Chapter 5: Integration

Autumn 2016

Department of MathematicsHong Kong Baptist University

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§5.1 Sums and Sigma Notation

Definition (Sigma Notation)

If m and n are integers with m ≤ n, and if f is a function definedat the integers m,m + 1, . . . , n, the symbol

∑ni=m f (i) represents

the sum of the values of f at those integers:

n∑i=m

f (i) = f (m) + f (m + 1) + · · ·+ f (n).

The explicit sum appearing on the right side of the equation is theexpansion of the sum represented in sigma notation on the leftside. m is the lower limit, and n is the upper limit.

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Theorem (Summation Formulas)

(a)n∑

i=1

i = 1 + 2 + · · ·+ n =n(n + 1)

2.

(b)n∑

i=m

i = m + (m + 1) + · · ·+ n =(n −m + 1)(n + m)

2.

(c)n∑

i=1

i2 = 12 + 22 + · · ·+ n2 =n(n + 1)(2n + 1)

6.

(d)n∑

i=1

i3 = 13 + 23 + · · ·+ n3 =n2(n + 1)2

4.

(e)n∑

i=1

r i−1 = 1 + r + r2 + · · ·+ rn−1 =rn − 1

r − 1.

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§5.2 Areas as Limits of Sums

This section aims to find the area of a region R lying under thegraph y = f (x) of a nonnegative-valued, continuous function f ,above the x-axis and between the vertical lines x = a and x = b.

- x

6y

y = f (x)

a b

R

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To compute the area, we divide the intervals [a, b] into nsubintervals using division points:

a = x0 < x1 < x2 < · · · < xn−1 < xn = b.

Denote by ∆xi the length of the ith subinterval [xi−1, xi ]:

∆xi = xi − xi−1, i = 1, . . . , n.

The area of the ith rectangle is f (xi )∆xi . The sum of all theseareas is

Sn =n∑

i=1

f (xi )∆xi .

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It is often convenient to choose the points xi so that ∆xi areall equal. In this case we have

∆xi = ∆x =b − a

nor xi = a +

i

n(b − a).

Sn is an approximation to the area of the region R, and theapproximation gets better as n increases. That is,

Area of R = limn→∞

Sn

= limn→∞

b − a

n

n∑i=1

f (xi ).

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Example 1:

Find the area of the region bounded by the parabola y = x2 andthe straight lines y = 0, x = 0, and x = b, where b > 0.

Solution: Let f (x) = x2. We use equal subintervals, each oflength b/n and xi = ib/n. Thus,

Sn =b − 0

n

n∑i=1

f (xi ) =b3

n3

n∑i=1

i2 =b3(n + 1)(2n + 1)

6n2.

Therefore, the required area A is

A = limn→∞

Sn

= limn→∞

b3(n + 1)(2n + 1)

6n2

=b3

3.

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§5.3 The Definite Integral

In this section we introduce the definite integral of a continuousfunction f on an closed, finite interval [a, b]. We no longer assumethat the values of f are nonnegative.

Definition

Let a = x0 < x1 < · · · < xn−1 < xn = b be a partition of theinterval [a, b]. Then a Riemann sum of f on [a, b] is a sum of theform

n∑i=1

f (ci )∆xi ,

where each ci is a point belonging to the interval [xi−1, xi ].

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Example 2:

Calculate two Riemann sums for the function f (x) = x2 on auniform partition of the interval [0, b], with ci being the left andright end-points of the subintervals respectively.

Solution: With equal length subintervals, we have ∆x = b/n andthe division points are xi = ib/n for i = 0, 1, . . . , n.

If ci = xi−1 is the left end-point, then the Riemann sum is

S(1)n =

n∑i=1

(xi−1)2∆x =(n − 1)(2n − 1)b3

6n2.

If ci = xi is the right end-point, then the Riemann sum is

S(2)n =

n∑i=1

(xi )2∆x =

(n + 1)(2n + 1)b3

6n2.

Note that limn→∞

S(1)n = lim

n→∞S(2)n =

b3

3.

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Theorem

Let f be continuous on the closed, bounded interval [a, b]. For theuniform partition a = x0 < x1 < · · · < xn = b, choose the pointsci ∈ [xi−1, xi ] and define the Riemann sum

Sn =n∑

i=1

f (ci )∆xi .

Then there is a unique real number I , independent of the choice ofci , such that limn→∞ Sn = I . The number I is called the definiteintegral of f on [a, b] and is denoted by I =

∫ ba f (x)dx .

