28. definite integrals – classworkmar 16, 2018 · the definite integral represents the limit of...

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28. Definite Integrals – Classwork The course started off with defining the “slope of the tangent line to a function.” We developed a term for that concept: the derivative. We will do the same for areas under curves. We define the term for “the area between the function f x( ) , the x-axis, and the lines x = a and x = b ” as the definite integral. We write this as

f x( )a

b

∫ dx and read it as “the definite integral between x = a and x = b of f x( )dx ” or “the definite integral of

f x( )dx between x = a and x = b .” We call f x( ) the integrand and a and b the limits of integration. This expression might not make much sense, but there is a reason for it. It stems back to Riemann sums, which represented a sum of rectangles. In the figure to the left, we examine one Riemann sum rectangle (it makes no difference whether it is a LRS or RRS). Its height is f x( ) and its base (or width) is Δx . From our discussion of differentials, we know that Δx = dx . So f x( ) ⋅dx is just another way of specifying the area of a rectangle. The definite integral represents the limit of the sum of these areas (the integral sign looks like a flattened S for sum) as the number of rectangles approaches infinity,

which is saying that dx approaches 0. So f x( )a

b

∫ dx simply means the sum of the

areas of infinitely thin rectangles, starting at a and ending at b, which gives the figure to the right.

When you see a definite integral f x( )a

b

∫ dx , you immediately think: “area under the curve f x( ) .” However, we

have to be careful. Areas are always positive. Definite integrals can be both positive and negative. • When a < b , the integral is being analyzed from left to right. So dx is a positive number. If f x( ) is above

the x-axis (positive), our f x( )a

b

∫ dx is a summation of positive times positive numbers which is positive.

• When a > b , the integral is being analyzed from right to left. So dx is a negative number. If f x( ) is above

the x-axis (positive), our f x( )a

b

∫ dx is a summation of positive times negative numbers which is negative.

f x( ) > 0 curve above axis( ) +( ) f x( ) < 0 curve below axis( ) –( )a < b⇒ left to rightmeaning that dx > 0 +( ) f x( )

a

b

∫ dx > 0 +( ) +( ) = + f x( )a

b

∫ dx < 0 –( ) +( ) = –

a > b⇒ right to leftmeaning that dx < 0 –( ) f x( )

a

b

∫ dx < 0 +( ) –( ) = – f x( )a

b

∫ dx > 0 –( ) –( ) = +


Here are 5 simple rules, which should make total sense to you if you think of f x( )a

b

∫ dx as an area.

1. f x( )dx = 0a

a

∫ . If we start at a and end at a, all we have is a vertical line which has no area.

2. f x( )dx =b

a

∫ − f x( )a

b

∫ dx . From b to a gives a number. This will be the negative of the value if you start

at a and end at b. So you may switch the limits of integration as long as you negate the integral.

3. k ⋅ f x( )a

b

∫ dx = k f x( )a

b

∫ dx . You can move a constant in and out a definite integral.

4. f x( )± g x( )⎡⎣ ⎤⎦a

b

∫ dt = f x( )dx ± g x( )dxa

b

∫a

b

∫ . The integral of a sum is the sum of the integrals.

5. f x( )dx +a

b

∫ f x( )dx =b

c

∫ f x( )dxa

c

∫ . Add the area from a to b to the area from b to c and you get the area

from a to c. Ex 1) Below you are given the graph of f x( ) formed by lines and a semicircle. Find the value of the definite integrals.

