mistake. J

Example 2.3.3. Find the derivative of y =5

(x2 − 1)2.

Solution: First, we rewrite y with properties of exponents as y = 5(x2−1)−2. Secondly,we need to find the decomposition of f = y(u). Here y(u) = 5u−2 and u(x) = x2 − 1.Next, we apply the Chain Rule:

dy

dx=dy

du· dudx

= −10u−3 · 2x= −20x(x2 − 1)−3

20

Thusdy

dx= −20x(x2 − 1)−3. In some circumstances it may be more useful to write this

asdy

dx=−20x

(x2 − 1)3. J

We work one last problem.

Example 2.3.4. Find the slope of the tangent line at x = 1 on the function h(x) =2(x4 − 2x2)5.

Solution: First, we need to find the decomposition of h = y(u). Here y(u) = 2u5 andu(x) = x4 − 2x2. Next, we apply the Chain Rule:

dy

dx=dy

du· dudx

= 10u4 · (4x3 − 4x)

= 10(x4 − 2x2)4 · (4x3 − 4x)

= 10(4x3 − 4x)(x4 − 2x2)4

Thusdy

dx= 10(4x3 − 4x)(x4 − 2x2)4. Next we need to evaluate the derivative at x = 1:

dy

dx

∣∣∣x=1

= 10[4(1)3 − 4(1)

] [(1)4 − 2(1)2

]4= 10(−1)4(4− 4)

= 0

So the slope of the tangent line is zero, concluding our problem. J

As we conclude this section students should keep in mind that the Chain Rule is a veryuseful tool throughout calculus. There are many situations where it can be used in someform or fashion that ends by making complex problems easier.

2.4 Differentiation of Product and Quotient Functions

In this section we are interested in differentiating functions that can be written as prod-ucts and quotients of other functions. We first motivate differentiation of a functionmade of the product of two functions.

Proposition 2.4.1: (Product Rule) Let u and v be differentiable functions such thaty(x) = u(x) · v(x), then

dy

dx=d(uv)

dx= v

du

dx+ u

dv

dx

Proof. We begin by assuming that y(x) = u(x) · v(x), then we compute its derivative

21

directly using the definition.

dy

dx=d(uv)

dx= lim

h→0

y(x+ h)− y(x)

h

= limh→0

u(x+ h) · v(x+ h)− u(x)v(x)

h

= limh→0

u(x+ h) · v(x+ h)− v(x+ h)u(x) + v(x+ h)u(x)− u(x)v(x)

h

= limh→0

v(x+ h)[u(x+ h)− u(x)] + u(x)[v(x+ h)− v(x)]

h

= limh→0

[v(x+ h)[u(x+ h)− u(x)]

h+u(x)[v(x+ h)− v(x)]

h

]= lim

h→0

v(x+ h)[u(x+ h)− u(x)]

h+ limh→0

u(x)[v(x+ h)− v(x)]

h

= v(x) limh→0

u(x+ h)− u(x)

h+ u(x) lim

h→0

v(x+ h)− v(x)

h

= vdu

dx+ u

dv

dx

as desired.

Example 2.4.1. Differentiate y = 3x4√

2− 3x.

Solution: We proceed with the Product Rule with u(x) = 3x4 and v(x) =√

2− 3x.

dy

dx= v

du

dx+ u

dv

dx

=√

2− 3x(12x3) + 3x4

(1

2(2− 3x)−

12 (−3)

)= 12x3

√2− 3x− 9

2x4(2− 3x)−

12

= 12x3√

2− 3x− 9x4

2√

2− 3x

Note that the final answer could take on several different forms. J

There is a similar formula for differentiating function that can be written as the quotient

of two functions. For example: f(x) =u(x)

v(x)or

u

v.

Proposition 2.4.2: (Quotient Rule) Let u and v be differentiable functions such

that y(x) =u(x)

v(x)with v(x) 6= 0 , then

dy

dx=d(u/v)

dx=vu′ − uv′

v2

using prime notation.

22

The proof of the Quotient Rule is very similar to the Product Rule, thus we omit it here.

Example 2.4.2. Take the derivative of y =2x3

4x+ 1.

Solution: We proceed with the Product Rule with u(x) = 2x3 and v(x) = 4x+ 1.

dy

dx=vu′ − uv′

v2

=(4x+ 1)(6x2)− 2x3(4)

(4x+ 1)2

=6x2(4x+ 1)− 8x3

(4x+ 1)2

=24x3 + 6x2 − 8x3

(4x+ 1)2

=16x3 + 6x2

(4x+ 1)2

Note again that the final answer could take on several useful forms. J

Students are cautioned that with the Product Rule the order makes no difference: uv′+vu′ = vu′+uv′, however, with the Quotient Rule the order of the terms in the numeratoris significant because subtraction is not commutative.

2.5 Implicit Differentiation

Up to now, the variable in the function has been the same variable that we take thederivative with respect to, as in the following examples (in terms of the Chain Rule):

Example 2.5.1. We find each derivative and use the Chain Rule even though it is notneeded here:

1.d

dx(x3) = 3x2dx

dx= 3x2

2.d

dt(t4) = 4t3

dt

dt= 4t3

What if we are differentiating a function that depends on one variable with respect toanother variable? Like the next example:

Example 2.5.2. Findd

dx(2t3). Assume that t depends on x.

Solution: Note here that we are differentiating with respect to x, but the functiondepends on t, which in turn depends on x. To work this we must apply the Chain Rule:

d

dx(2t3) = 6t2

dt

dx

Students may find the form of our answer disturbing, but we can go no further becausewe do not know the exact relationship between t and x. J

23

Example 2.5.3. Find the derivatived

dt(2y − t3). Assume that y depends on t.

Solution: Here note that we are taking the derivative with respect to t and that ydepends on t.

d

dt(2y − t3) = 2

dy

dt− 3t2

Concluding our problem. J

Normally we think of y as the dependent variable but in the next problem we think ofx as the dependent variable.

Example 2.5.4. Find the derivativedx

dy, for x = y2 − 3y + 2.

Solution: In this problem we are taking the derivative with respect to y and note herethat x depends on y.

dx

dy= 2y − 3


We now turn our attention to Implicit Functions or functions in which the dependentvariable is not isolated on one side of the equation. In many cases Implicit Functionshave dependent variables that cannot be isolated in any way. Conveniently, there is amethod in calculus for differentiating such functions. We now discuss this highly generalprocess and note that we will make use of the concepts above.

Example 2.5.5. Differentiate x2 + y2 = y3 − x with respect to x. Assume that ydepends on x.

Solution: In this problem we are taking the derivative with respect to x and note thaty depends on x.

2x+ 2ydy

dx= 3y2 dy

dx− 1

Next, just solve the equation fordy

dxand we are done.

2ydy

dx− 3y2 dy

dx= −2x− 1

(2y − 3y2)dy

dx= −2x− 1

dy

dx=

2x+ 1

3y2 − 2y


24

Example 2.5.6. Find the slope of the tangent line at x = 0 on the ellipse(x4

)2+(y

9

)2= 1

Solution: We begin by rewriting our implicit function:

x2

16+y2

81= 1

81x2 + 16y2 = 1296 Multiply both sides by 1296

In this problem we are taking the derivative with respect to x and assume that y dependson x.

162x+ 32ydy

dx= 0

32ydy

dx= −162x

dy

dx=−162x

32y

dy

dx= −81x

16y

As we can see from the derivative above we need an x value and a y value. So we useone of the two original equations to find the y value.

x2

16+y2

81= 1

(0)2

16+y2

81= 1

y2

81= 1

y2 = 81

Which tells us that there are two y values for x = 0: y = 9 and y = −9. So now we cango back to our derivative formula and substitute. First, we use x = 0 and x = 9:

dy

dx= −81x

16y

= − 81(0)

16(−9)

= − 0

−144

= 0

25

Secondly, we use x = 0 and y = −9:

dy

dx= −81x

16y

= −81(0)

16(9)

= − 0

144= 0

So there are two tangents lines at (0, 9) and (0,−9), each with slope zero. See the graphbelow for a look at the graph and the tangent lines. The tangent lines are the dashedlines, which clearly have a slope of zero. J

−8 −4 4 8

−8

−4

4

8 (x4

)2+(y9

)2= 1

x

y

Figure 2.2: Illustration of Previous Problem

Differentials

In the final part of this section we discuss differentials. Informally, a differential, dy, isa very small change in y. So dx is a very small change in x. More formally,

Definition 2.5.1: Let y be a differentiable function of x, then

dy = f ′(x)dx

is called the differential of y spoken ”d y”. /

26

y = f(x)

P

Slope = dydx

dx

dy

x

y

Figure 2.3: Differentials dy and dx

For purposes that will become apparent later in the course we need to be able to identifydy. The process is generally, quite simple. We treat dx as if it were a variable and solvethe equation for dy. One can also just use the definition to write dy.

Example 2.5.7. Find the differential dy, if y = 6x4 − 2x2 − 1.

Solution: We begin by findingdy

dx, then to finish, solve for dy:

dy

dx= 24x3 − 4x

dy = (24x3 − 4x)dx


Example 2.5.8. Find the differential dy of the implicit function x2 + y2 = y3 − x.Assume y depends on x.

Solution: Note that we have found this implicit derivative earlier:

dy

dx=

2x+ 1

3y2 − 2y

dy =

(2x+ 1

3y2 − 2y

)dx


This brings us to the end of other section.

2.6 Higher-Order Differentiation

It certainly makes sense that we could differentiate iteratively. In other words, takethe derivative of the derivative or we could take the derivative of the derivative of the

27

derivate (as many times as we like). The derivative of the derivative is called the secondderivative. The derivative of the derivative of the derivative or the derivative of thesecond derivative is called the third derivative. Our notation for such higher-orderderivatives follows:

d2y

dx2or y′′ or f ′′(x) or D2y

We will use many of these notations as practice throughout this section.

Example 2.6.1. Find the first four derivatives of f(x) = 3x5 + 2x4 − 5x3 − x2 + 7.

Solution: We list the first four derivatives by differentiating f four times and recordingthe result at each step:

df

dx= 15x4 + 8x3 − 15x2 − 2x

d2f

dx2= 60x3 + 24x2 − 30x− 2

d3f

dx3= 180x2 + 48x− 30

d4f

dx4= 360x+ 48

Students should be clear that each of the derivatives above is our answer in this case, asthe problem asks for each of the first four derivatives. J

Example 2.6.2. Find the third derivatives of G(r) =√r+ 3√r. Leave your answer as

a rational exponent.

Solution: First, we rewrite G(r) in terms of rational exponents so that we can use thepower rule.

G(r) = r12 + r

13

Next, we find the first and second derivatives so that we can find the third:

G′(r) =1

2r−

12 +

1

3r−

23

G′′(r) = −1

4r−

32 − 2

9r−

53

G(3)(r) =3

8r−

52 +

10

27r−

83

Thus the third derivative of G(r) is3

8r−

52 +

10

27r−

83 . J

This brings our section on Higher-Order Derivatives to a close and our chapter on anintroduction to Differentiation.

28

3 Differentiation of the TranscendentalFunctions

In general it is not so easy to differentiate transcendental functions, however in this sec-tion we will focus on some of the ”elementary” transcendental functions: Trigonometric,Logarithmic, and Exponential Functions.

3.1 Differentiation of Sine and Cosine Functions

As we begin to discuss the calculus of the Trigonometric Functions we must recall some-thing related to them. In particular, there are many methods of measuring angles, butfor the purpose of calculus we will use radian measure only from this point on. Thusthe argument of a Trigonometric Function will always be in units of radians. We willnot review these functions further as students should have encountered these functionsin a previous course.

First, we state a theorem and two lemmas and will prove only the lemmas. These arenecessary to derive the derivative of the Sine Function.

Theorem 3.1.1: (The Sandwich Theorem) Let I be an interval having the pointa as a limit point. Let f , g, and h be functions on I, except possibly at a and supposethat for every x in I not equal to a, we have:

g(x) ≤ f(x) ≤ h(x)

and also suppose thatlimx→a

g(x) = limx→a

h(x) = L

thenlimx→a

f(x) = L

/

We do not prove the Sandwich Theorem because the required theory is beyond the scopeof this course.

Lemma 3.1.1: If x is a small angle (measured in radians), then limx→0

sinx

x= 1.

Proof. In proving the Lemma above we will make use of the diagram below.

29

O A

B

D

C

1

xcosx

sinx ta

nx

Let x be a small angle, in radians, in a circle of radius 1 (see figure above), in which

sinx =BC

1= BC

cosx =OC

1= OC

tanx =AD

1= AD

It is easy to see that the area of the sector OAB is greater than the area of triangleOBC but less than the area of triangle OAD. Now noting that

area of triangle OBC =1

2BC ·OC =

1

2sinx cosx

and

area of triangle OAD =1

2OA ·AD =

1

2tanx

both, by the area of a triangle formula.

Further, by the formula for finding the area of a sector of a circle:

area of sector OAB =1

2r2x =

x

2

So we now have that1

2sinx cosx <

x

2<

1

2tanx

which by multiplying everything by2

sinxwe arrive at

cosx <x

sinx<

1

cosx

If we now let everything go to zero we know that limx→0

(cosx) = 1 and limx→0

(1

cosx

)= 1

and thus by the Sandwich Theorem limx→0

( x

sinx

)= 1 and finally by reciprocating we can

conclude that limx→0

(sinx

x

)= 1, proving the lemma.

30

We can also see this informally by look at the graph of f(x) =sinx

x:

−10 −5 5 10

1

f(x) =sinx

x

x

y

Figure 3.1: Graph of f(x) =sinx

x

Even though the function f is not defined at zero we can see that both the left and righthand limits approach one.

Lemma 3.1.2: If x is a small angle (measured in radians), then limx→0

cos(x)− 1

x= 0.

Proof. We begin by noting that

(1− cosx)(1 + cosx) = sin2 x

so

limx→0

cos(x)− 1

x= lim

x→0

− sin2 x

x(1 + cosx

= limx→0

−(sinx)(sinx)/x1

1+cosx

=−0(1)

1/(1 + 1)Direct Substitution and Lemma 3.1.1

= 0

Concluding our proof.

We now state and prove the derivative of the Sine Function.

Proposition 3.1.1: Let x be an angle measured in radians and f(x) = sinx, thendf

dx= cosx.

31

Proof. We use the definition of the derivative here.

df

dx= lim

h→0

f(x+ h)− f(x)

h

= limh→0

sin(x+ h)− sinx

h

= limh→0

[sinx cosh+ cosx sinh]− sinx

hSum angle formula

= limh→0

cosx sinh− sinx+ sinx cosh

h

= limh→0

cosx sinh− sinx(1− cosh)

h

= limh→0

cosx sinh

h− limh→0

sinx(1− cosh)

h

= (cosx) limh→0

sinh

h− (sinx) lim

h→0

(1− cosh)

h

= (cosx)(1)− (sinx)(0) Lemmas 3.1.1 and 3.1.2

= cosx

giving us the desired conclusion.

Now that we have seen considerable theory and derived the derivative of the Sine Func-tion. We will point out that the derivative of the Cosine Function is developed in asimilar fashion and simply give the derivative.

Proposition 3.1.2: Let x be an angle measured in radians and f(x) = cosx, thendf

dx= − sinx.

We now move to working a few examples.

Example 3.1.1. Differentiate y = 3x2 − cos(4x).