Remark: The definite integral of f (x) over [a, b] is a number butnot a function of x . Replacing x with another variable does notchange the value of the integral, e.g.,

∫ ba f (x)dx =

∫ ba f (t)dt.

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The various parts of the symbol

∫ b

af (x)dx have their own names:

(i)∫

is called the integral sign, it resembles the letter S since itrepresents the limit of a sum.

(ii) a and b are called the limits of integration; a is the lowerlimit and b is the upper limit.

(iii) The function f is the integrand and x is the variable ofintegration.

(iv) dx is the differential of x . It replaces ∆x in the Riemannsums.

The definite integral can be interpreted as the signed areabetween the curve y = f (x) and the x-axis.

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For continuous functions on [a, b], the limit limn→∞ Sn alwaysexists and does not depend on how one chooses theevaluation point ci in each subinterval. Such functions arecalled Riemann integrable.

For discontinuous functions, this limit may not exist, or it mayyield different values depending on the choice of ci . Suchfunctions are called non-integrable. See the textbook (§5.3and Appendix IV) for more details.

The connection between Riemann sums and definite integralsis useful for solving problems involving areas, volumes, centresof mass, etc.

We will see from §5.5 that definite integrals can be evaluatedby reversing the differentiation process.

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Example 3:

Let R be the planar region bounded above by y = 4− x2 andbelow by y = x2 − 2x . Estimate its area using a Riemann sum,then express the exact area as a definite integral.

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Example 3:

Let R be the planar region bounded above by y = 4− x2 andbelow by y = x2 − 2x . Estimate its area using a Riemann sum,then express the exact area as a definite integral.

Solution: The two curves intersect at two points, given by

4− x2 = x2 − 2x =⇒ x = −1 or x = 2.

Decompose the interval [−1, 2] into n subintervals [x0, x1], . . . ,[xn−1, xn] with ∆xi = 3/n. The height of the ith strip is

Hi = H(ci ) = 4− c2i − (c2i − 2ci ).

This gives the Riemann sumn∑

i=1

(4 + 2ci − 2c2i )∆xi . Letting n

tend to infinity leads to the exact area

Area =

∫ 2

−1(4 + 2x − 2x2) dx .

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§5.4 Properties of the Definite Integral

Theorem

Let f and g be integrable on an interval containing the points a, b,and c . Then

(a) An integral over an interval of zero length is zero,∫ a

af (x)dx = 0.

(b) Reversing the limits of integration changes the sign of theintegral, ∫ a

bf (x)dx = −

∫ b

af (x)dx .

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Theorem

(c) If A and B are constants, then∫ b

a(Af (x) + Bg(x)) dx = A

∫ b

af (x)dx + B

∫ b

ag(x)dx .

(d) An integral depends additively on the interval of integration,∫ b

af (x)dx +

∫ c

bf (x)dx =

∫ c

af (x)dx .

(e) If a ≤ b and f (x) ≤ g(x) for a ≤ x ≤ b, then∫ b

af (x)dx ≤

∫ b

ag(x)dx .

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Theorem

(f) If a ≤ b, then ∣∣∣∣∫ b

af (x)dx

∣∣∣∣ ≤ ∫ b

a|f (x)|dx .

(g) If f is an odd function, i.e., f (−x) = −f (x), then∫ a

−af (x)dx = 0.

(h) If f is an even function, i.e., f (−x) = f (x), then∫ a

−af (x)dx = 2

∫ a

0f (x)dx .

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Theorem (The Mean-Value Theorem for Integrals)

If f is continuous on [a, b], then there exists a point c in [a, b] suchthat ∫ b

af (x)dx = (b − a)f (c).

Proof: Let f (l) be the minimum value and f (u) be the maximumvalue of f on [a, b]. By the property (e), We have

f (l) =1

b − a

∫ b

a

f (l)dx ≤ 1

b − a

∫ b

a

f (x)dx ≤ 1

b − a

∫ b

a

f (u)dx = f (u).

Then by the Intermediate-Value Theorem, there exists a number cbetween l and u such that

f (c) =1

b − a

∫ b

af (x)dx .

This proves the theorem.23 / 46

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Average Value of a Function

Definition

If f is integrable on [a, b], then the average value or mean valueof f on [a, b], denoted by f̄ , is

f̄ =1

b − a

∫ b

af (x)dx .

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Piecewise Continuous Functions

Example 4:

Find

∫ 3

0f (x)dx , where f (x) =

√

1− x2 if 0 ≤ x ≤ 12 if 1 < x ≤ 2x − 2 if 2 < x ≤ 3.