a) f t( )4

4

∫ dt b) f t( )0

1

∫ dt

c) f t( )1

3

∫ dt d) f t( )0

3

∫ dt

e) f t( )3

6

∫ dt f) f t( )6

3

∫ dt g) f t( )0

6

∫ dt h) f t( )6

10

∫ dt

i) f t( )10

6

∫ dt j) f t( )0

10

∫ dt k) f t( )10

0

∫ dt l) f t( )−1

0

∫ dt

m) f t( )−3

0

∫ dt n) f t( )0

−3

∫ dt o) f t( )−4

−3

∫ dt p) f t( )−4

0

∫ dt

q) f t( )−4

10

∫ dt r) f t( )−4

10

∫ dt s) f t( )−4

10

∫ dt t) −2 f t( )1

−1

∫ dt


2) Suppose f x( )dx =18, −2

5

∫ g x( )dx = 5, −2

5

∫ h x( )dx = −11, −2

5

∫ and f x( )dx = 0, find −2

8

∫

a) f x( )+ g x( )⎡⎣ ⎤⎦−2

5

∫ dx b) f x( )− g x( )− h x( )⎡⎣ ⎤⎦−2

5

∫ dx c) 4g x( )5

−2

∫ dx

d) g x( )+ 2⎡⎣ ⎤⎦−2

5

∫ dx e) f x( )− 6⎡⎣ ⎤⎦−2

5

∫ dx f) 12h x( )+ 5⎡

⎣⎢⎤⎦⎥

−2

5

∫ dx

g) h x − 2( )0

7

∫ dx h) g x + 2( )−4

3

∫ dx i) f x( )5

8

∫ dx

j) 1− f x( )⎡⎣ ⎤⎦8

5

∫ dx k) f x − 3( )+ 3⎡⎣ ⎤⎦1

8

∫ dx l) 2 f x − 4( )− 6⎡⎣ ⎤⎦2

9

∫ dx

3) Evaluate the following by making a sketch of the function.

a) x + 2( )dx−3

3

∫ b) x + 2 dx−3

3

∫ c) x + 2 dx−3

3

∫ d) 2 − x dx−3

3

∫


28. Definite Integrals – Homework 1. In the graph to the right, y = f x( ) consists of

lines and a quarter circle. Find the values of the following definite integrals.

a. f x( )0

1

∫ dx b. f x( )2

4

∫ dx

c. f x( )1

4

∫ dx d. f x( )5

5

∫ dx

e. f x( )4

5

∫ dx f. f x( )5

6

∫ dx g. f x( )4

6

∫ dx h. f x( )0

6

∫ dx

i. f x( )3

2

∫ dx j. f x( )5

0

∫ dx k. f x( )6

0

∫ dx l. f x( )−3

0

∫ dx

m. f x( )0

−3

∫ dx n. f x( )−3

2

∫ dx o. f x( )4

−3

∫ dx p. f x( )−6

−3

∫ dx

q. f x( )−3

−6

∫ dx r. f x( )−6

0

∫ dx s. f x( )6

−6

∫ dx t. −2 f x( )−1

−2

∫ dx

u. f x( )−2

1

∫ dx v. f x( )−2

1

∫ dx w. f x( )−6

6

∫ dx x. f x−6

6

∫ dx

2. Evaluate the following by making a sketch of the function.

a. 2x −1( )dx−4

4

∫ b. 2x −1 dx−4

4

∫ c. 2x −1( )dx−4

4

∫ d. 2x −1 dx−4

4

∫


3. Suppose f x( )dx = 2, 0

2

∫ f x( )dx = −1, and 1

2

∫ f x( )dx = 7, 2

4

∫ evaluate the following:

a. f x( )dx 1

4

∫ b. 3 f x( )dx 0

4

∫ c. f x( )dx 0

1

∫

d. f x +1( )dx 0

1

∫ e. f x( )+ 3⎡⎣ ⎤⎦dx 0

2

∫ f. 2 f x − 2( )dx 2

4

∫

4. If f x( )dx−3

5

∫ = 9 and g x( )dx−3

5

∫ = −2 , find

a. f x( )+ g x( )⎡⎣ ⎤⎦dx−3

5

∫ b. g x( )− f x( )⎡⎣ ⎤⎦dx5

−3

∫ c. 2 f x( )− 12g x( )+ 1

4⎡⎣⎢

⎤⎦⎥dx

−3

5

∫

5. The graph of f x( ) , defined on −1.25,k[ ] , is shown on the figure on the right

with the areas of the shaded regions indicated by A, B, C and D. Find the following definite integrals in terms of A, B, C and D.

a. f x( )dx−1.25

2

∫ b. f x( )dx0

2

∫ − f x( )dx−1.25

0

∫

c. f x( ) dx−1.25

2

∫ d. f x( )dx2

−1.25

∫ − f x( ) dx2

−1.25

∫ e. f x( )−1( )dx−1

1

∫

f. If f x( )dx−1.25

k

∫ = 0 , find f x( )dx2

k

∫ , k > 2


29. The Accumulation Function – Classwork

Now that we understand that the concept of a definite integral is nothing more than an area, let us consider a very special type of problem. Suppose we have a function f x( ) = 2 . We know that this is a horizontal line as shown in the figure to the right. Note: realize that the equation of the graph f x( ) = 2 is the same as f t( ) = 2 or f k( ) = 2 . Whether we use x, t, k or any other variable does not

matter. The graph is still a horizontal line. We are used to using x, but we will see that there is a good reason that we will occasionally use another variable.