Solution: We use the previous rules in concert with the proposition immediately aboveto differentiate.

dy

dx= 6x+ sin(4x)× 4

= 6x+ 4 sin(4x)

giving us our solution. J

Example 3.1.2. Differentiate y = − sin(6x3) cos(2x). Do not simplify beyond cleaningup signs.

Solution: We use the previous rules in concert with the propositions above to differ-entiate.

dy

dx= cos(2x)(− cos(6x3)(18x2) + sin(6x3) sin(2x)(2)

= −18x2 cos(2x) cos(6x3) + 2 sin(6x3) sin(2x)

32

giving us the solution. J

This concludes our introductory section on Differentiation of the Trigonometric Func-tions.

3.2 Differentiation of the Tangent, Cotangent, Secant, andCosecant Functions

In this section we complete the differentiation formulas for the Trigonometric Functions.We will derive the derivative of the Tangent Function and leave the rest for students toderive.

Proposition 3.2.1: Let x be an angle measured in radians and f(x) = tanx, thendf

dx= sec2 x.

Proof. To find the derivative of the Tangent Function we begin by noting that

tanx =sinx

cosx

then using the Quotient Rule we have

dy

dx=

cos2 x+ sin2 x

cosx

=1

cos2 x= sec2 x

which concludes our proof.

The rest of the differentiation formulas are listed in the table below.

d

dx[secx] = secx tanx

d

dx[cscx] = − cscx cotx

d

dx[cotx] = − csc2 x

Example 3.2.1. Differentiate y = 3 tan(x2). Do not simplify beyond cleaning upsigns.

Solution: Using the formulas and note we need the Chain Rule here we have

dy

dx= 3 sec2(x2)(2x)

= 6x sec2(x2)

giving us the solution. J

Example 3.2.2. Differentiate y = 4 sec3(x). Do not simplify beyond cleaning up signs.

33

Solution: First we should note that sec3 x = (secx)3 thus we need the Chain Rule.

dy

dx= 4× 3(secx)2 × secx tanx

= 12 tanx sec3 x

giving us the derivative. J

Example 3.2.3. Using implicit differentiation, find the derivative, dydx where xy +

y cotx = 0.

Solution: Here we work very carefully using various rules and using implicit differen-tiation:

1y + xdy

dx+dy

dxcotx− y csc2 x = 0 Chain Rule

xdy

dx+dy

dxcotx = y csc2(x)− y

dy

dx(x+ cotx) = y csc2(x)− y

dy

dx=y csc2(x)− yx+ cotx

giving us the derivative. J

This concludes our section and brings us to the end of our formal discussion of derivativesof the Trigonometric Functions.

3.3 Differentiation of Exponential and Logarithmic Functions

The course textbook will break this section into separate sections, however, in the noteswe will treat Exponential and Logarithmic Functions as they naturally exist, as twosides of the same mathematical object. It should therefore come as no surprise thattheir derivatives are intertwined as well. We will begin by finding the derivative of aspecial exponential function, then use this to find the derivative of Logarithmic Functionsand from there find the derivative of all Exponential Functions.

The Special Exponential Function

We begin this section by reminding ourselves about just exactly what an exponentialfunction is.

Definition 3.3.1: Let a be a positive real number and n be any real number, thenany function of the form

f(x) = ax

is called a Simple Exponential Function. /

34

We could certainly create more complex exponential functions by adding, subtracting,multiplying, and dividing with other functions. We could also increase the complexityby taking roots or further exponentiating the simple exponential function. None-the-lessonce we learn to differentiate f(x) = ax we will be able to differentiate another morecomplex exponential function via our previous rules for differentiating sums, differences,products, quotients, and composite functions.

Of course this being a calculus course we will naturally ask the question: How do we takethe derivative of f(x) = ax? To answer this question we resort back to our definition ofthe derivative.

df

dx= lim

h→0

f(x+ h)− f(x)

h

= limh→0

a(x+ h)− ax

h

= limh→0

axah − ax

h

= ax limh→0

ah − 1

h

Note that this last expression implies that the derivative of ax depends on itself, ax. Weshould also be aware at this point that we have not generated a simple formula for thederivative of f(x) = ax.

We now propose that there is a special choice of a in f(x) = ax such that limh→0

ah − 1

h= 1.

If we can find that special choice of a, then we have found a special function whosederivative is the same as the function itself!

Taking a look at the graph below we have used a computer to plot a graph of the limitabove.

1 2 3

1

2

3

a

L

Figure 3.2: Graph of L = limh→0

ah − 1

h

35

Simulation and the graph above should verify to the student that this special choice isapproximately a ≈ 2.718. You will recognize this number from a previous course. It issometimes called Euler’s Number.

Definition 3.3.2: Euler’s Number, denoted e ≈ 2.718, is an irrational number that

when substituted in for a, solves the equation limh→0

ah − 1

h= 1 /

Thus, limh→0

eh − 1

h= 1. The student should verify this limit equation is true by creating

a table of values with h approaching zero.

In conclusion to all of the above we have a new differentiation formula:

d

dxex = ex

Thus the function f(x) = ex is the function who is its own derivative. Believe it ornot this simple differentiation formula will allow us to differentiate a huge collection oftranscendental functions.

Differentiation of Logarithmic Functions

We now move on to Logarithmic Functions. Logarithmic Functions are the inverses ofExponential Functions.

Definition 3.3.3: Let a be a positive real number, then f(x) = loga x is called thebase a Logarithmic Function. /

An example of a logarithmic function is f(x) = loge x which is sometimes called theNatural Logarithmic Function because it has Euler’s Number as its base. We usuallywrite it y = lnx. It is the inverse function of y = ex discussed earlier. This means thaty = loge x is the same as x = ey. As a result of this fact we can find the derivative off(x) = y using implicit differentiation.

x = ey

1 = eydy

dxDifferentiating both sides

1

ey=dy

dx1

x=dy

dx

Thus

d

dxlnx =

1

x

We can further differentiate all logarithmic functions with any base a using the sameidea.

36

Proposition 3.3.1: The derivative of y = loga x is y′ =1

x ln a.

Proof. Note here that y = loga x⇔ x = ay. Thus

x = ay

1 = eln ay Law of Logarithms

1 = ey ln a Another Law of Log’s

1 = ey ln a · ln adydx

1

ey ln a · ln a=dy

dx1

eay ·ln a=dy

dx1

x ln a=dy

dx

giving us the desired conclusion.

Thus

d

dx(loga x) =

1

x ln a

Differentiation of Exponential Functions

As our final bit of theory for this section we derive a formula for differentiating thesimple exponential function for any base a. Thus we want to differentiate y = ax.

We begin by noting that by the definition of logarithms y = ax ⇔ x = loga y. We willuse the right hand side and differentiate implicitly.

x = loga y

1 =1

y ln a

dy

dxProposition above

y ln a =dy

dxMultiply both sides by y ln a

ax ln a =dy

dxSubstitute y from above

So

d

dx(ax) = ax ln a

We now have a complete list of differentiation formulas for the elementary transcendentalfunctions. The rest of the section will be spent working through a few examples.

Example 3.3.1. Differentiate y = log3(2x3 + 2x).

37

Solution: We differentiate here using proposition 3.3.1 and the chain rule.

y′ =1

(2x3 + 2x) ln 3(6x2 + 2)

=6x2 + 2

(2x3 + 2x) ln 3

concluding our problems. J

Example 3.3.2. Differentiate y = ex · 33x+1.

Solution: We differentiate here using the product and the chain rules. We let u = ex

and v = 33x+1

y′ = u′v + uv′

= ex33x+1 + 3ex · 33x+1 ln 3

= (1 + 3 ln 3)ex · 33x+1 Factor

is our derivative. Either of the last two lines would be an acceptable expression for thederivative. J

Example 3.3.3. Differentiate f(x) = ln(lnx).

Solution: We differentiate here using the chain rules.

f ′(x) =1

lnx· 1

lnx

=1

(lnx)2

concluding our problems. J

We now come to the end of the formal discussion of differentiation.

38

4 Introduction to Integration

In this chapter we begin discussion of the other major tier of calculus, integration. Ifyou treat Differentiation and Integration as functions we can easily show that they areinverses.

4.1 The Indefinite Integral

To understand the Integral we must first discuss the rather intuitive idea of an an-tiderivative.

Definition 4.1.1: An antiderivative of an expression is a new expression which, ifdifferentiated, gives the original expression. /

After reading this definition one should be thinking that the derivative and antiderivativeare inverse operations, for lack of a better word. In other words, the derivative of theantiderivative of a function is the function.

Example 4.1.1. The derivative of x4 is 4x3, thus an antiderivative of 4x3 is x4.

Referring to the example above we should also not that there are infinitely many an-tiderivatives of any expression or function.

Example 4.1.2. The derivative of x4 +3 is 4x3, thus an antiderivative of 4x3 is x4 +3,because the derivative of a constant is zero.

In general, the antiderivative of 4x3 is x4 + c where c is any constant real number.

Let us be a bit more rigorous now. If F (x) + c is a function, then its derivative is

d

dx[F (x) + c] = F ′(x) + 0 = F ′(x)

Further, and because dx is very small (but not zero) we can multiple both sides by dxand convert the above into differential form which gives us

d [F (x) + c] = F ′(x)dx

Thus the differential of F (x) + c is F ′(x)dx. Conversely, the antiderivative of F ′(x)dxis F (x) + c. Mathematicians have shortened this notation somewhat by replacing thephrase ”antiderivative of” with the integral sign

∫. So

Definition 4.1.2: Let F (x) be a differentiable function, then the Indefinite Integralof F ′ with respect to x is ∫

F ′(x) dx = F (x) + c

39

The expression or function that we are finding the antiderivative of is called the Inte-grand. In this case the integrand is F ′(x). /

Example 4.1.3. Find the indefinite integral of g(x) = 2x.

Solution: To solve this problem we need to think of a function whose derivative is 2x.We should know that as x2. Thus ∫

2x dx = x2 + c

So x2 + c is the general antiderivative of 2x. J

Before we move on we should say that it is customary to replace F ′(x) with f(x) in thedefinition of the integral. Thus F ′(x) = f(x) which allows us to rewrite the definitionas follows

Definition 4.1.3: Let F (x) be a differentiable function whose derivative is f(x), thenthe Indefinite Integral of f with respect to x is∫

f(x) dx = F (x) + c

/

From this point on we will use this convention.

We should also mention that the constant c is usually refereed to as the constant ofintegration.

Some Integration Rules

Here we introduce some of the very basic rules and properties of integrals.

Property 4.1.1: ∫dx = x+ c

The next few properties are some fairly standard properties that should be no surpriseabout integrals.

Property 4.1.2: ∫af(x) dx = a

∫f(x) dx+ aF (x) + c

Example 4.1.4. [1, p.884]Find

∫3kx2 dx, if k is a constant.

Solution: We already know that the antiderivative of 3x2 is x3. We simply apply theproperty above to pull the k out of the picture.∫

3kx2 dx = k

∫3x2 dx

= kx3 + c

40


Example 4.1.5. Find

∫7 dx.

Solution: Again 7 is a constant so pull it out in front of the integral sign.∫7 dx = 7

∫dx

= 7x+ c


The following property is also true for differences.

Property 4.1.3:∫[f1(x) + f2(x) + · · · ] dx =

∫f1(x) dx+

∫f2(x) dx+ · · ·+ c

The next property is just the power rule from our discussion of differentiation in reverse.

Property 4.1.4: Let n be any real number except for −1, then∫xn dx =

xn+1

n+ 1+ c

Example 4.1.6. Find

∫1

3x5 dx.

Solution: Here we use a combination of properties.∫1

3x5 dx =

1

3

∫x5 dx

=1

3

x6

6+ c

=1

18x6 + c

=x6

18+ c

which is our answer. J

Example 4.1.7. [1, p.887] Find∫x5−2x3+5x

x dx.

Solution: Here again, we use a combination of properties.∫x5 − 2x3 + 5x

xdx =

∫ (x4 − 2x2 + 5

)dx

=

∫x4 − 2

∫x2 dx+ 5

∫dx

=x5

5− 2x3

3+ 5x+ c

41

which is our solution. J

Example 4.1.8. [1, p.887] Find

∫x3 − x2 + 5x− 5

x− 1dx.

Solution: In this problem we can use long division to simplify the quotient, thenintegrate.

x2+ 5

x− 1)x3−x2+5x−5

x3−x2

0 +5x−5

5x−5

Thus x3−x2+5x−5x−1 = x2 + 5, which can easily be integrated.∫

x3 − x2 + 5x− 5

x− 1dx =

∫x2 + 5 dx

=x3

3+ 5x+ c


This concludes our introductory section on integration.

4.2 Basic Integration Rules

In this section we add some of the basic integration methods derived from differentiation.First, we will take a look at reversing the chain rule. With respect to integration this isreferred to as the Substitution Rule or u-Substitution.

We know from the chain rule that if f , y, and u are differentiable functions such thatf(x) = y(u(x)), then

dy

dx=dy

du· dudx

We learned this in a previous chapter. If we write the above in prime notation andconvert to differential form we have

dy = y′(u) · u′(x) dx

and now integrate both sides∫y′(u) · u′(x) dx = y(u(x)) + c

The above derivation is the Substitution Method. In less mathematical terms, it saysthat, we can integrate a function if the integrand contains:.

• a composite function of the form y(u(x)),

42

• the derivative of u(x), u′(x).

In using the Substitution Method students are highly encouraged to clearly state u,du

dx

and to convertdu

dxto differential form.

Example 4.2.1. Find

∫2x√

1 + x2 dx.

Solution: In this example note that the integral is not straightforward. But noticethat there is a composite function:

y(u) =√u where u = 1 + x2

and further,

du = 2x dx

It should be clear after looking at the above, for a moment, that our integrand is of theform y′(u) · u′(x). So by substituting from above we have∫

2x√

1 + x2 dx =

∫ √u du

=

∫u

12 du

=2

3u

32 + c

=2

3(1 + x2)

32 + c

=2

3

√(1 + x2)3 + c

which is our answer. Students should verify that this is correct by differentiating. Donot forget to replace u each time you use the Substitution Method. J

Example 4.2.2. Find

∫ √2x+ 1 dx.

Solution: In this example note that the integral is not straightforward. But noticethat there is a composite function:

y(u) =√u where u = 2x+ 1

and further,

du = 2 dx anddu

2= dx

43

In this cause we do not see 2 in the integrand, but let us see what we can do∫ √2x+ 1 dx =

∫ √u

2du

=1

2

∫u

12 du

=1

3u

32 + c

=1

3(2x+ 1)

32 + c

=1

3

√(2x+ 1)3 + c

which is our answer. Again, students should verify that this is correct by differentiatingand don’t forget to replace u. J

Example 4.2.3. Find

∫x3 cos(x4 + 2) dx.

Solution: Notice again that there is a composite function:

y(u) = cosu where u = x4 + 2

and further,

du = 4x3 dx anddu

4= x3 dx

and so ∫x3 cos(x4 + 2) dx =

∫cosu

4du

=1

4

∫cosu du

=1

4sin(u) + c

=1

4sin(x4 + 2) + c

which is our answer. J

At this point students should see that the Substitution Method can be employed as longas the derivative of u with respect to the variable and exponent are the same. Coefficientsdo not have to match, as they can be adjusted by solving for dx.

Miscellaneous Integration Rules Derived from Differentiation

Next, we generate a small list of integration rules that come from reversing rules thatwe have used while taking the derivative.