Solution: By definition, the value of the integral is∫ 3

0f (x)dx =

∫ 1

0

√1− x2dx +

∫ 2

12dx +

∫ 3

2(x − 2)dx

=π

4+ 2 +

1

2

=π + 10

4.

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§5.5 The Fundamental Theorem of Calculus

The following Fundamental Theorem of Calculusdemonstrates the relationship between the definite integraldefined in §5.3 and the indefinite integral (or antiderivative)introduced in §2.10.

Both conclusions of the Fundamental Theorem are useful:

(i) Part I concerns the derivative of an integral; it tells youhow to differentiate a definite integral with respect to itsupper limit.

(ii) Part II concerns the integral of a derivative; it tells youhow to evaluate a definite integral if you can find anantiderivative of the integrand.

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Theorem (The Fundamental Theorem of Calculus)

Suppose that f is continuous on an interval I containing point a.

Part I. Let the function F be defined on I by

F (x) =

∫ x

af (t)dt.

Then F is differentiable on I , and F ′(x) = f (x) there. Thus,F is an antiderivative of f on I :

d

dx

∫ x

af (t)dt = f (x).

Part II. If G is any antiderivative of f on I , so that G ′(x) = f (x) on I ,then for any b in I we have∫ b

af (x)dx = G (b)− G (a).

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Proof: [Part I] Using the definition of the derivative, we have

F ′(x) = limh→0

F (x + h)− F (x)

h

= limh→0

1

h

(∫ x+h

af (t)dt −

∫ x

af (t)dt

)= lim

h→0

1

h

∫ x+h

xf (t)dt.

By the Mean-Value Theorem for Integrals, there exists a pointc ∈ [x , x + h] such that∫ x+h

xf (t)dt = hf (c).

This leads to

F ′(x) = limh→0

f (c) = limc→x

f (c) = f (x).

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[Part II] If G (x) is any antiderivative of f (x) so that G ′(x) = f (x),then (F (x)− G (x))′ = F ′(x)− G ′(x) = 0. This leads to

F (x) = G (x) + C .

Hence, ∫ x

af (t)dt = F (x) = G (x) + C .

Letting x = a, we have 0 = G (a) + C and so C = −G (a). Now welet x = b, then∫ b

af (t)dt = G (b) + C = G (b)− G (a).

Note that t can be replaced by x (or any other variable) as thevariable of integration on the left-hand side. This proves Part II.

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Definition

To facilitate the evaluation of definite integrals using theFundamental Theorem of Calculus, we define the evaluationsymbol:

F (x)∣∣∣ba

= F (b)− F (a).

Remark: By definition, we have∫ b

af (x)dx =

(∫f (x)dx

)∣∣∣∣ba

,

where∫f (x)dx denotes the indefinite integral or general

antiderivative of f . When evaluating a definite integral this way,we can omit the constant of integration from the indefinite integralsince

(F (x) + C )∣∣∣ba

= F (b) + C − (F (a) + C ) = F (x)∣∣∣ba.

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Example 5:

Find the integral of

∫cosh(x)dx , where cosh(x) =

1

2(ex + e−x).

Solution: We have∫cosh(x)dx =

∫1

2(ex + e−x)dx

=1

2(

∫exdx +

∫e−xdx)

=1

2(ex − e−x) + C

= sinh(x) + C ,

where sinh(x) =1

2(ex − e−x).

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Example 6:

Find the following integrals:

(a)

∫ π

0sin(x)dx = − cos(x)|π

0= 2

(b)

∫ 0

−πsin(x)dx = − cos(x)|0−π

= −2

(c)

∫ π

−πsin(x)dx = − cos(x)|π−π

= 0

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Example 7:

Find the area A of the plane region lying above the x-axis andunder the curve y = 3x − x2.

Solution: Noting that y = 3x − x2 = x(3− x), the two interceptson the x-axis are (0, 0) and (3, 0). The area of the region is thengiven by

A =

∫ 3

0(3x − x2)dx

=

(3

2x2 − 1

3x3)∣∣∣∣3

0

=27

2− 27

3− (0− 0)

=9

2.

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Example 8:

Using the conclusion in Part I of the Fundamental Theorem to findthe derivatives of the following functions:

(a) F (x) =

∫ 3

xe−t

2dt, (b) G (x) = x2

∫ x

−4e−t

2dt.

Solution:

(a) Note that F (x) = −∫ x3 e−t

2dt. By Part I we have

F ′(x) = − d

dx

∫ x

3e−t

2dt = −e−x2 .