• Consider the expression f t( )0

x

∫ dt . This is a function of x. It represents the area under the f t( ) curve shown

above starting at t = 0 and ending at t = x. So, as x changes, f t( )0

x

∫ dt changes as well. Complete the chart:

It should be apparent that as x gets larger, f t( )0

x

∫ dt increases as well .

We are accumulating area and we call the expression f t( )0

x

∫ dt the

accumulation function. It should be apparent why we use f t( ) = 2 rather than f x( ) = 2 . f x( )0

x

∫ dx would

be confusing. • To the right is the function f x( ) = 2x .

Let’s calculate the value of the accumulation function from x = 0 to 4.

Now plot those values on the same graph. And connect them with a smooth curve. The curve should be familiar. What is it? ____________

Now, let’s take the derivative of this function: ddx

f t( )0

x

∫ dt⎡

⎣⎢⎢

⎤

⎦⎥⎥= . This leads us to a very

important result. ddx

f t( )0

x

∫ dt⎡

⎣⎢⎢

⎤

⎦⎥⎥= f x( ) . We will study this formula in great detail later in the course. But

what this says is that the slope of the tangent line to the accumulation of a function is function itself. If we start with a function, accumulate it, then take its derivative and we end up back with the function, it should lead you to believe that the accumulation function and derivatives have an inverse relationship. This is all building to the Fundamental Theorem of Calculus, which will be examined in the next chapter.

x 0 1 2 3 4 5

f t( )0

x

∫ dt

x 0 1 2 3 4

f t( )0

x

∫ dt


1) Let F x( ) = f t( )0

x

∫ dt where the graph of f x( ) is to the right.

Remember, f x( ) is the same as f t( ) . To make this easier to understand, think of f t( ) as the rate of snow accumulation over an 8-hour period of time. For instance, at t = 1, snow is falling at the rate of 3 inches per hour. At t = 3, it is not snowing, and at t = 4, snow is melting at 4 inches per hour. F x( ) then represents the accumulation of snow on the ground at any point in time x.

a) Complete the table for the given times x.

x 0 1 2 3 4 5 6 6.5 7 8

F x( ) = f t( )dt0

x

∫

b) We have seen that ddx

f t( )0

x

∫ dt⎡

⎣⎢⎢

⎤

⎦⎥⎥= f x( ) . So since F x( ) = f t( )

0

x

∫ dt , it follows that ′F x( ) = f x( ) . With

that in mind, complete the table. x 0 1 2 3 4 5 6 6.5 7 8

′F x( ) = ddx

f t( )dt0

x

∫⎡

⎣⎢⎢

⎤

⎦⎥⎥= f x( )

c) Remember that we can determine information for when F x( ) increases/decreases as well as relative extrema

by looking at ′F x( ) . Using the table above, make a sign chart of ′F x( ) and determine intervals where F increases, F decreases, and when F has a relative maximum and minimum and what those values are.

d) Remember we can determine information about the concavity of F x( ) by looking at ′′F x( ) . Since ′F x( ) = f x( ) , it follows that ′′F x( ) = ′f x( ) . Looking at the original curve, which is f x( ) , construct a sign

chart for ′′F x( ) and determine intervals where F is concave up, concave down, and inflection pts.

e) Sketch a graph of F x( )


2) Let F x( ) = f t( )0

x

∫ dt where f is the

function graphed to the right. f consists of lines and a semi-circle. Our goal is to sketch F x( ) .