44

Property 4.2.1: If x 6= 0, then ∫dx

x= ln |x|+ c

Property 4.2.2: If x is an angle in radians, then∫cosx dx = sin(x) + c

Property 4.2.3: If x is an angle in radians, then∫− sinx dx = cos(x) + c

Property 4.2.4: If x is an angle in radians, then∫sec2 x dx = tan(x) + c

Property 4.2.5: If x is a real number, then∫ex dx = ex + c

Of course this list can be extended for other derivatives that we know or have a formulaof. We work one more example for this section.

Example 4.2.4. Find

∫4z

3z2 − 5dz.

Solution: In this case it may be difficult to see but there is a composite functionswhose derivative (with respect to variable and exponent) exists within the integrand.

y(u) =1

uwhere u = 3z2 − 5

and further,

du = 6z dz anddu

6= z dz

and so ∫4z

3z2 − 5dz =

∫4

6

1

udu

=2

3

∫1

udu

=2

3ln |u|+ c

=2

3ln |3z2 − 5|+ c

which is our integral. J

This brings us to the end of another section.

45

4.3 Constant of Integration

Here we will be taking a quick look at applications of integration that will allow usto identify specific functions from a given derivative and other information. It is notdifficult to see that in the previous sections we have been looking at entire families offunctions as a result of integrating. The family of functions results from the lack ofclarity about the constant c.

In this section we will be dealing what are called Differential Equations. Theseequations come up in many applications, across many fields of inquiry.

Definition 4.3.1: A Differential Equation is an equation that contains a derivative./

Example 4.3.1. Graph the function whose derivative is dydx = 2x.

Solution: As we know there are infinitely many functions whose derivative is 2x. Wecan find the family of functions by taking the integral of the given differential equation.

dy

dx= 2x

dy = 2x dx Separate Variables∫dy =

∫2x dx

y = x2 + c

Note that the second line above is called Separation of Variables, this will come uplater.

We will now graph y above for a few integer values of c, but note that c can be any realnumber.

−4 −2 2 4

−4

−2

2

4

x

y

Figure 4.1: Graph of y = x2 + c family of functions

This concludes our problem. J

46

In application, we are often interested in a specific function, not a family. In order tospecify a particular function we need more than just the derivative to find c, we needa boundary condition or initial condition. The phrase initial condition is usually(but not always) reserved for when the independent variables is time.

Example 4.3.2. Find the function whose derivative is dydx = 2x and passes through

the point (2, 9).

Solution: We have already found the family of functions in the previous problem.

y = x2 + c

Next, we just need to substitute the boundary condition into the function.

y = x2 + c

9 = 22 + c

5 = c

Thus our desired function is y = x2 + 5. J

Example 4.3.3. Find the function whose second derivative is d2ydx2

= 4 and passesthrough the point (2, 6) and is parallel to y = 3x.

Solution: Here we will need to integrate twice.

y′′ = 4

y′ =

∫4 dx

= 4x+ c1

At this point we can use the fact that the slope (first derivative) of the function we arelooking for is 3 because the slope of y = 3x is 3 at the point we are interested in.

3 = 4(2) + c1

−5 = c1

Thus we have more information about our function: its derivative function is y = 4x−5.Now we can integrate again.

y′ = 4x− 5

y =

∫4x− 5 dx

= 2x2 − 5x+ c2

Finally, we can use the boundary condition to find c2.

y = 2x2 − 5x+ c2

6 = 2(2)2 − 5(2) + c2

6 = −2 + c2

8 = c2

47

So our final answer is the function y = 2x2 − 5x + 8 whose second derivative is 4 andsatisfies the conditions above. J

This brings our section to a close.

4.4 The Definite Integral

In this section we define what is called the definite integral from the indefinite integral.

In the previous section we discussed finding the indefinite integral. Here we expand onthis in terms of an integral function. Consider

F (x) =

∫f(x) dx

We could evaluate this function just like any previously studied function. We would firsthave to find the indefinite integral and then substitute in whatever x is required. Let uslook at a specific example.

Example 4.4.1. Compute F (1) =

∫2x dx.

Solution: To find F (1) we must first find the indefinite integral.

F (x) =

∫2x dx

= x2 + c

So F (x) = x2 + c, which can now be evaluated.

F (1) = (1)2 + c

= 1 + c

We can do nothing else until we know more about c under these circumstances. J

Now that we understand that we can evaluate an integral function at a particular x, wecan talk about adding, subtracting, multiplying, or dividing such functions easily.

The Definite Integral

Let us consider evaluating the integral function F (x) =

∫f(x) dx at x = a and x = b.

This would give us F (a) and F (b). Further the result in each case would be some numberplus c. If we then subtracted, note that the two resulting constants would cancel awayleaving just the difference of two numbers. This is the definite integral.

Definition 4.4.1: Let f be an integrable function on the interval [a, b] with a ≤ b,then ∫ b

af(x) dx = F (b)− F (a)

is called the Definite Integral from a to b. /

48

The rest of this section is spent practicing the definite integral.

Example 4.4.2. Compute

∫ 2

0x3 dx.

Solution: To begin we use the definition of the definite integral and find the indefiniteintegral of x3. ∫ 2

0x3 dx = F (2)− F (0)

=1

4(2)4 − 1

4(0)4

= 4− 0

= 4

Thus

∫ 2

0x3 dx = 4. J


∫ 3

−1(3x− 2)4 dx.

Solution: In this case we may want to find the indefinite integral of (3x − 2)4 first.

We can use u-substitution, with u = 3x− 2 and du = 3 dx which isdu

3= dx.∫

(3x− 2)4 dx =

∫u4

3du

=1

3

∫u4 du

=1

3

(u5

5+ c

)=

(3x− 2)5

15+ c

Note that once we introduce the definite integral, the constant c will disappear. Further,∫ 3

−1(3x− 2)4 dx =

(3(3)− 2)5

15− (3(−1)− 2)5

15

=75

15− (−5)5

15

=6644

5

Thus

∫ 3

−1(3x− 2)4 dx =

6644

5. J


∫ 1

0ze(z2) dz.

49

Solution: Again, we begin by finding the indefinite integral of zez2

with respect to z.

We can use u-substitution, with u = z2 and du = 2z dz which isdu

2= z dz. We will

work slowly and carefully here.∫ze(z2) dz =

∫e(x2)(z dz)

=

∫eu

2du

=1

2

∫eu du

=1

2(eu + c)

=eu

2+ c

=e(z2)

2+ c

Note that once we introduce the definite integral, the constant c will disappear. Further,∫ 1

0ze(z2) dz =

e(12)

2− e(02)

2

=e

2− 1

2

=e− 1

2

Thus

∫ze(z2) dz =

e− 1

2. J

This concludes our section on the definite integral.

4.5 Area Under the Curve of a Function

In a previous course you would have learn the theory about how the definite integralcorresponds to the area under the curve from x = a to x = b. Here we again formalizethis and practice a few problems.

Theorem 4.5.1: Let f be an integrable function on the interval [a, b] with a ≤ b, thenthe exact area under the curve of f is given by the definite integral

A =

∫ b

af(x) dx

where the area of the region is further bound by the x axis, and the lines x = a andx = b. /

50

We illustrate this with the next diagram.

∫ b

a

f(x) dx

a b

x

f(x)

Figure 4.2: Area under the Curve of f(x)

Looking at the final problem in the last section by rephrasing:

Example 4.5.1. Find the area between the curve f(x) = (3x − 2)4, the x axis, andthe vertical lines x = −1 and x = 3.

Solution: To solve this problem we must compute:∫ 3

−1(3x− 2)4 dx

We found in the previous section that this definite integral evaluates to 66445 . Thus

Area under the curve =

∫ 3

−1(3x− 2)4 dx =

6644

5= 1328.8 units2

Concluding the problem. J

Example 4.5.2. Find the area between the curve y = ze(z2), the x axis, and thevertical lines z = 0 and z = 1.

Solution: To solve this problem we must compute:∫ 1

0ze(z2) dz

We found in the previous section that this definite integral evaluates to e−12 . Thus


∫ 1

0ze(z2) dz =

e− 1

2≈ 0.859 units2


51

Example 4.5.3. Find the area under the curve of the function g(x) = sinx from x = 0to x = π.

Solution: To solve this problem we must compute:∫ π

0sinx dx

To begin we must find the indefinite integral corresponding the above definite integral.∫sinx dx = − cos(x) + c

Next evaluate the definite integral∫ π

0sinx dx = − cos(π)− (− cos 0)

= 1 + 1

= 2

Thus


∫ π

0sinx dx = 2 units2


52

5 Integration Methods

In this chapter we focus on various integration methods. We begin with some simpleintegration formulas for Transcendental Functions and then move on to methods thatare somewhat more tedious than the methods previously discussed in the course.

5.1 Integrals of Exponential and Logarithmic Functions

Here we look at methods for integrating Exponential and Logarithmic Functions. Wealready know the integral for f(x) = ex.∫

ex dx = ex + c

Further, we can use u Substitution to make the formula above more versatile.

Here we are more interested in integrating a more general class of Exponential Function.Consider the function y = ax and the question of what is its integral. We begin byrewriting ax with properties of exponents and use u substitution.∫

ax dx =

∫eln ax dx

=

∫ex ln a dx

Next, let u = x ln a and du = ln a dx, furtherdu

ln a= dx. So continuing from above∫

eu

ln adu =

1

ln a

∫eu du

=1

ln a(eu + c)

=eu

ln a+ c

=eln ax

ln a+ c

=ax

ln a+ c

So

53

Property 5.1.1: Let a be an positive real number, then∫ax dx =

ax

ln a+ c

Example 5.1.1. Find

∫18x2

(42x3

)dx.

Solution: In this example we use substitution and the property above with u = 2x3,

du = 6x2 dx, anddu

6= x2 dx. So∫

18x2(

42x3)dx = 18

∫ (42x3

)x2 dx

= 18

∫4u

6du

=18

6

∫4u du

= 3

(4u

ln 4+ c

)=

3 · 4u

ln 4+ c

concluding our problem. J

Next, we move to integrating logarithms. We give the next formula without proof forthe time being. The proof will be given in a coming section.

Property 5.1.2: Let x be a positive real number, then∫lnx dx = x ln(x)− x+ c

Example 5.1.2. Find

∫x ln

(3x2)dx.

Solution: In this example we use substitution again and the property above with

u = 3x2, du = 6x dx, anddu

6= x dx. So∫

x ln(3x2)dx =

∫ln(3x2)x dx

=1

6

∫lnu du

=1

6(u ln(u)− u+ c)

=3x2 ln(3x2)− 3x2

6+ c

=x2 ln(3x2)− x2

2+ c


54

Finally, we use the previously unproven property to see how to integrate a generallogarithmic expression. We use a property of logarithms that you should be familiarwith from a previous algebra class. It is called the Change of Base Formula.

Property 5.1.3: Let a and x be positive real numbers, then∫loga x dx =

x ln(x)− xln a

+ c

Proof. We begin by assuming that a and x are positive real numbers, then∫loga x dx =

∫lnx

ln adx Change of Base

=1

ln a

∫lnx dx Integral of lnx

=x ln(x)− x

ln a+ c

as desired.

Example 5.1.3. Find

∫log(3x− 7) dx.

Solution: Again we use substitution and the property above with u = 3x − 7, du =

3 dx, anddu

3= dx. So∫

log(3x− 7) dx =

∫log u

3du

=1

3

∫log u du

=1

3

(u ln(u)− u

ln 10+ c

)=

(3x− 7) ln(3x− 7)− 3x+ 7

3 ln 10+ c

We can leave the answer in this form. J

This concludes our section.

5.2 Trigonometric Integrals

Here we look at integration of the remainder of the Transcendental Functions for thiscourse. Some of these integrals that are trivial to find with the Substitution Method,others will require theory from a later section. We present them here with proof for thefirst three.

Property 5.2.1: Let x be some real number, then∫sinx dx = − cos(x) + c

55

Proof. Begin with the assumptions in the property statement, we rewrite the integral as∫sinx dx = −

∫− sinx dx

= − cos(x) + c


Property 5.2.2: Let x be some real number, then∫cosx dx = sin(x) + c

Proof. The integral should be obvious here becaused

dx(sin(x) + c) = cosx.

Property 5.2.3: Let x be some real number, then∫tanx dx = − ln | cosx|+ c

Proof. To prove this we use a trigonometric identity and substitution with u = cosx,du = − sinx dx, and − du = sinx dx.∫

tanx dx =

∫sinx

cosxdx

=

∫1

udu

= − ln |u|+ c

= − ln | cosx|+ c


Property 5.2.4: Let x be some real number, then∫cotx dx = ln | sinx|+ c

Property 5.2.5: Let x be some real number, then∫secx dx = ln | secx+ tanx|+ c

Property 5.2.6: Let x be some real number, then∫cscx dx = ln | cscx+ cotx|+ c

All of the rules above are applied in similar fashion. In many cases student will need toapply u substitution.

56

Example 5.2.1. Find

∫ 1

0x sinx2 dx.

Solution: Here we have a definite integral. We must begin by finding the indefinite

integral. We can us Substitution with u = x2, du = 2x dx, anddu

2= x dx. So we have

∫ 1

0x sinx2 dx =

∫ u1

u0

sinu

2du

=1

2

∫ u1

u0

sinu du

= −1

2[cosu]u1u0

= −1

2

[cosx2

]10

= −1

2[cos 1− cos 0]

≈ 0.2298


We come to the end of another section.

5.3 Average and Root Mean Square Values

We have reached a point were we can take a look at a simple application of integration,now that we have learned how to integrate Trigonometric Functions.

[3, page 451]It is easy to calculate the mean (average) value of finitely many numbersy1, y2, . . . , yn:

Average of y’s =y1 + y2 + · · ·+ yn

n

But how do we compute the average temperature during a day if infinitely many tem-perature readings are possible?

57

6 12 18 24

5

10

15

Tave

T (t)

t

T

Figure 5.1: Daily Temperature at Time t (in hours)

Perhaps we could divide the interval [a, b] of our function into n equal subintervals eachwith length ∆x = (b− a)/n. Then choose points inside each subinterval x∗1, x∗2, x∗1,. . . ,x∗n and calculate the mean of the numbers f(x∗1), f(x∗2), . . . , f(x∗n), which would be:

f(x∗1) + f(x∗2) + · · ·+ f(x∗n)

n

but remember from above that

∆x = (b− a)/n

b− a∆x

= n

So

f(x∗1) + f(x∗2) + · · ·+ f(x∗n)

n=f(x∗1) + f(x∗2) + · · ·+ f(x∗n)

b−a∆x

=1

b− a[f(x∗1)∆x+ f(x∗2)∆x+ · · ·+ f(x∗n)∆x]

=1

b− a

n∑i=1

f(x∗i )∆x

As you may expect, this is only going to give us an approximation of the average value,so we do what any good calculus student would do and let n get very large or take thelimit as n goes to infinity.

limn→∞

1

b− a

n∑i=1

f(x∗i )∆x =1

b− a

∫ b

af(x) dx

Note that as n gets larger and larger, the interval ∆x must get smaller and smaller, thusbecoming dx.

58

Definition 5.3.1: Let f be an integrable function over the interval (a, b), then theAverage Value of f or Average Ordinate is given by the integral

1

b− a

∫ b

af(x) dx

/

Example 5.3.1. [1, page 1000] Find the average ordinate of a half-cycle of the sinu-soidal voltage

v = V sin θ (volts)

Solution: First, a half cycle has length π. So we can use the interval from a = 0 tob = π and compute the average value.