(b) By the Product Rule and Part I, we have

G ′(x) = 2x

∫ x

−4e−t

2dt + x2

d

dx

∫ x

−4e−t

2dt

= 2x

∫ x

−4e−t

2dt + x2e−x

2.

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By building in the Chain Rule into the conclusion in Part I ofthe Fundamental Theorem, we have the following formula:

d

dx

∫ g(x)

af (t)dt = f (g(x))g ′(x).

More generally, if the lower limit is also a function of x , then

d

dx

∫ g(x)

h(x)f (t)dt = f (g(x))g ′(x)− f (h(x))h′(x).

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Example 9:

Find the derivatives of the following functions:

(a) F (x) =

∫ 5x

−4e−t

2dt, (b) G (x) =

∫ x3

x2e−t

2dt.

Solution:

(a) Let the integrand be f (t) = e−t2

and the upper limit beg(x) = 5x . Then

F ′(x) = f (g(x))g ′(x) = e−(5x)2(5) = 5e−25x

2.

(b) Note that G (x) =∫ x3

0 e−t2dt −

∫ x2

0 e−t2dt. We have

G ′(t) = e−(x3)2(3x2)− e−(x

2)2(2x)

= 3x2e−x6 − 2xe−x

4.

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§5.6 The Method of Substitution

In this section and in Chapter 6 we will develop sometechniques of integration, that is, methods for findingantiderivatives of functions.

Let’s begin by assembling a table of some known indefiniteintegrals. These results have all emerged during ourdevelopment of differentiation formulas for elementaryfunctions and we should memorize them.

Note that formulas 1-6 are special cases of formula 7. Thefollowing linearity formula also makes it possible to integratesums and constant multiples of functions:∫

(Af (x) + Bg(x))dx = A

∫f (x)dx + B

∫g(x)dx .

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When an integral cannot be evaluated by inspection, werequire one or more special techniques. The most importantof these techniques is the method of substitution, theintegral version of the Chain Rule.

If we rewrite the Chain rule,d

dxf (g(x)) = f ′(g(x))g ′(x), in

integral form, we obtain∫f ′(g(x))g ′(x)dx = f (g(x)) + C .

[Another approach to the above formula] Let u = g(x). Thendu/dx = g ′(x), or equivalently, du = g ′(x)dx . Thus∫

f ′(g(x))g ′(x)dx =

∫f ′(u)du = f (u) +C = f (g(x)) +C .

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Example 10:

Find the indefinite integral I =

∫ex√

1 + exdx .

Solution: Let u = 1 + ex . Then du = exdx . Using the method ofsubstitution,

I =

∫ √udu

=2

3u3/2 + C

=2

3(1 + ex)3/2 + C .

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Example 11:

Find the indefinite integral I =

∫1√

e2x − 1dx .

Solution: An appropriate substitution seems not obvious here.Note however that

I =

∫1

ex√

1− e−2xdx =

∫e−x√

1− e−2xdx .

Let u = e−x . Then du = −e−xdx . Using the method ofsubstitution,

I = −∫

1√1− u2

du

= − sin−1(u) + C

= − sin−1(e−x) + C .

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Theorem (Substitution of Definite Integrals)

Suppose that g is a differentiable function on [a, b] that satisfiesg(a) = A and g(b) = B. Also suppose that f is continuous on therange of g . Then∫ b

af (g(x))g ′(x)dx =

∫ B

Af (u)du.

Proof: Let F be an antiderivative of f , that is, F ′(u) = f (u).Then

d

dxF (g(x)) = F ′(g(x))g ′(x) = f (g(x))g ′(x).

Thus,∫ b

af (g(x))g ′(x)dx = F (g(x))

∣∣∣ba

= F (g(b))− F (g(a))

= F (B)− F (A) =

∫ B

Af (u)du.

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Example 12:

Find the integral of

∫ 3

2

x2

x3 − 1dx .

Solution: Let u = x3 − 1. Then

du = 3x2dx .

If x = 2, then u = 7; if x = 3, then u = 26. Using the method ofsubstitution, ∫ 3

2

x2

x3 − 1dx =

1

3

∫ 26

7

1

udu

=1

3ln(u)

∣∣∣∣267

=1

3ln(

26

7).

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Example 13:

Evaluate the integral I =

∫ 8

0

cos√x + 1√

x + 1dx .

Solution: Let u =√x + 1. Then

du =1

2√x + 1

dx .

If x = 0, then u = 1; if x = 8, then u = 3. Using the method ofsubstitution,

I = 2

∫ 3

1cos(u)du

= 2 sin(u)∣∣∣31

= 2 (sin(3)− sin(1)) .

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