Find the following: a) F 0( ) b) F 2( ) c) F 4( ) d) F 6( ) e) F −1( ) f) F −2( ) g) F −3( ) h) F −4( ) i) ′F 4( ) j) ′F 2( ) k) ′F 6( ) l) ′F 5( ) m) ′′F 2( ) n) ′′F 5( ) o) ′′F −0.5( ) p) ′′F 0( ) q) On what subintervals of −4,6[ ] is F increasing and decreasing? Justify your answer. r) Find the absolute minimum and maximum values of x on the interval. Justify your answer. s) On what subintervals of −4,6[ ] is F concave up/concave down? What are the inflection points? Justify. t) Sketch the function F.


29. The Accumulation Function – Homework

1. Let F x( ) = f t( )0

x

∫ dt where the graph of f x( ) is to the right

consisting of lines and a quarter-circle. a. Complete the table.

x 0 1 2 3 4 5 7 8

F x( )′F x( ) = f x( )

b. On what subintervals of 0,8[ ] is F increasing and decreasing. Justify your answer. c. Where on 0,8[ ]does F attain its minimum and maximum values? What are these values? Justify. d. Find the concavity of F and find any inflection points. Justify your answer. e. Sketch a graph of F. 2. “Your call will be answered in the order that it was

received” has been heard by anyone who has been put on hold. You call electronically goes into a queue that gets longer when calls are received and gets shorter when calls are answered. Over a 10-minute period, the rate of change of the length of the queue is given by the continuous function q t( ) , as shown by the table to the right. Using a trapezoidal rule, approximate the length of the queue after 10 minutes and determine at what value t the queue is the longest. Assume the queue is empty at t = 0.

t 0 1 2 3 4 5 6 7 8 9 10q t( ) 6 5 3 0 −3 −2 0 5 4 0 −4


3. Let F x( ) = f t( )0

x

∫ dt where the graph of f x( ) is below consisting of lines and a semi-circle.

a. Complete the table. If values are not exact, approximate them. x −4 −3 −2 −1 0 1 2 3 4 5 6 7 8

F x( )′F x( )

b. On what subintervals of −4,8[ ] is F increasing and decreasing. Justify your answer. c. Where on −4,8[ ]does F attain its minimum and maximum values? What are these values? Justify. d. Find the concavity of F and find any inflection points. Justify your answer. e. Sketch a graph of F.


4. You have bought 100 shares of Newton stock and decide to keep it for 15 days. Below is a graph that represents the change in price of a share of stock on each day. For instance, at the end of day 1, your stock has increased by $1.50 a share. At the end of day 4, the value of the stock has not changed. At the end of day 6, it has gone down by $3.50 a share. The graph to the right represents f t( )made up of straight lines. Remember that the function represents the change in the value of your stock and not

the value. Let F x( ) = f t( )dt0

x

∫ .

a. Complete the table. If values are not exact, approximate them. x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15f x( ) 0 1.5 3 5 0 −2 −3.5 −1 −1 0 4 2.5 0 0.5 0 −5F x( )

b. On what subintervals of [0, 15] is F increasing and decreasing. Justify your answer.

c. Where on [0, 15] does F attain its minimum and maximum values? What are these values? Justify. d. Find the concavity of F and find any inflection points. Justify your answer. e. Sketch a graph of F. f. How much money did you make on Newton stock during the 15-day period? g. What day did the stock’s value have the biggest rise? h. Between what two days did the stock’s value have the steepest rise? i. What day did the stock’s value have the biggest decline? k. What day do you wish you sold the stock? j. Between what two days did the stock’s value have l. What day (after 2 days) are you most glad the steepest decline?

you didn’t sell your stock?


30. The Fundamental Theorem of Calculus – Classwork So far in calculus, we have learned 4 major concepts: 1) limits, 2) derivatives, 3) indefinite integrals, and 4) definite integrals. We have seen that a definite integral represents an area under a curve. The question is: what do definite integrals have to do with derivatives? The answer is that the Fundamental Theorem of Calculus (FTC) is a theorem that links the concept of the derivative of a function with the concept of the function’s indefinite integral. It also states that the definite integral of a function can be computed using any of its infinitely many anti-derivatives. The Fundamental Theorem of Calculus (FTC) simply states that if f and F are continuous functions defined on a

closed interval a,b[ ] and if ′F x( ) = f x( ) , then f x( )dx = F x( )⎤⎦ab = F b( )− F a( )

a

b

∫ .