Vave =V

π − 0

∫ π

0sin θ dθ

=V

π[− cos θ]π0

=V

π[− cosπ + cos 0]

=2V

π≈ 0.637V volts


In physics and probability theory there is application for what is called the Root MeanSquare Value of a Function. In higher mathematics it is called the L2-Norm. Wewill not derive this quantity as it is similar to the previous derivation.

Definition 5.3.2: Let f be an integrable function on the interval (a, b), then theContinuous Root Mean Square of f is given by

rms =

√1

b− a

∫ b

a[f(x)]2 dx

/

There is nothing special with either of the formulas in this section, we simply substituteand integrate.

Example 5.3.2. [1, page 1001] Find the rms value for the sinusoidal voltage in theprevious example.

Solution: We simply need to substitute into the equation above and integrate. Be

59

aware that the integration here could been done with several combinations of steps.

rms =

√1

b− a

∫ b

a[f(θ)]2 dθ

=

√1

π − 0

∫ π

0V 2 sin2 θ dθ

=

√V 2

π

∫ π

0sin2 θ dθ

=

√V 2

π

∫ π

0

1

2(1− cos(2θ)) dθ Half Angle Formula

Next, we use u substitution on the Cosine portion of the integrand, with u = 2θ, du =

2 dθ, anddu

2= dθ.√

V 2

π

∫ π

0

1

2− cos(2θ)

2dθ =

√V 2

π

∫ π

0

1

2dθ −

∫ π

0

cos(2θ)

2dθ

=

√V 2

π

[[θ

2

]π0

−∫ π

0

cosu

4du

]

=

√√√√V 2

π

[[θ

2

]π0

−[

sinu

4

]uπu0

]

=

√V 2

π

[[θ

2

]π0

−[

sin 2θ

4

]π0

]

=

√V 2

π

[π2− 0]

=

√V 2

π· π

2

=

√V 2

2

=V 2

√2

≈ 0.707V (volts)

As an aside students should be aware that the integral above may come up in otherplaces such as homework, quizzes, or tests. This concludes our problem. J

60

5.4 Integration by Parts

In this section we take a look at integration technique that is essentially the reverseprocess for the product rule from differentiation. We know the product rule for theproduct function, u · v)(x), as

d

dx(uv) = u

dv

dx+ v

du

dx

Multiplying both sides by dx we have

d(uv) = u dv + v du

Rearranging givesu dv = d(uv)− v dv

Next, integrating both sides we have∫u dv =

∫d(uv)−

∫v du

After simplification we have the formula

Property 5.4.1: (Integration by Parts) Let u and v be integrable functions, then∫u dv = uv −

∫v du

Integration by parts is a tool that will allow us to integrate an even more general classof function. Next, we give some helpful information about integration by parts:

• In problems in which you are asked to evaluate an integral the integrand in yourproblem corresponds to the left hand side of the equation in Property 5.4.1.

• The question is: what do we let u and dv be? We describe a priority system afterthe first example.

• After assigning the above components always find du and v by differentiation andintegration respectively.

• Steps can sometimes become confusing because of notation. For example we useu in the Integration by Parts process and at the same time we may have to use usubstitution to integrate some intermediate step. Perhaps use another variable toavoid confusion.

• Integration by Parts is not always a trivial process and sometimes requires finesseand creativity.

Example 5.4.1. Find

∫x cos(3x) dx.

61

Solution: Note here that the integrand is a product with which we have no previoustools to integrate. We will use integration by parts and use the rule of thumb above.Let u = x and dv = cos(3x) dx. Next, we find du and v. With du = dx and find vby u substitution (use another variable, perhaps w). We have w = 3x, dw = 3 dx, anddw3 = dx.

v =

∫cos(3x) dx

=

∫cos(w)

3dw

=1

3

∫cos(w) dw

=1

3(sin(u) + c1)

=1

3sin(3x) + c1

So we now have u, v, du, and dv. Substitute into the Parts formula.∫u dv = uv −

∫v du∫

x cos(3x) dx = x

(1

3sin(3x) + c1

)−∫

1

3sin(3x) + c1 dx

=x

3sin(3x) + xc1 −

∫1

3sin(3x) + c1 dx

=x

3sin(3x) + xc1 −

1

3

∫sin(3x) dx− c1

∫dx

=x

3sin(3x) + xc1 −

1

3

∫sin(3x) dx− c1

∫dx

Next, we have to integrate by u substitution, we use w again and retrieve the samevalues as w from our work above, so continuing from the last line from above we have

=x

3sin(3x) + xc1 −

1

3

∫1

3sin(w) dx− c1

∫dx

=x

3sin(3x) + xc1 +

1

9

∫cos(w) dx− c1x

=x

3sin(3x) + xc1 +

1

9cos(3x)− xc1 + c2

=x

3sin(3x) +

1

9cos(3x) + c Notice xc1’s canceled

This is our solution. Students should verify that this in fact the correct integral bydifferentiation. J

62

Note that the constants resulting from the integration of v disappeared, this will alwayshappen like this, so it may be helpful for some students to ignore it and only use aconstant of integration at the end of the problem.

To say a little more about our choice of u in the original problem with the Parts Formula,there are some exceptions for taking u outside of the rule of thumb for taking dv as themore complicated piece.

Take u in this priority (LIATE):

1. Logarithmic Function

2. Inverse Trigonometric Function

3. Algebraic Function

4. Trigonometric Function

5. Exponential Function

After choosing u using the priority system above, let everything else be dv. Further,Students are highly recommended to memorize this priority system as it will not beprovided on the tests.

We now use this to prove a previously unproven rule:

Proposition 5.4.1: Let x be a positive real number, then∫lnx dx = x ln(x)− x+ c

Proof. We integrate by parts. Using the priority system above we should let u = lnxand dv = dx. This gives us du = 1

x dx and v = x. Further,∫u dv = uv −

∫v du∫

lnx dx = x ln(x)−∫x

xdx

= x ln(x)−∫

dx

= x ln(x)− x+ c

as desired.

Example 5.4.2. Find

∫xex dx.

63

Solution: Using the priority system above we choose u = x because it is algebraic, sodv = ex dx, which gives du = dx and it is easy to see that v = ex. Next, we substituteinto the Parts formula: ∫

u dv = uv −∫v du∫

xex dx = xex −∫ex dx

= xex − ex + c

which is our solution. Again, students should differentiate to check your answer. J

It is legitimate to ask what would have happened if we had let u = ex and dv = x dx,and so on. Students are encouraged to try this way, however, the resulting Integrationby Parts steps result in an integral that is more difficult to integrate than the originalproblem. If this happens in practice, consider reversing u and dv.

In some cases we have to Use the Parts Formula more than once.

Example 5.4.3. Integrate

∫x2ex dx.

Solution: Here we use the priority system and choose u = x2 because it is algebraicand dv = ex dx. With this choice we have du = 2x dx and v = ex. Substitution into theParts Formula gives ∫

u dv = uv −∫v du∫

x2ex dx = x2ex −∫

2xex dx

Notice again that we have a product function as an integrand on the far right side, so nowwe use Integration by Parts again with u1 = 2x because it is algebraic and dv1 = ex dx.This gives us du1 = 2 dx and v1 = ex. So we now apply this to the integral on the farright of the above using the Integration by Parts Formula.∫

x2ex dx = x2ex −∫

2xex dx

= x2ex −[u1v1 −

∫v1 du1

]= x2ex −

[2xex −

∫2ex dx

]= x2ex − 2xex + 2

∫ex dx

= x2ex − 2xex + 2ex + c


This brings us to the end of our section.

64

5.5 Integrating Rational Functions

We have looked at many integration methods. In this section we expand our integrationtools to include a broader selection of Rational Functions (functions with fractions). Wewill use the method of Partial Fraction Decomposition in addition to other tools fromLinear Algebra to rewrite Rational Expressions in a more useful form.

To begin there are two types of Rational Function (expression): Proper and Improper.

Definition 5.5.1: A Proper Rational Expression is a Rational Expression wherethe degree of the numerator is less than the degree of the denominator. /

and

Definition 5.5.2: An Improper Rational Expression is a Rational Expressionwhere the degree of the numerator is greater than the degree of the denominator. /

To be clear, students should also be aware that a Function or Expression is not Rationalunless the both the numerator and denominator contain polynomials.

Integrating Improper Rational Functions

When integrating improper rational expressions, if we are lucky, the denominator willbe a factor of the numerator. In this case we simply divide out the common factor andintegrate. As an example we look at a problem worked earlier in the course.

Example 5.5.1. [1, p.887] Find∫x5−2x3+5x

x dx.

Solution: We begin as discussed above and cancel the common factor between thenumerator and denominator.∫

x5 − 2x3 + 5x

xdx =

∫ (x4 − 2x2 + 5

)dx

=

∫x4 − 2

∫x2 dx+ 5

∫dx

=x5

5− 2x3

3+ 5x+ c


However, as you may expect, this is a special case scenario and does not happen often.In general, we must use long division from algebra to convert the rational expressioninto a potentially more integrable form. Consider the following:

Example 5.5.2. [1, page 1005] Find

∫x3 − 2x2 − 5x− 2

x+ 1dx.

Solution: It may not always be possible to factor and cancel as in the previous problem

65

(it actually is here). None-the-less, we use long division.

x2− 3x− 2

x+ 1)x3−2x2−5x−2

x3+ x2

−3x2−5x

−3x2−3x

−2x−2

−2x−2

Thus ∫x3 − 2x2 − 5x− 2

x+ 1dx =

∫(x2 − 3x− 2) dx

=

∫(x2 − 3x− 2) dx

=x3

3− 3x2

2− 2x+ c

which brings us to the end of the problem. J

Integrating Proper Rational Functions with Partial Fractions

We now deal with the more complex aspect of this section, integrating a proper rationalfunction. In this section students should be aware that problems can be tedious andmistake prone. As a consequence, students should take plenty of time on these problems,work carefully, checking your work over at regular intervals. Additionally, this sectionis likely to expose student deficiencies in algebra, not calculus, as this section is veryalgebra intensive.

Consider the rational expressions

3

x+ 1and

2

x− 4

Note that the sum of these two expressions is

3

x+ 1+

2

x− 4=

3x− 12 + 2x+ 2

(x+ 1)(x− 4)

=5x− 10

x2 − 3x− 4

The above computation is a standard algebraic operation on two rational expressions.The reverse of the above process is called Partial Fraction Decomposition. In otherwords,

3

x+ 1and

2

x− 4

66

are the Partial Fractions of5x− 10

x2 − 3x− 4

As you may suspect, reversing this process is not trivial.

Rules to Keep in Mind for Partial Fraction Decompositions

1. Factor the denominator completely to find the set of all possible factors.

2. For each linear factor ax+ b in the denominator, there will be a partial fraction

Aiax+ b

3. For the repeated factors (ax+ b)n, there will be n partial fractions of the form

n∑i=1

Ai(ax+ b)i

4. For each quadratic factor ax2 + bx+ c, there will be a partial fraction

Aix+Biax2 + bx+ c

5. For the repeated quadratic factors (ax2 + bx+ c)n, there will be n partial fractionsof the form

n∑i=1

Aix+Bi(ax2 + bx+ c)i

In this course students will rarely have to deal with issues related to items 4 and 5 above.

Arguably the easiest way to learn Partial Fraction Decomposition is to practice it. Sowe begin by working the previous problem backwards.

Example 5.5.3. Find the partial fraction decomposition of 5x−10x2−3x−4

.

Solution: We already know the answer, thus we will use this as a good learning op-portunity. We begin by factoring the denominator completely. Note here that there isno repeated factor.

5x− 10

x2 − 3x− 4=

5x− 1

(x+ 1)(x− 4)

Secondly, we break the right hand side of the above into two fractions with a differentunknown numerator. The denominators will be the unique factors.

5x− 10

(x+ 1)(x− 4)=

A

x+ 1+

B

x− 4

67

Our goal here is to find A and B. To start this we multiply both sides by the factored(or unfactored, it does not matter) denominator.

5x− 10

(x+ 1)(x− 4)=

A

x+ 1+

B

x− 4

(x+ 1)(x− 4)5x− 10

(x+ 1)(x− 4)= (x+ 1)(x− 4)

[A

x+ 1+

B

x− 4

]5x− 10 =

A(x+ 1)(x− 4)

x+ 1+B(x+ 1)(x− 4)

x− 4

5x− 10 = A(x− 4) +B(x+ 1)

5x− 10 = Ax− 4A+Bx+B

Next, we regroup all of the like terms in terms of the variable x and factor out.

5x− 10 = Ax− 4A+Bx+B

5x− 10 = Ax+Bx− 4A+B

5x− 10 = (A+B)x− 4A+B

Now, looking at the last equation above very carefully note that{A+B = 5

−4A+B = −10

Which is the a system of equations in two variables. We can solve this with severalmethods. We use substitution here.

A+B = 5

A = 5−B

Now, by substituting into the other equation we retrieve

−4(5−B) +B = −10

−20 + 4B +B = −10

5B = 10

B = 2

We can now back-substitute int the system to find A.

A = 5−B= 5− 2

= 3

This verifies that the partial fraction decomposition is 3x+1 and 2

x−4 . J

68

Students are again reminded that these problems are time consuming and are very easyto make mistakes in.

We have yet to see what this has to do with integration. Let us look at an example thatuses the very problem we just worked.

Example 5.5.4. Find

∫5x− 10

x2 − 3x− 4dx.

Solution: Note here that as we may try this problem is not integrable with any ofthe methods that we have discussed thus far. So we convert the integrand to its partialfraction decomposition (which we just found).∫

5x− 10

x2 − 3x− 4dx =

∫3

x+ 1+

2

x− 4dx

= 3

∫1

x+ 1dx+ 2

∫1

x− 4dx

= 3 ln |x+ 1|+ 2 ln |x− 4|+ c

Generally the answer above is acceptable, however, your textbook will combine thelogarithms into a single logarithm. Thus we show the work here.

= 3 ln |x+ 1|+ 2 ln |x− 4|+ c

= ln |x+ 1|3 + ln |x− 4|2 + c

= ln∣∣(x+ 1)3(x− 4)2

∣∣+ c


What you have seen in the above example is essentially the gist of using partial frac-tion decomposition to integrate proper fractions. In fact there are many integral tableformulas derived from this process. We work one more example for practice.

Example 5.5.5. Find

∫4x3 − 20x2 + 35x− 24

x(x− 2)3dx.

Solution: So again we work through the partial fraction decomposition because inte-grating the proper fraction cannot be completed by previous methods. Notice that thereis a repeated linear factor so

4x3 − 20x2 + 35x− 24

x(x− 2)3=A

x+

B

x− 2+

C

(x− 2)2+

D

(x− 2)3

We need to find A, B, C, and D. To begin, multiply both sides by the denominator onthe left hand side.

x(x− 2)3 4x3 − 20x2 + 35x− 24

x(x− 2)3= x(x− 2)3

[A

x+

B

x− 2+

C

(x− 2)2+

D

(x− 2)3

]4x3 − 20x2 + 35x− 24 = A(x− 2)3 +Bx(x− 2)2 + Cx(x− 2) +Dx

69

Continuing to multiply out we have

= Ax3 − 6Ax2 + 12Ax− 8A+Bx3 − 4Bx2 + 4Bx+ Cx2 − 2Cx+Dx

Next, we collect like terms in terms of powers of x which gives

4x3 − 20x2 + 35x− 24 = (A+B)x3 + (−6A− 4B + C)x2 + (12A+ 4B − 2C +D)x+ (−8A)

This generates the system of equations:A+B = 4

−6A− 4B + C = −20

12A+ 4B − 2C +D = 35

−8A = −24

We can solve the last equation easily for A, giving us A = 3. Next, use this solution tosolve for B in equation 1, giving us B = 1. Further, we solve the second equation byusing our solutions above.

−6A− 4B + C = −20

−6(3)− 4(1) + C = −20

−22 + C = −20

C = 2

Next, solving the third equation similarly we get D = −1. We now have the decompo-sition by substitution of these values and thus we can attempt to integrate.∫

4x3 − 20x2 + 35x− 24

x(x− 2)3dx =

∫3

x+

1

x− 2+

2

(x− 2)2− 1

(x− 2)3dx

= 3

∫1

xdx+

∫1

x− 2dx+ 2

∫1

(x− 2)2dx−

∫1

(x− 2)3dx

= 3 ln |x|+ ln |x− 2| − 2

x− 2+

1

2(x− 2)2

This concludes a very long problem. Note that the two integrals on the far right arefound by u Substitution. J

Students are welcome to leave their answers in uncombined fashion, but be remindedthe textbook will often times attempt to combine logarithms. This brings us to the endof another section.

5.6 Integration by Trigonometric Substitution

In this section we take advantage of trigonometry to help us integrate a few specialproblems. Consider a situation where we are trying to find the area under a curvedescribed by an ellipse or circle. This may have the form∫ √

a2 − x2 dx

70

The method of Trigonometric Substitution will help us to solve this problem. The basicidea behind Trigonometric Substitution is similar to u Substitution and the idea can beexpanded to integrate a wider variety of functions and expressions.

We begin with a most basic idea from a previous course.

Theorem 5.6.1: (Pythagorean Theorem) Let x and y be legs of a right triangleand r be the hypotenuse, then

x2 + y2 = r2

/

θ

x (Adjacent)

y (Opposite)

r(H

ypot

enuse

)

Figure 5.2: An angle in standard position

We can rewrite the above in terms of the angle θ and Trigonometric Functions:

sin2 θ + cos2 θ = 1

This should be a familiar identity. In fact, there are many identities that you may recallthat can be derived from this. Many of these will be useful at some point in this section.

sec2 θ = 1 + tan2 θ csc2 θ = 1 + cot2 θ csc θ = 1sin θ

sec θ = 1cos θ cot θ = 1

tan θ sin2 θ =1

2(1− cos(2θ))

cos2 θ =1

2(1 + cos(2θ)) sin(2θ) = 2 sin θ cos θ cos(2θ) = cos2 θ − sin2 θ

Table 5.1: Some Trigonometric Identities

71

Students should also be proficient at converting each of these into a slightly differentform for versatility.

The next question is: What substitution can be made to simplify these special inte-grands? To answer this we must consider what would simplify an integral of the formdiscussed at the beginning of this section:∫ √

a2 − x2 dx

Considering many of the Trigonometric Identities have a term of 1, it might be a goodidea to eliminate the a2 in the integrand above. Thus if x contained a, maybe x = ab asa factor we could factor out a2 and

√a2 = |a|, which could then be taken front of the

integral sign. This would leave

a

∫ √1− b2 dx

The next concern would be b. What should it be? Well note that from the originaltrigonometric identity

sin2 θ + cos2 θ = 1

cos2 θ = 1− sin2 θ

cos θ =√

1− sin2 θ

The right hand side of the bottom equation is exactly what we are now looking at afterthe a has been removed. Note that x in the original integrand has already been squared,thus the correct value for b is sinθ and an appropriate Trigonometric Substitution forthis integrand would be x = a sin θ. As can be seen from the equation we could replacethe above integrand, temporarily, with an integrand involving cos θ, which will turn outin many cases to be far easier to integrate than the original integrand.

As a side note, we must restrict θ such that −π2 ≤ θ ≤

π2 . This point will essentially be

irrelevant for our purposes.

Let us verify that this actually works

Example 5.6.1. Find

∫dx

x2√

4− x2.

Solution: Note here that the integrand contains a term of the form a2−x2 with a = 2,thus we will make the substitution discussed above. So we let x = a sin θ = 2 sin θ, furtherwe will need to know what dx is so dx = 2 cos θ dθ.

Next, we rewrite the integral above in terms of the trigonometric substitution and sim-

72

plify using trigonometric identities.∫dx

x2√

4− x2=

∫2 cos θ dθ

4 sin2 θ√

4− 4 sin2 θ

=1

2

∫cos θ dθ

sin2 θ√

4(1− sin2 θ)

=1

4

∫cos θ dθ

sin2 θ√

1− sin2 θTake out

√4

=1

4

∫cos θ dθ

sin2 θ cos θConvert root with Trig. Id.

=1

4

∫dθ

sin2 θ

=1

4

∫csc2 θ dθ Trig. Id.

= −1

4

∫− csc2 θ dθ

d

dθ(cot θ) = − csc2 θ

= −1

4cot θ + c

As with u Substitution, we must remember to take out the θ. This is easily accomplishedby drawing the right triangle for θ from the original substitution. Remember thatx = 2 sin θ, which means x

2 = sin θ, which implies that x is the opposite side and 2 is thehypotenuse. Further, using the Pythagorean Theorem or other identities, the adjacentside is

√4− x2.

θ√

4− x2

x2

Figure 5.3: Right Triangle for our Substitution

73

From this triangle we can find the cot θ:

cot θ =Adjacent

Opposite

=

√4− x2

x

Thus ∫dx

x2√

4− x2= −1

4cot θ + c

= −√

4− x2

4x+ c

This concludes our problem. Students should verify this result by differentiation. Becareful as you will need to use the chain rule in conjunction with the quotient rule andthen do some simplification to get the original integrand. J

At this point we have one substitution technique, where the substitution is trigonometric.Next, we list all of the standard trigonometric substitutions that we will use in thiscourse.

Expression Substitution Identity√a2 − x2 x = a sin θ 1− sin2 θ = cos2 θ√a2 + x2 x = a tan θ 1 + tan2 θ = sec2 θ√x2 − a2 x = a sec θ sec2 θ − 1 = tan2 θ

Table 5.2: Trigonometric Substitutions

The next example illustrates why it is a good idea to be proficient with Table 5.1.

Example 5.6.2. Find

∫dx

(x2 + 1)2.

Solution: According to Table 5.2 we should use the substitution x = 1 tan θ = tan θ.Further, dx = sec2 θ dθ.

74

∫dx

(x2 + 1)2=

∫sec2 θ dθ

(tan2 θ + 1)2

=

∫sec2 θ dθ

(sec2 θ)2

=

∫sec2 θ dθ

sec4 θ

=

∫dθ

sec2 θ

=

∫cos2 θ dθ Trig. Id.

=

∫1− sin2 θ dθ Trig. Id.

=

∫dθ −

∫sin2 θ dθ

= θ −[∫

1

2dθ −

∫cos(2θ)

2dθ

]= θ −

[1

2θ −

∫cos(2θ)

2dθ

]= θ − 1

2θ +

1

2

∫cos(2θ) dθ

=1

2θ +

1

4sin(2θ) + c u Substi.

=1

2θ +

1

4(2 sin θ cos θ) + c Trig. Id.

Finally, we need to draw our triangle to remove θ.

θ

1

x√ x2 +

1


75

This gives us

=1

2θ +

1

4(2 sin θ cos θ) + c

=1

2tan−1 x+

1

2

x√x2 + 1

· 1√x2 + 1

+ c

=1

2tan−1 x+

x

2x2 + 2+ c

Completing a very long problem. This problem illustrates how integration, in general,can be quite complicated and often impossible. J

We work one more problem involving a definite integral. The only difference with adefinite integral is that you substitute the limits of integration in at the very end afterthe indefinite integral has been evaluated.

Example 5.6.3. Find

∫ 4

√7

dx

x2√x2 − 7

.

Solution: Here we use the substitution x =√

7 sec θ which gives dx =√

7 sec θ tan θ dθ.We begin by evaluating the indefinite integral.∫

dx

x2√x2 − 7

=

∫ √7 sec θ tan θ dθ

7 sec2 θ√

7 sec2 θ − 7

=

√7

7

∫tan θ dθ√

7 sec θ√

sec2 θ − 1

=

√7

7

∫tan θ dθ√

7 sec θ√

sec2 θ − 1

=

√7

7√

7

∫tan θ dθ

sec θ tan θ

=1

7

∫dθ

sec θ

=1

7

∫cos θ dθ

=1

7sin θ

Next, we remove θ by drawing the appropriate triangle.

76

θ√

7

√x2 − 7

x


This gives us

=1

7sin θ

=1

7·√x2 − 7

x+ c

=

√x2 − 7

7x+ c

Finally, we can substitute our limits of integration, remembering to ignore the constantof variation. ∫ 4

√7

dx

x2√x2 − 7

=

[√x2 − 7

7x

]4

√7

=

√42 − 7

7(4)−

√√7

2 − 7

7√

7

=3

28− 0

7√

7

=3

28

which is our final answer. J

This concludes our section and chapter on integration techniques. From this point on,in these notes, we will be focused more on applications of calculus. It would be an idealtime to review all differentiation and integration techniques.

77

6 Differential Equations

In this chapter we focus on an introduction to differential equations and some methodsof solving such equations. A Differential Equation is an equation that contains oneor more derivatives. When we solve a simple linear equation in algebra we are typicallyexpecting a number as a result. For example, when we solve the equation: 2x + 1 = 5we get the solution x = 2. Our solution is a number. But as you should recall from thiscourse and previous courses the solution to a differential equation is a family of functions,in general. If we are given a Boundary Value Problem -a differential equation witha boundary condition, then the solution may be a specific function.

We begin this chapter with a classification system for differential equations.

6.1 Notation and Classification

We begin with notation. Differential equations may be given in several forms thatcorrespond to the different expressions used to denote a derivative.

Example 6.1.1. Some common differential equation notation.

a)dy

dx− 3 = 4xy

b) y′′ − 6y′ + 2xy = 0

c) Dy − 3 = 4xy

d) D2y − 6Dy + 2xy = 0

In this document we will typically (but not always) use the Leibniz notation,dy

dx. We

prefer the Leibniz notation because it is intuitive to convert to differential form-we simplymultiply both sides by dx.

dy

dx− 3 = 4xy

dy − 3 dx = 4xy dx

Differential form will be useful when we start to solve these equations.

Next, we turn our attention to the classification of differential equations.

78

Ordinary versus Partial Differential Equations

There are two major classes of differential equations: Ordinary and Partial

Definition 6.1.1: An Ordinary Differential Equation, or ODE, is a differentialequation that contains one or more derivatives of an unknown function that depends ona single independent variable. /

Definition 6.1.2: A Partial Differential Equation, or PDE, is a differential equa-tion that contains one or more derivatives of an unknown function that depends on morethan one independent variable. /

The examples listed above are all Ordinary Differential Equations. In this course wewill be interested only in ODE’s as Partial Differential Equations are complex to solveand often times require numerical methods and a computer. None-the-less we list someexamples of PDE’s to show their notation.

Example 6.1.2. Some common PDE notation.

a)∂y

∂t= 3

∂x

∂t

b)∂2y

∂x2− 2

∂y

∂x− 4y = 0

The symbol ∂ is used to denote a partial derivative, which we do not cover in thisdocument.

Order of a Differential Equation

Definition 6.1.3: The Order of a differential equation is the order of the highestorder derivative in the equation. /

Example 6.1.3. Find the order of each of the ODE’s:

a)dy

dx− 3 = 4xy

b) y′′ − 6y′ + 2xy = 0

c) −4y′′′ + 6y′′ = 7xy

Solution: Here we simply find the highest order derivative.

a) In our first example we can see that the highest order derivative is a first derivativeso the degree of this ODE is one.

b) Next, we can see that the highest order derivative is a second derivative so thedegree of this equation is two.

c) Finally, we can see that the highest order derivative here is a third derivative sothe degree of this equation is three.


79

Degree of a Differential Equation

Here we will need to discuss two details before looking at examples:

Definition 6.1.4: The Degree of a Derivative is the power to which that derivativeis raised. /

and

Definition 6.1.5: The Degree of a Differential Equation is the degree of thehighest order derivative in the equation. /

Example 6.1.4. [1, page 1031]

a) (y′′)3 − 5(y′)4 = 7 is a third-degree, second-order ODE

b) To find the degree of the the ODE

x√y′ − 2

= 1

we must get the derivative out of the denominator and out from under the radical.To achieve this, multiply both sides by the denominator on the left and squareboth sides. This gives us

y′ − 2 = x2

which makes this a first-degree, first-order ODE.

Solving Simple Differential Equations

When we are given a differential equation the first method of solving should be tomultiply both sides by dx and then integrate both sides.

Example 6.1.5. Solve the equationdy

dx= 5x+ 3.

Solution: We begin as suggested by multiplying both sides by dx.

dy

dx= 5x+ 3

dy = (5x+ 3) dx∫dy =

∫(5x+ 3) dx

y =5

2x2 + 3x+ c

Completing our problem. J

80

Checking a Solution

To check a possible solution to a differential equation we simply substitute the functionand its derivatives into the equation.

Example 6.1.6. [1, page 1032] Is the function y = e2x a solution of the differentialequation y′′ − 3y′ + 2y = 0?

Solution: Begin by finding the necessary derivatives of y.

• y = e2x

• y′ = 2e2x

• y′′ = 4e2x

Next, substitute into the differential equation and verify that the equation balances.

y′′ − 3y′ + 2y = 0

4e2x − 3(2e2x

)+ 2

(e2x)

= 0

4e2x − 6e2x + 2e2x = 0

−2e2x + 2e2x = 0

0 = 0

Which certainly balances, so y = e2x is a solution to the above ODE. J

One may ask if the above solution is the only solution to the given ODE. It turns outit is not. In fact the above solution is a Particular Solution. There is a GeneralSolution. Students should consider testing the function y = C1e

2x + C2ex to see if it

solves the ODE above.

This brings us to the end of another section.

6.2 Separation of Variables

Certain First-Order Differential Equations, f(x, y) are solvable by separation of vari-ables. This method essentially implies a process of multiplying through to convert theequation to differential form, then rearranging the equation so that x’s are collected onone side and y’s on the other.

Definition 6.2.1: A Separable Equation is a first-order differential equation inwhich the expression for dy

dx can be factored as a function of x times a function of y-meaning it can be written in the form

dy

dx= g(x)f(y)

/

81

Example 6.2.1. Solve the differential equationdy

dx=x2

y2and find the solution that

satisfies the boundary condition y(0) = 1.

Solution: We begin by rewriting the equation in differential form and attempt tocollect the x’s on one side and y’s on the other. After-which we will integrate both sides.

dy

dx=x2

y2

y2 dy = x2 dx∫y2 dy =

∫x2 dx

1

3y3 =

1

3x3 + c

y =3√x3 + c

Note that would could have used constants on both sides on line 4 above, however, wewould be required to subtract the constant on the left from the constant on the right toisolate y. This would have just given us yet a third constant, thus it will save time byjust using one constant on either side.

Next, we must use the boundary condition to find c.

y =3√x3 + c

1 =3√

03 + c

1 = 3√c

1 = c

So our solution is y = 3√x3 + 1. J


dx=

6x2

2y + cos y.

Solution: We begin by rewriting the equation in differential form and attempt tocollect the x’s on one side and y’s on the other. After-which we will integrate both sides.

dy

dx=

6x2

2y + cos y

(2y + cos y) dy = 6x2 dx∫(2y + cos y) dy =

∫6x2 dx

y2 + sin y = 2x3 + c

Here we leave the solution in implicit form because it is impossible to express y explicitlyas a function of x. J

82


dx= xy2.