We will not formally prove the FTC (most calculus books and websites on the topic have proofs). What is below is an informal proof that shows the link between the definite integral and derivatives.30a

We start with a continuous function f x( ) as shown on the figure on the right. Each value of x has a corresponding area function F x( ) , representing the area under f x( ) between 0 and x. The function F x( )may not be known, but it is given that it represents the area under the curve. The area under the curve between x and x + h (the light colored area) could be computed by finding the area between 0 and x + h, then subtracting the area between 0 and h. In other words, the area of this thin sliver is F x + h( )− F x( ) . There is another way to approximate the area of this same sliver. If h is multiplied by f x( ) we have the area of a rectangle that is approximately the same size of this light colored sliver. So F x + h( )− F x( ) ≈ f x( )h . This approximation becomes a perfect equality if we add the “excess” area as shown in the figure. So it follows: F x + h( )− F x( ) ≈ f x( )h + Excess( ) .

Rearranging terms, we get f x( ) = F x + h( )− F x( )h

−Excess( )h

. As h approaches 0, the fraction Excess( )h

goes

to zero. This occurs because the excess area is less than the area of the small rectangle. That small rectangle has

area of height • h and height ⋅hh

= height . As h approaches 0, the height goes to zero as well.

So, f x( ) = limh→0

F x + h( )− F x( )h

. This we know is the derivative of F so f x( ) = ′F x( ) . That is, the derivative of

the area function F x( ) is the original function f x( ) ; or the area function is simply an anti-derivative of the original function. “Finding the area under a curve” and finding a derivative are inverse operations. This is the crux of the Fundamental Theorem of Calculus.

Excess

Area = F x( )


Find the following definite integrals:

1) x2 dx1

2

∫ 2) 3x2 − 2x −1( )−1

3

∫ dx 3) x0

9

∫ dx

4) 23x2

dx5

10

∫ 5) 2x −1( )2

−1

1

∫ dx 6) x − 2x3dx

1

8

∫

7) 2 + sin x( )dx0

π

∫ 8) 4x + sec2 x( )dx0

π 4

∫ 9) 4 − 2x dx0

5

∫

Your TI-84 calculators are capable of approximating the values of definite integrals and can do it from the home screen or the graphing screen (where you actually see the area under the curve). Place the function in Y1 or one of the Y= equations. Use Math 9:FnInt(Y1,X,lower limit, upper limit). Y1 is found in the VARS YVARS menu.

For instance to find x2 −1( )dx0

2

∫ , use the upper right. You can even change the answer

into a fraction for relatively simple functions. To actually see the area, create a window that shows the function within the limits given and press 2nd Calc ∫ f(x) dx. You are asked to input the lower limit and upper limit. How does the calculator find the definite integral. Like finding the numerical derivative, the calculator approximates the definite integral. It uses a method to (but superior to)

Riemann sums. But it is not exact. For instance, it can easily be shown that cosdx0

π

∫ = 0 .

But the calculator, as shown to the right, gets an answer which is essentially, but not exactly zero. That is because the calculator is not actually integrating but approximating.


30. The Fundamental Theorem of Calculus – Homework Find the following definite integrals. Confirm by calculator.

1. 3x dx0

1

∫ 2. x − 5( )dx−2

3

∫ 3. x2 + 2x −1( )dx−1

4

∫

4. 2x − 5( )2 dx0

2

∫ 5. 4x2

+1⎛⎝⎜

⎞⎠⎟ dx

2

3

∫ 6. x − 1x2

⎛⎝⎜

⎞⎠⎟ dx

−2

−1

∫

7. x − 2x

⎛⎝⎜

⎞⎠⎟dx

1

9

∫ 8. x3 dx−2

2

∫ 9. t2 3 − t1 3( )dt0

1

∫

10. x − 2 dx0

3

∫ 11. cos x dx−π 2

π 2

∫ 12. 2x − sin x( )dx0

π

∫

13. 3sin x − 2cos x( )dx0

π 2

∫ 14. x − sec2 x( )dx0

π 4

∫ 15. secθ tanθ( )dθ0

π 3

∫

28. definite integrals – classworkmar 16, 2018 · the definite integral represents the limit of...

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