Solution: Before we begin, students should note that y = 0 will solve the abovedifferential equation, we call this the trivial solution. It is advisable to always look outfor the trivial solution.

We begin by rewriting the equation in differential form and attempt to collect the x’son one side and y’s on the other. After-which we will integrate both sides.

dy

dx= xy2

1

y2dy = x dx∫

1

y2dy =

∫x dx

−1

y=

1

2x2 + c

y = − 2

x2 + c

So our general solution is y = 0 and y = − 2

x2 + c. J


dx=

y

5− x.

Solution: Note again, that y = 0 will solve the above DE. But we also need to findthe non-trivial solution.

We begin by rewriting the equation in differential form and attempt to collect the x’son one side and y’s on the other. After-which we will integrate both sides.

dy

dx=

y

5− x(5− x) dy = y dx

1

ydy =

1

5− xdx∫

1

ydy =

∫1

5− xdx

ln |y| = − ln |5− x|+ c

ln |y|+ ln |5− x| = c

ln [y(5− x)] = c

eln[y(5−x)] = ec

y(5− x) = ec

y =ec

5− x

83

So our general solution is y =ec

5− xwhich can also be written as y =

k

5− x. J

This concludes our section on on separable equations.

6.3 Applications of Differential Equations

In this section we take a brief look at some applications of differential equations derivedfrom RL and RC circuit theory. Here will focus our attention on the derivation of someuseful formulas in circuit theory.

RL Circuit

We begin our application discussion with RL circuits.

Definition 6.3.1: A resistor-inductor circuit or RL Circuit or RL Filter or RLNetwork, is an electric circuit composed of resistors and inductors driven by a voltageor current source. /

Here we consider only the simplest case of a first order RL circuit of one resistor andone inductor. Note that voltage across an inductance L is given by the equation

vL = Ldi

dt

E

R

L

Figure 6.1: Simple RL Circuit

In the above diagram we are interested in the change of current, i, over time t. As aresult of having the inductor at L, in the circuit, it takes time for the current to stabilizein the event we closed a switch above the battery at E. As the inductor at L will ”pushback” against the changing current.

To describe the change in current over time we will make use of Kirchhoff’s Loop Lawwhich, in simple terms, says that when we go through the circuit we must loose allvoltage. So we have

E −Ri− Ldidt

= 0

84

which is a first order, ordinary differential equation which happens to be separable.Continuing we have

E −Ri = Ldi

dtdt

L=

di

E −Ri∫dt

L=

∫di

E −Rit

L+ c = − 1

Rln |E −Ri|

−RtL

+ c = ln |E −Ri|

e−RtL

+c = eln(E−Ri)

ke−RtL = E −Ri

E − ke−RtL = Ri

E

R− ke−

RtL

R= i

We can now find k with the boundary condition of i(0) = 0 because the current at time0 is zero. This gives

E

R− ke−

R(0)L

R= 0

E

R=ke−

0L

RE = ke0

E = k

Substitution of E in for k we retrieve:

i =E

R− Ee−

RtL

R

=E

R

(1− e−

RtL

)which is the formula for the Current in a Charging Inductor.

The graph of the changing current is depicted below.

85

ER

t

i

Figure 6.2: Current Over Time in a Charging Inductor

It will likely come as no surprise that the term ER is called the steady-state current.

Further, from the voltage across an inductance, Kirchhoff’s Loop Law, and the currentin a charging inductor from above (multiply both sides by R and distribute E) we have

vL = Ldi

dt= E −Ri

= E −(E − Ee−

RtL

)= Ee−

RtL

which is the Voltage Across a Charging Inductor. See the graph below.

E

t

vL

Figure 6.3: Voltage Over Time in a Charging Inductor

86

RC Circuit

We now turn our attention to RC Circuits.

Definition 6.3.2: An RC Circuit is an electric circuit composed of resistors andcapacitors driven by a voltage or current source. /

A first order RC circuit is composed of one resistor and one capacitor and is the simplesttype of RC circuit. The diagram below presents a simple first order RC circuit.

E

R

C

Figure 6.4: Simple RL Circuit

[1, page 1058] Consider a situation with a fully charged capacitor is discharged by throw-ing the switch at t = 0. We will write an expression for the voltage across the capacitorand the current i. Again we use Kirchhoff’s Loop Law and the laws of physics and notethat the voltage across a resistor at any instant must be equal to the voltage across thecapacitor, but of opposite sign. Further, the current through the resistor, giving

− vR

= Cdv

dt

We can again separate variables and solve

dv

v= − dt

RC∫dv

v= −

∫dt

RC

ln v = − t

RC+ c

We know the boundary condition because at t = 0 the voltage across the capacitor isthe same as the battery voltage, E. Thus v(0) = E, which gives:

ln v = − t

RC+ c

lnE = c

87

Thus the solution is

ln v = − t

RC+ lnE

ln v − lnE = − t

RC

ln( vE

)= − t

RCv

E= e−

tRC

v = Ee−tRC

which is the formula for the Voltage Across a Discharging Capacitor.

Further, we get the current through the resistor and the Current Through the Dis-charging Capacitor by dividing the voltage v by R.

i =E

Re−

tRC

This concludes our analysis of circuits.

6.4 Second-Order Homogeneous Differential Equations (rightside zero)

The Differential Equations that we focus on in this section have many applications inphysics and engineering. Further, the solutions make heavy use of topics from earliercalculus and elementary algebra.

Definition 6.4.1: Any equation of the form

P (x)d2y

dx2+Q(x)

dy

dx+R(x)y = G(x)

orP (x)y′′ +Q(x)y′ +R(x)y = G(x) (6.1)

where P , Q, R, and G are continuous functions, is called a Second Order LinearDifferential Equation /

In this course we will only be interested in the case where P , Q, R, and G are constant,however there is considerable theory on solving the more general case. Further, in thissection we will be focusing on the case where G(x) = 0 (right side zero). When theright hand side of the equations above is zero we call the equation Homogeneous. IfG(x) 6= 0, then we call the equation non-homogeneous.

Clearly we wish to be able to solve these equations. There are two ideas that arefoundation to being able to solve Second-Order Homogeneous Differential Equations.The first we state as a theorem and prove.

88

Theorem 6.4.1: If y1(x) and y2(x) are solutions of the second-order homogeneousdifferential equation (6.1) and c1 and c2 are constants, then the function

y(x) = c1y1(x) + c2y2(x)

is also a solution of equation 6.1. /

Another way of stating the above theorem is: if we know two solutions of equation (6.1),then the linear combination y = c1y1 + c2y2 is also a solution.

Proof. Since y1 and y2 are solutions to equation (6.1), we have

P (x)y′′1 +Q(x)y′1 +R(x)y1 = 0

andP (x)y′′2 +Q(x)y′2 +R(x)y2 = 0

Further, by using the rules of differentiation, we generate the following

P (x)y′′ +Q(x)y′ +R(x)y

= P (x)[c1y1(x) + c2y2(x)]′′ +Q(x)[c1y1(x) + c2y2(x)]′ +R(x)[c1y1(x) + c2y2(x)]

= P (x)[c1y′′1(x) + c2y

′′2(x)] +Q(x)[c1y

′1(x) + c2y

′2(x)] +R(x)[c1y1(x) + c2y2(x)]

= P (x)c1y′′1(x) + P (x)c2y

′′2(x) +Q(x)c1y

′1(x) +Q(x)c2y

′2(x) +R(x)c1y1(x) +R(x)c2y2(x)

= c1[P (x)y′′1(x) +Q(x)y′1(x) +R(x)y1(x)] + c2[P (x)y′′2(x) +Q(x)y′2(x) +R(x)y2(x)]

= c1[0] + c2[0]

= 0

Thus the linear combination of solutions is also a solution to the second-order homoge-neous differential equation (6.1).

The other theorem is more abstract and will not be proven, as proof would requireconsiderable material from a more advanced course. Simply put it says that the generalsolution is a linear combination of two linearly independent solutions y1 and y2.Linearly independent means that neither of these solutions are constant multiples of theother. For example: y1(x) = 5x3 and y2(x) = 10x3 are not linearly independent becausewe can multiply y1 by a constant to generate y2; but y1(x) = ex and y2(x) = xex areindependent.

Before we move on, let us recall that in this course we will only be interested in thecase were equation (6.1) will have constant coefficients. As a result we will rewrite thisequation as

ay′′ + by′ + cy = 0 (6.2)

89

Theorem 6.4.2: If y1 and y2 are linearly independent solutions of equation 6.1, andP (x) 6= 0, then the general solution is given by

y(x) = c1y1(x) + c2y2(x)

where c1 and c2 are constants. /

Moving forward now that we have some theory that says that there exists solutions andwe have a little information about the structure of the solution. We need to think of thetype function that might satisfy equation (6.2). To make it easier think carefully, we arelooking for a function y such that a constant times its second derivative plus anotherconstant times the first derivative plus a third constant times the function is zero. Weknow that the function y = erx (with constant r) is a function that has the propertythat its derivative is a constant multiple of itself: y′ = rerx and y′′ = r2erx. Let us nowsubstitute these expressions into equation (6.2).

ay′′ + by′ + cy = 0

ar2erx + brerx + cerx = 0

(ar2 + br + c)erx = 0

Students should recall from algebra that erx is never 0. Thus, y = erx is a solution ofequation (6.2) if r is a root of the quadratic equation

ar2 + br + c = 0 (6.3)

Equation (6.3) is called the Characteristic Equation or Auxiliary Equation of thedifferential equation ay′′ + by′ + cy = 0.

As a result of the above work, students should note that the roots r1 and r2 can befound by factoring or in general by the quadratic formula.

r1 =−b+

√b2 − 4ac

2a

and

r2 =−b−

√b2 − 4ac

2aFrom the above we must discuss three possible cases based on the value of the discrimi-nant b2 − 4ac.

1. Case 1: b2 − 4ac > 0

In this case the roots r1 and r2 of the characteristic equation are real and distinct,so y1 = er1x and y2 = er2x are two linearly independent solutions of equation (6.2).Thus by the previous theorem

Property 6.4.1: If the roots r1 and r2 of the equation ar2 + br + c = 0 are realand distinct, then the general solution of ay′′ + by′ + cy = 0 is

y(x) = c1er1x + c2e

r2x

90

2. Case 2: b2 − 4ac = 0

In this case we have a double root or repeated root (r1 = r2). Further, the rootsof the characteristic equation are real and equal, we will denote the common rootsas r. Checking this with the quadratic formula we have

r = − b

2aso 2ar + b = 0 (6.4)

Keep the above equations in mind. We will use them shortly.

Next, we already know that y1 = erx is a solution to equation (6.3). We will nowverify that y2 = xerx is a solution to equation (6.3) as well.

ay′′2 + by′2 + cy2 = a(2rerx + r2xerx) + b(erx + rxerx) + cxerx

= 2arerx + ar2xerx + berx + brxerx + cxerx

= (2ar + b)erx + (ar2 + br + c)xerx

= 0(erx) + 0(xerx)

= 0

In the next to last line in the equations above we know that the first term is zerobecause of equation 6.4 above and the second term is zero because r is a root ofthe characteristic equation.

All of this tells us that y2 = xerx is a solution to equation (6.3). Thus we have

Property 6.4.2: If r is a real, double root of the equation ar2 + br + c = 0 ,then the general solution of ay′′ + by′ + cy = 0 is

y(x) = c1erx + c2xe

rx

3. Case 3: b2 − 4ac < 0

In this final case the roots r1 and r2 of the characteristic equation are complexnumbers. Thus we can write

r1 = α+ iβ and r2 = α− iβ

with

α = − b

2aand β =

√b2 − 4ac

2a

Next, we can use Euler’s equation (which we will not derive here) from ComplexAnalysis

eiθ = cos θ + i sin θ

91

we can write the solution to the differential equation as

y = C1er1x + C2e

r2x

= C1e(α+iβ)x + C2e

(α−iβ)x

= C1eαx (cosβx+ i sinβx) + C2e

ax (cosβx− i sinβx)

= eαx [(C1 + C2) cosβx+ i(C1 − C2) sinβx]

= eαx(c1 cosβx+ c2 sinβx)

with c1 = C1 + C2 and c2 = i(C1 − C2). This gives all solutions (real or complex)of the differential equation. We summarize with the following property.

Property 6.4.3: If the roots of the characteristic equation ar2 + br + c = 0 arethe complex numbers r1 = α + iβ and r2 = α − iβ, then the general solution ofay′′ + by′ + cy = 0 is

y = eαx (c1 cosβx+ c2 sinβx)

This concludes the derivation of the different solutions of the second-order homogeneousdifferential equation with right side zero. We now look at an example of each case.

Example 6.4.1. Solve the equation 4y′′ + 12y′ + 9y = 0.

Solution: We begin by solving the characteristic equation 4r2 + 12r+ 9 = 0. We mayfactor or use the quadratic formula. We factor and note that

(2r + 3)2 = 0

So we have a double root of r = −32 . So by Case 2 and Property 6.4.2 we have the

general solution as

y = c1e− 3x

2 + c2xe− 3x

2


Example 6.4.2. Solve the equation y′′ − 6y′ + 13y = 0.

Solution: We begin by solving the characteristic equation r2 − 6r + 13 = 0. Here wemust use the quadratic formula.

r =6±√

36− 52

2

=6±√−16

2= 3± 2i

So we have complex roots with α = 3 and β = 2. So by Case 3 and Property 6.4.3 wehave the general solution

y = e3x (c1 cos 2x+ c2 sin 2x)


92

Example 6.4.3. Solve the equation 3d2y

dx2+dy

dx− y = 0.

Solution: We begin by solving the characteristic equation 3r2 + r − 1 = 0. Here wemust use the quadratic formula as the left hand side will not factor.

r =−2±

√13

6

So we have unique real roots r1 =−2 +

√13

6and r2 =

−2−√

13

6. So by Case 1 and

Property 6.4.1 we have the general solution

y = c1e(−1+

√13)x

6 + c2e(−1−

√13)x

6


This brings our section to a close.

6.5 Second Order Nonhomogeneous Differential Equations

Before we begin, much of the theory behind the methods discussed within this sectiondepend on higher mathematics, thus we will omit this theory and focus on procedure.

In this section we take a look at the nonhomogeneous case of the second order lineardifferential equation. In other words, how do we solve the equation discussed at thebeginning of the previous section when the right hand side is not zero?

To narrow our problem and stay within the scope of this course we will again be interestedonly the case where the left hand side as only constant coefficients, but the right handside need not be constant. Further, there are several methods for solving the differentialequations of this section, some quite powerful, such as Variation of Parameters, butwould require much more time than we have. We will make use of the Method ofUndetermined Coefficients.

Recall that in the previous section we were interested in differential equations of theform

P (x)y′′ +Q(x)y′ +R(x)y = G(x) (6.5)

where P , Q, and R were constant and G(x) = 0. In this section we will be looking atthe case where G(x) 6= 0. In fact G could itself be a function. Further, we will needto make direct use of our work in the previous section on homogeneous equations tofind the general solution to the nonhomogeneous equations. This is so significant to thissection that for the rest of this section we will refer to

P (x)y′′ +Q(x)y′ +R(x)y = 0 (6.6)

as the Associated Homogeneous Differential Equation to Equation 6.5 above.

To begin to understand how to solve Nonhomogeneous Equations let us take a look ata theorem.

93

Theorem 6.5.1: Suppose Y1(x) and Y2(x) are two solutions to Equation 6.5 and thaty1(x) and y2(x) are a fundamental set of solutions of the associated homogeneous differ-ential equation (6.6) then,

Y1(x)− Y2(x)

is a solution to Equation 6.6 and can be expressed as

Y1(x)− Y2(x) = c1y1(x) + c2y2(x)

/

As this section is relatively tedious we will avoid the proof of this theorem, however itis straightforward.

So how does this theorem help us? Well, let’s assume that y(x) is the general solution toequation 6.5 and that yp(x) is any solution to Equation 6.5 that we can somehow find.Then using the second part of the theorem above we have

y(x)− yp(x) = c1y1(x) + c2y2(x)

Solving for y(x) we have

y(x) = c1y1(x) + c2y2(x) + yp(x)

We will letyc(x) = c1y1(x) + c2y2(x)

and call it the Complementary Solution and yp(x) will be called the ParticularSolution. So the general solution to Equation 6.5 is given by

y(x) = yc(x) + yp(x)

The reasoning above gives us a rough means of finding the general solution. We nowoutline the Method of Undetermined Coefficients.

1. Find the complementary solution yc(x), by solving the associated homogeneousdifferential equation.

2. Write the particular solution yp(x). It should be the sum of terms from the righthand side of equation 6.5 (the term G(x)) as well as the first and higher derivativesof each term of G(x) (exclude coefficients and use undetermined coefficients, A, B,C, etc). Ignore any duplicates.

3. If a term in yp(x) is a duplicate of one in yc(x), multiply that term in yp(x) by xn,using the lowest power n that eliminates the duplication and any new duplicationwith other terms of yp(x).

4. Find y′p(x) and y′′p(x).

5. Substitute yp(x), y′p(x), and y′′p(x) into the differential equation.

94

6. Carefully solve the equation for the undetermined coefficients in step 2.

7. Combine yc(x) and yp(x) to obtain the general solution.

Let us take a look at an example.

Example 6.5.1. Solve the equation y′′ − 2y′ − 3y = 3e2x.

Solution: We use the step by step process above.

1. Find the complementary solution yc(x), by solving the associated homogeneousdifferential equation: Students should verify that given the characteristic equationr2 − 2r − 3 = 0 we have r1 = 3 and r2 = −1 giving us

yc(x) = c1e3x + c2e

−x

as our complementary solution.

2. Write the particular solution yp(x). Note that the right hand side of our equationhas 3e2x, which has higher derivatives 6e2x, 12e2x. . . . Notice the repetition of e2x.Thus it is reasonable to conclude that

yp(x) = Ae2x

3. If a term in yp(x) is a duplicate of one in yc(x), multiply that term in yp(x) by xn,using the lowest power n that eliminates the duplication and any new duplicationwith other terms of yp(x). There is no duplication between yp(x) and yc(x). Sowe move on with our previous choice of yp(x).


y′p(x) = 2Ae2x

y′′p(x) = 4Ae2x


y′′ − 2y′ − 3y = 3e2x

4Ae2x − 2(2Ae2x

)− 3

(Ae2x

)= 3e2x

6. Carefully solve the equation for the undetermined coefficient A.

4Ae2x − 2(2Ae2x

)− 3

(Ae2x

)= 3e2x

4Ae2x − 4Ae2x − 3Ae2x = 3e2x

−3Ae2x = 3e2x

−3A = 3

A = −1

So yp(x) = −e2x

95

7. Combine yc(x) and yp(x) to the obtain general solution. This gives us the generalsolution y(x) = c1e

3x + c2e−x − e2x.


Example 6.5.2. Solve the equation y′′ − y′ − 6y = 36x+ 50 sinx.


1. Find the complementary solution yc(x), by solving the associated homogeneousdifferential equation: Students should verify that given the characteristic equationr2 − r − 6 = 0 we have r1 = 3 and r2 = −2 giving us

yc(x) = c1e3x + c2e

−2x


2. Write the particular solution yp(x). Note that the right hand side of our equationhas 36x+ 50 sinx, which has higher derivatives 2x+ 3 cosx, 2− 3 sinx, 3 cosx. . . .We will have more terms this time.

yp(x) = A+Bx+ C sinx+D cosx

3. If a term in yp(x) is a duplicate of one in yc(x), multiply that term in yp(x) by xn,using the lowest power n that eliminates the duplication and any new duplicationwith other terms of yp(x). Again there is no duplication, so we move on.


y′p(x) = B + C cosx−D sinx

y′′p(x) = −C sinx−D cosx


y′′ − y′ − 6y = 36x+ 50 sinx

−C sinx−D cosx− (B + C cosx−D sinx)− 6 (A+Bx+ C sinx+D cosx)

= 36x+ 50 sinx

6. Carefully solve the equation for the undetermined coefficient A. Collecting termsand equating coefficients gives

−6B = 36

−B − 6A = 0

D − 7C = 50

−7D − C = 0

from which A = 1, B = −6, C = −7, and D = 1. So yp(x) = 1−6x−7 sinx+cosx

96


3x + c2e−2x + 1− 6x− 7 sinx+ cosx.


Example 6.5.3. Solve the equation y′′ + 3y′ + 2y = 5e−2x.


1. Find the complementary solution yc(x), by solving the associated homogeneousdifferential equation: Students should verify that given the characteristic equationr2 + 3r + 2 = 0 we have r1 = −1 and r2 = −2 giving us

yc(x) = c1e−x + c2e

−2x


2. Write the particular solution yp(x). Note that the right hand side of our equationhas 5e−2x, which has higher derivatives −10e−2x, −20e−2x, −40e−2x. . . . So yp(x)might take the form

yp(x) = Ae−2x

3. If a term in yp(x) is a duplicate of one in yc(x), multiply that term in yp(x) by xn,using the lowest power n that eliminates the duplication and any new duplicationwith other terms of yp(x). Note here that e−2x is a term in yc(x). So we multiplyby x in yp(x) giving

yp(x) = Axe−2x


y′p(x) = Ae−2x − 2Axe−2x = Ae−2x(1− 2x)

y′′p(x) = −4Ae−2x + 4Axe−2x = 4Ae−2x(x− 1)

5. Substitute yp(x), y′p(x), and y′′p(x) into the left hand side of the differential equa-tion.

y′′ − 3y′ − 2y = 4Ae−2x(x− 1) + 3Ae−2x(1− 2x) + 2Axe−2x

= Ae−2x(4x− 4 + 3− 6x+ 2x)

= −Ae−2x = 5e−2x

6. Carefully solve the equation for the undetermined coefficient A. Looking at theabove we have A = −5 So yp(x) = −5xe−2x


−x + c2e−2x − 5xe−2x.


Students are highly recommended to practice with several of these problems. Mistakesare easy to make. This concludes the section.

97

6.6 RLC Circuits

In a previous section we studied RL and RC circuits. Each gave rise to a first orderdifferential equation. We now take a look at RLC circuits. RLC circuits are thoseconsisting of a resistor, an inductor, and a capacitor, connected in series or in parallel.

E

R

LC

Figure 6.5: RLC Circuit

According to the physics of electronics the sum of the voltage drops must equal theapplied voltage, so

Ri+ Ldi

dt+q

C= E

with q =∫i dt. Substitution of q and differentiating retrieves

Rdi

dt+ L

d2i

dt2+

i

C= 0

As we can see this is a second order homogeneous differential equation with characteristicequation

Lr2 +Rr +1

C= 0

We must use the quadratic formula to solve for r.

r =−R±

√R2 − 4L/C

2L

= − R

2L±√

R2

4L2− 1

LC

In the above expression the square root of 1LC is called the Resonant Frequency of

RLC Circuit with dc Source or

ωn =

√1

LC

and let

a2 =R2

4L2

98

we have our roots of the characteristic equation as

r = −a±√a2 − ω2

n

or if a2 < ω2n

m = −a± iωdwhere ω2

d = ω2n − a2. From this we have three possible cases. See table below.

Roots ofCharacteristic Equation Type of Solution Relation

Nonreal Underdamped a < ωnNo damping a = 0

Real, equal Critically damped a = ωnReal, unequal Overdamped a > ωn

Table 6.1: RLC Circuit Conditions

We leave further investigation of RLC Circuits with the student. The course textbookpresents a good amount of material on this topic for those interested.

This concludes our chapter on an introduction to differential equations.

99

7 The Laplace Transform

In this chapter we focus on development of the Laplace Transform as a tool for solv-ing Differential Equations. As has been discussed in the past functions are difficult tointegrate in general and broadening our list of tools could be useful. Students are en-couraged to look back at various topics from this course and algebra before embarkingon this chapter. Recommended topics for review include: limits, completing the square,rationalizing the numerator/denominator, factorization, and partial fraction decompo-sition.

We develop the Laplace transform slowly over the chapter with several sections colli-mating at the end with its application to solve differential equations.

7.1 Improper Integrals

Before we can understand the Laplace Transform we must first take a short time todiscuss improper integrals.

Definition 7.1.1: An Improper Integral is an integral where at least one of thelimits are at infinity or negative infinity. As a result there are few types of improperintegral. We list them below.

1. If f is continuous on [a,∞), then∫ ∞a

f(x) dx = limb→∞

∫ b

af(x) dx

2. If f is continuous on (−∞, b], then∫ b

−∞f(x) dx = lim

a→−∞

∫ b

af(x) dx

3. If f is continuous everywhere, then∫ ∞−∞

f(x) dx =

∫ b

−∞f(x) dx+

∫ ∞b

f(x) dx

In each case above, if the limit is finite we say that the improper integral convergesand that the limit is the value of the integral. If the any limit fails to exist, then theintegral is said to diverge. /

We now look at some examples.

100


∫ ∞1

1

x2dx.

Solution: We begin by applying the first property in the definition above.∫ ∞1

1

x2dx = lim

b→∞

∫ b

1x−2 dx

= limb→∞

−1

x

∣∣∣b1

= limb→∞

(−1

b+ 1

)= −0 + 1

= 1



∫ ∞3

1

xdx.

Solution: Again, we begin by applying the first property in the definition above.∫ ∞3

1

xdx = lim

b→∞

∫ b

3

1

xdx

= limb→∞

lnx∣∣∣b3

= limb→∞

(ln b− ln 3)

=∞− ln 3

=∞

Thus this integral diverges. J

Example 7.1.3. [2, p.546]Compute

∫ ∞−∞

1

1 + x2dx.

Solution: Here we apply the third property in the definition.∫ ∞−∞

1

1 + x2dx =

∫ b

−∞

1

1 + x2dx+

∫ ∞b

1

1 + x2dx

= lima→−∞

∫ 0

a

1

1 + x2dx+ lim

c→∞

∫ c

0

1

1 + x2dx

Next, each integral can be evaluated by Trigonometric Substitution with x = tan θ anddx = sec2 θ dθ. Conveniently they are the same with the exception of the limits of

101

integration.

= lima→−∞

∫ θ0

θa

sec2 θ dθ

1 + tan2 θ+ limc→∞

∫ θc

θ0

sec2 θ dθ

1 + tan2 θ

= lima→−∞

∫ θ0

θa

sec2 θ dθ

sec2 θ+ limc→∞

∫ θc

θ0

sec2 θ dθ

sec2 θ

= lima→−∞

∫ θ0

θa

dθ + limc→∞

∫ θc

θ0

dθ

= lima→−∞

θ∣∣θ0θa

+ limc→∞

θ∣∣θcθ0

Further, converting the integrals back in terms of x with θ = tan−1x, we have

= lima→−∞

[tan−1 x]0a + limc→∞

[tan−1 x]c0

= lima→−∞

[tan−1 0− tan−1 a] + limc→∞

[tan−1 c− tan−1 0]

= lima→−∞

[− tan−1 a] + limc→∞

[tan−1 c]

Finally, note that if a gets large negative, then tan−1 a approaches −π2 and if c gets

large, then tan−1 c approaches π2 . This gives us

= −(−π

2

)+π

2= π


This brings us to the end of this section and takes us to the first section introducing theLaplace Transform.

7.2 The Laplace Transform of a Function

In this section we formally introduce the Laplace Transform. To begin let us consider aPower Series, which we will discuss more in later chapters.

Definition 7.2.1: Let a(n) be a discrete function of a positive integer n, then thePower Series associated with a(n) is given by

∞∑n=0

a(n)xn

with the real variable x. /

With the above in mind, let us consider replacing a(n) (a function of a discrete variable)with a continuous function of a real variable, f(t). In Analysis (the theory of calculus)this becomes

102

∫ ∞0

f(t)xt dt

we continue to modify this by changing the base of xt∫ ∞0

f(t)(elnx

)tdt

Next, it is easy to see that this improper integral will likely diverge unless we put inplace some restrictions and even then it may diverge. For example, keeping lnx < 0 andrewriting as −s = lnx gives us exactly the Laplace Transform.

Definition 7.2.2: Let f(t) be an integrable function on the interval [0,∞), then theLaplace Transform is defined by

L[f(t)] =

∫ ∞0

f(t)e−st dt

/

In other words, we can think of the Laplace Transform as the continuous analog of thediscrete power series.

The resulting transformed function is of a complex variable s, which we will call F (s).So

L[f(t)] = F (s)

We can find the Laplace Transform of a function directly using the above definition. Westart with a simple example.

Example 7.2.1. If f(t) = 1, then find L[f(t)].

Solution: Here we apply the definition above.

L[f(t)] = L[1]

=

∫ ∞0

1e−st dt

=

∫ ∞0

e−st dt

=

[−1

se−st

]∞0

= −1

s

[e−st

]∞0

= −1

s(0− 1)

=1

s


103

In fact it easy to see that as a result of the Laplace Transform being an integral we cansimply pull constants and factors not involving t in front of of integral sign and thus

Property 7.2.1: Let a be some constant real number, then

L[af(t)] = aL[f(t)]

Example 7.2.2. If f(t) = 7, then find L[f(t)].

Solution: Using the above property and the previous example we have

L[f(t)] = L[7]

= 7L[1]

= 7 · 1

s

=7

s

The process in this problem is critical to general use of the Laplace Transform. Further,students should consider working this example out using the integral definition above toconvince themselves of the truth of the property. J

Example 7.2.3. If f(t) = t, then find L[f(t)].

Solution: We begin with f(t) = t.

L[f(t)] = L[t]

=

∫ ∞0

te−st dt

=

[e−st

s2(−st− 1)

]∞0

Integration by Parts

= 0−(− 1

s2

)=

1

s2


There are many other elementary functions that we could find the Laplace Transformfor, however many have already been computed in a Table of Laplace Transforms. Thereis such a table in your textbook and one will be provided on your exam.

We continue with a useful property that again can be derived easily by looking at thedefinition of the Laplace Transform.

Property 7.2.2: Let ai be some constant real number for all i and f be an integrablefunction on the interval [0,∞), then

L

[n∑i=1

aifi(t)

]=

n∑i=1

aiL[fi(t)]

104

This property simply says that we can ”distribute” a Laplace Transform over a sum ordifference and still pull out all constants.

Moving along, we will need to be able to find the Laplace Transform of derivatives andintegrals. We now derive these.

Example 7.2.4. Find the Laplace Transform of f ′(t).

Solution: The meaning of the above is

L[f ′(t)] =

∫ ∞0

f ′(t)e−st dt

Here we will need Integration by Parts. We choose u = e−st and dv = f ′(t) which givesdu = −se−st dt and v =

∫f ′(t)dt = f(t). Thus

L[f ′(t)] =[f(t)e−st

]∞0

+ s

∫f(t)e−st dt

= 0− f(0) + sL[f(t)]

= sL[f(t)]− f(0)

giving us the formula. J

Similarly, we can derive the Laplace Transform for the Second Derivative. We give theformula and leave the derivation to the student.

Property 7.2.3: Let f be an integrable function on the interval [0,∞), then

L[f ′′(t)] = s2L[f(t)]− sf(0)− f ′(0)

We conclude with the Laplace Transform of an integral.

Example 7.2.5. Find the Laplace Transform of∫ t

0 f(t) dt.

Solution: The meaning of the above is

L[∫ t

0f(t) dt

]=

∫ ∞0

[∫ t

0f(t) dt

]e−st dt

Here again we will need Integration by Parts. We choose u =∫ t

0 f(t) dt and dv = e−st dt

which gives du = f(t) dt and v = − e−st

s . Thus

L[∫ t

0f(t) dt

]=

[−1

se−st dt

∫ t

0f(t) dt

]∞0

+

∫ ∞0

f(t)e−st dt

= −1

sL[f(t)]


At this point we have derived the Laplace Transform of many functions. We will nowgive a table of those above and many others not derived here. Students should be awarethat some of the functions used in the table may need special attention. An examplemay be the hyperbolic trigonometric functions: sinhx and coshx.

105

Table of Laplace Transforms

f(t) L[f(t)] = F (s)

11

s(1)

eatf(t) F (s− a) (2)

tnf(t) (−1)ndnF (s)

dsn(3)

f ′(t) sF (s)− f(0) (4)

f ′′(t) s2F (s)− sf(0)−

f ′(0) (5)

tnn!

sn+1(6)

sin ktk

s2 + k2(7)

cos kts

s2 + k2(8)

eat1

s− a(9)

sinh ktk

s2 − k2(10)

cosh kts

s2 − k2(11)

aeat − bebt

a− bs

(s− a)(s− b)(12)

e−at − e−bt b− a(s+ a)(s+ b

(13)

ae−at − be−bt s(a− b)(s+ a)(s+ b)

(14)

f(t) L[f(t)] = F (s)

1− cos ata2

s(s2 + a2)(15)

at− sin ata3

s2(s2 + a2)(16)

teat1

(s− a)2(17)

tneatn!

(s− a)n+1(18)

eat sin ktk

(s− a)2 + k2(19)

eat cos kts− a

(s− a)2 + k2(20)

eat sinh ktk

(s− a)2 − k2(21)

eat cosh kts− a

(s− a)2 − k2(22)

t sin kt2ks

(s2 + k2)2(23)

t cos kts2 − k2

(s2 + k2)2(24)

t sinh kt2ks

(s2 − k2)2(25)

t cosh kts2 − k2

(s2 − k2)2(26)

sin at

tarctan

a

s(27)

106

Some research will reveal more formula for Laplace Transforms. This is certainly not acomplete list.

Next, we will practice using the previous table with problems.

Example 7.2.6. Find L[f(t)] where f(t) = 6t+ 4t2 − 2t3.

Solution: Here we use the fact that L is a linear operator, giving

L[f(t)] = L[6t+ 4t2 − 2t3]

= 6L[t] + 4L[t2]− 2L[t3]

=6

s2+ 4

2!

s3− 2

3!

s4Transforms 1 and 6

=6

s2+

8

s3− 12

s4

=6s2 + 8s− 12

s2

Often we will need to combine our fractions for solving boundary value problems. J

Example 7.2.7. Find L[f(t)] where f(t) = 4e3t + 3 cos 4t− 2t3e6t.

Solution: Here we use the fact that L is a linear operator, giving

L[f(t)] = L[4e3t + 3 cos 4t− 2t3e6t]

= 4L[e3t] + 3L[cos 4t]− 2L[t3e6t]

= 41

s− 3+ 3

s

s2 + 42− 2

3!

(s− 6)4Transforms 8, 9, and 14

=4

s− 3+

3s

s2 + 16− 12

(s− 6)4

Here we leave this as our answer. Students should feel free to add and subtract thefractions. J

As we conclude this sections students should be aware that the Laplace Transform Tableon the previous page is not only giving us useful information for the Laplace Transformbut also the Inverse Laplace Transform. As a result we will find a transform table usefulin the next couple of sections.

7.3 Inverse Laplace Transform

In this section we simply take the topic of the previous section on the Laplace Transformand reverse it. More specifically we will start with a function of s, F (s) and transform itto a function of t, f(t). With respect to the table on page 106 we will be trying to modifyour given function of s so that it matches one of the entries in the column of F (s)’s in thetable. Once we have found the correct function (modified with the appropriate constantif necessary) we will then convert to a function of t by looking at the associated functionin the f(t) column.

107

We should finally note that as a result of L being a linear operator, so is its inverse L−1.Thus

L−1[aF (s) + bG(s)] = aL−1[F (s)] + bL−1[G(s)]

.

Example 7.3.1. Find the Inverse Laplace Transform of F (s) =3

s− 2

s− 4+

6

s2.

Solution: We are reminded that our goal is to use any mathematical tool to makethe above function look (or pieces of it) look like something in the F (s) column of thetransform table. We begin by making use of the previous fact that L−1 is a lineartransformation.

L−1[F (s)] = L−1

[3

s− 2

s− 4+

6

s2

]= 3L−1

[1

s

]− 2L−1

[1

s− 4

]+ 6L−1

[1

s2

]At this point we will use transform numbers (1), (9) with a = 4, and (6) with n = 1respectively.

= 3(1)− 2e4t + 6t

= 3− 2e4t + 6t


Example 7.3.2. Find the Inverse Laplace Transform of G(s) =2s+ 3

s2 + 25.

Solution: Note in this problem using the fact that L−1 is a linear transformationdoes nothing for us. Instead we will make a mathematical adjustment from rationalexpressions by rewriting the function as the sum of two rational functions.

L−1[G(s)] = L−1

[2s+ 3

s2 + 25

]= L−1

[2s

s2 + 25

]+ L−1

[3

s2 + 25

]= 2L−1

[s

s2 + 25

]+ 3L−1

[1

s2 + 25

]At this point we will use transform numbers (8) and (7) with k = 5 in both casesrespectively. Note here that in the right hand side, transform number (7) does not fitperfectly. We are off by a constant of 5 in the numerator. To fix this we multiply by 1in the form of the needed numerator, so 1 = 5

5 .

= 2L−1

[s

s2 + 25

]+

3

5L−1

[5

s2 + 25

]= 2 cos 5t+

3

5sin 5t


108

Example 7.3.3. Find the Inverse Laplace Transform ofH(s) =1

(s+ 1)4+

s− 3

(s− 3)2 + 6.

Solution: Note again in this problem using the fact that L−1 is a linear transformationdoes little to help. However we will note that transform numbers (14) with n = 3 anda = −1 and (16) with k =

√6 and a = 3. Further we will need to adjust the numerator

of the left hand inverse transform by multiplying by 3!3! . So we have

L−1[H(s)] = L−1

[1

(s+ 1)4+

s− 3

(s− 3)2 + 6

]= L−1

[1

(s+ 1)4

]+ L−1

[s− 3

(s− 3)2 + 6

]=

1

3!L−1

[3!

(s+ 1)4

]+ L−1

[s− 3

(s− 3)2 + 6

]=

1

3!t3e−t + e3t cos (t

√6) Transforms (14) and (16)

=1

6t3e−t + e3t cos (t

√6)


Students should be aware that the types of ”mathematical tricks” seen in the previousproblems are common when trying to solve differential equations using many methods.This concludes our section on the Inverse Laplace Transform.

7.4 Boundary Value Problems and the Laplace Transform

Now that we understand Laplace and Inverse Laplace Transforms we will use them tolook at some examples of solving boundary value problems. The method illustrated inthis section is a valid alternative for solving such problems provided we know y(0), y′(0),. . . and we can find an appropriate transform.

First we present a rough procedure for solving these problems.

1. Take the Laplace Transform of both sides of the differential equation.

2. Institute the boundary condition(s).

3. Solve the resulting equation for L(y).

4. Take the Inverse Laplace Transform on both sides. The result is the solution.

Example 7.4.1. Solve the boundary value problem with differential equation y′′+4y =4 and boundary conditions y(0) = 0 and y′(0) = 0 by Laplace Transform.

109

Solution: Referring to the steps we begin by taking the Laplace Transform of bothsides of the given differential equation

L(y′′ + 4y

)= L (4)

L(y′′)

+ 4L (y) = 4L (1)

s2L(y)− sy(0)− y′(0) + 4L(y) =4

sSee Transform Table

s2L(y) + 4L(y) =4

sApply boundary conditions

(s2 + 4)L(y) =4

s

L(y) =4

s(s2 + 4)

Now that we have worked through the first 3 steps we apply the Inverse Laplace Trans-form.

L−1 (L(y)) = L−1

(4

s(s2 + 4)

)y = 1− cos 2t By Transform #13

This is the solution to our boundary value problem. J

Example 7.4.2. Solve the boundary value problem with differential equation y′+7y =e3t and boundary condition y(0) = 5 by Laplace Transform.

Solution: Again, referring to the steps we begin by taking the Laplace Transform ofboth sides of the given differential equation

L(y′ + 7y

)= L

(e3t)

L(y′)

+ 7L (y) =1

s− 3See Transform Table

sL(y)− y(0) + 7L(y) =1

s− 3See Transform Table

sL(y)− 5 + 7L(y) =1

s− 3Apply boundary conditions

sL(y) + 7L(y) =1

s− 3+ 5

(s+ 7)L(y) =1

s− 3+ 5

L(y) =1

(s− 3)(s+ 7)+

5

s+ 7

110

Next we apply the Inverse Laplace Transform:

L−1 (L(y)) = L−1

(1

(s− 3)(s+ 7)+

5

s+ 7

)y = L−1

(1

(s− 3)(s+ 7)

)+ L−1

(5

s+ 7

)=

1

10L−1

(7− (−3)

(s− 3)(s+ 7)

)+ 5L−1

(1

s+ 7

)See Transform table

=1

10

(e3t − e−7t

)+ 5e−7t See Transform table

=1

10e3t − 49

10e−7t

solving our boundary value problem. J

Example 7.4.3. Solve the boundary value problem with differential equation y′′+4y =4t and boundary conditions y(0) = 0 and y′(0) = 0 by Laplace Transform.

Solution: Taking Laplace Transforms on both sides we have

L(y′′ + 4y

)= L (4t)

L(y′′)

+ 4L (y) =4

s2See Transform Table

s2L(y)− sy(0)− y′(0) + 4L(y) =4

s2See Transform Table

s2L(y) + 4L(y) =4

s2

(s2 + 4)L(y) =4

s2

L(y) =4

s2(s2 + 4)

Next we apply the Inverse Laplace Transform:

L−1 (L(y)) = L−1

(4

s2(s2 + 4)

)y =

1

2L−1

(4(2)

s2(s2 + 4)

)See Transform #16

=1

2(2t− sin 2t)

solving our problem. J

This concludes our chapter on the Laplace Transform.

111

8 Infinite Series

In our final chapter we will be interested in infinite series and applications of seriesincluding convergence and divergence, Maclaurin series, Power series, Fourier series, andwaveform symmetries.

Sometimes it may be very difficult to differentiate or integrate a function in some ap-plication problem in which we may need to solve a differential or integral equation. Analternative to solving in this case is to convert the complicated function into an infiniteseries, solve, and settle for an approximate solution.

8.1 Introduction to Infinite Series

8.1.1 Sigma Notation

We begin with an introduction to Sigma series notation and look at some simple exam-ples.

Definition 8.1.1: (Sigma Notation.) Let ak be some real number for each k, then

n∑k=0

ak = a0 + a1 + a2 + · · ·+ an

k is called the index of summation and ak is called the term of summation. /

We use Sigma notation sometimes to shorten a summation.

Example 8.1.1. Find4∑

k=0

(2k + 1)

Solution: We simply use the definition above to evaluate.

4∑k=0

(2k + 1) = (2(0) + 1) + (2(1) + 1) + (2(2) + 1) + (2(3) + 1) + (2(3) + 1)

= 1 + 3 + 5 + 7 + 9

= 25

You will notice that this is simply the sum of the first five odd counting numbers. J

Example 8.1.2. Find

6∑k=1

(−1)k

112

Solution: Again we use the definition above.

6∑k=1

(−1)k = (−1)1 + (−1)2 + (−1)3 + (−1)4 + (−1)5 + (−1)6

= −1 + 1 + (−1) + 1 + (−1) + 1

= 0

In this problem, notice how the term of summation ultimately translates to a form ofsign switching. J

Example 8.1.3. Find1∑

k=−1

[(−1)k cos

(π2k)]

Solution: In this example we need to be careful.

1∑k=−1

[(−1)k cos

(π2k)]

=[(−1)−1 cos

(π2

(−1))]

+[(−1)0 cos

(π2

(0))]

+[(−1)1 cos

(π2

(1))]

= −1(0) + 1(0) + (−1)(0)

= 0

Here we should notice that the starting value can be any integer. J

8.1.2 Geometric Series

Next, we consider a special type of series. Consider the next example.

Example 8.1.4. Find 4 + 8 + 16 + 32 + 64

Solution: We can easily find the sum of this series, it is 124, but let us study thisexample closely. Notice it has a starting value of 4. Further note that there is a commonratio between any pair of adjacent terms. In other word, if we choose any pair of adjacentterms and divide the right hand number from the left hand number we always get thesame result, 2, in this case. We call this number the common ratio and as a consequenceof having a common ratio this series is called Geometric. Next, we generalize thegeometric series and try to understand it better. J

Definition 8.1.2: Let a and r be real numbers, then

n∑k=1

ark−1 = a+ ar + ar2 + ar3 + · · ·+ arn−1

is called the Partial Sum of a Geometric Series. The number a is called the startingvalue and r is the common ratio. /

Students should be warned at this point to not assume that a series is geometric. Thefirst few pair of terms should be verified to have a common ratio.

113

We now show how to find the partial sum of a geometric series.

Let us begin by denoting the partial sum of the geometric series with sn then we have

sn = a+ ar + ar2 + ar3 + · · ·+ arn−1

Further, we can multiply both sides of this equation by r giving us

rsn = ar + ar2 + ar3 + ar4 · · ·+ arn

Next, we subtract the bottom equation from the top.

sn = a+ar + ar2+ · · ·+ arn−1

rsn = ar + ar2+ · · ·+ arn−1+arn

With this we retrieve

sn − rsn = a− arn

sn(1− r) = a(1− rn) Factor out sn and a

sn =a(1− rn)

1− r

Thus

Property 8.1.1: The partial sum of a geometric series with starting value a andcommon ratio r is given by

sn =n∑k=1

ark−1 =a(1− rn)

1− r

114

Bibliography

[1] P. Calter, M. Calter, P. Wraight, D. Spencer, Technical Mathematics with Calculus,Wiley & Sons Canada Ltd., Mississauga Ontario, 2nd Edition, 2012.

[2] R. Finney, G.B. Thomas Jr., Calculus, Addison-Wesley Publishing, New York, 2ndEdition, 1994.

[3] J. Stewart, Single Variable Calculus - Early Transcendentals, Brooks/Cole CengageLearning, Toronto, 7th Edition, 2012.

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· contents 1 review of limits4 1.1 the geometry of limits. . . . . . . . . . . . . . . . . . . ....

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