chapter 6 the definite integral. antidifferentiation areas and riemann sums definite integrals...

Chapter 6

The Definite Integral

Antidifferentiation

Areas and Riemann Sums

Definite Integrals and the Fundamental Theorem

Areas in the xy-Plane

Applications of the Definite Integral

Chapter Outline

§ 6.1

Antidifferentiation

Antidifferentiation

Finding Antiderivatives

Theorems of Antidifferentiation

The Indefinite Integral

Rules of Integration

Antiderivatives in Application

Section Outline

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #5

Antidifferentiation

Definition Example

Antidifferentiation: The process of determining f (x) given f ΄(x)

If , then xxf 2

.2xxf



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find all antiderivatives of the given function.

The derivative of x9 is exactly 9x8. Therefore, x9 is an antiderivative of 9x8. So is x9 + 5 and x9 -17.2. It turns out that all antiderivatives of f (x) are of the form x9 + C (where C is any constant) as we will see next.

89xxf


Theorems of Antidifferentiation


The Indefinite Integral


Rules of Integration



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Determine the following.

Using the rules of indefinite integrals, we have

dx

xxx

3

12 2

dx

xdxxxdxdx

xxx

3

12

3

12 22

dxx

dxxxdx1

3

12 2

Cxxx

ln3

1

32

2

32



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the function f (x) for which and f (1) = 3.

The unknown function f (x) is an antiderivative of . One

antiderivative is . Therefore, by Theorem I,

xxxf 2

xxxf 2

3

2

3

233 xx

constant. a ,3

2

3

233

CCxx

xf

Now, we want the function f (x) for which f (1) = 3. So, we must use that information in our antiderivative to determine C. This is done below.

CCCf

13

12

3

1

3

12

3

113

233



So, 3 = 1 + C and therefore, C = 2. Therefore, our function is

CONTINUECONTINUEDD

.23

2

3

233

xx

xf



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

A rock is dropped from the top of a 400-foot cliff. Its velocity at time t seconds is v(t) = -32t feet per second.

(a) Find s(t), the height of the rock above the ground at time t.(b) How long will the rock take to reach the ground?

(c) What will be its velocity when it hits the ground?

(a) We know that s΄(t) = v(t) = -32t and we also know that s(0) = 400. We can now use this information to find an antiderivative of v(t) for which s(0) = 400.

The antiderivative of v(t) is To determine C, .00160400 2 CCCs

Therefore, C = 400. So, our antiderivative is s(t) = -16t2 + 400.

.1632 2 Cttdtts



s(t) = -16t2 + 400

CONTINUECONTINUEDD(b) To determine how long it will take for the rock to reach the ground, we simply need to find the value of t for which the position of the rock is at height 0. In other words, we will find t for when s(t) = 0.

This is the function s(t).

0 = -16t2 + 400 Replace s(t) with 0.

-400 = -16t2 Subtract.

25 = t2 Divide.

5 = t Take the positive square root since t ≥ 0.

So, it will take 5 seconds for the rock to reach the ground.



v(t) = -32t

CONTINUECONTINUEDD

This is the function v(t).

Replace t with 5 and solve.

So, the velocity of the rock, as it hits the ground, is 160 feet per second in the downward direction (because of the minus sign).

(c) To determine the velocity of the rock when it hits the ground, we will need to evaluate v(5).

v(5) = -32(5) = -160

§ 6.2

Areas and Riemann Sums

Area Under a Graph

Riemann Sums to Approximate Areas (Midpoints)

Riemann Sums to Approximate Areas (Left Endpoints)

Applications of Approximating Areas

Section Outline


Area Under a Graph

Definition Example

Area Under the Graph of f (x) from a to b: An example of this is shown to the right


Area Under a Graph

In this section we will learn to estimate the area under the graph of f (x) from x = a to x = b by dividing up the interval into partitions (or subintervals),

each one having width where n = the number of partitions that

will be constructed. In the example below, n = 4.n

abx

A Riemann Sum is the sum of the areas of the rectangles generated above.


Riemann Sums to Approximate Areas

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Use a Riemann sum to approximate the area under the graph f (x) on the given interval using midpoints of the subintervals

4,22;2 nxxxf

The partition of -2 ≤ x ≤ 2 with n = 4 is shown below. The length of each subinterval is

.1

4

22

x

-2 2x1 x2 x3 x4

x

x



Observe the first midpoint is units from the left endpoint, and the midpoints themselves are units apart. The first midpoint is x1 = -2 + = -2 + .5 = -1.5. Subsequent midpoints are found by successively adding

CONTINUECONTINUEDD

x.1x

2/x2/x

midpoints: -1.5, -0.5, 0.5, 1.5

The corresponding estimate for the area under the graph of f (x) is

xfxfxfxf 5.15.05.05.1

xffff 5.15.05.05.1

5125.225.025.025.2

So, we estimate the area to be 5 (square units).


Approximating Area With Midpoints of Intervals

CONTINUECONTINUEDD

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

-3 -2 -1 0 1 2 3



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Use a Riemann sum to approximate the area under the graph f (x) on the given interval using left endpoints of the subintervals

5,31;3 nxxxf

The partition of 1 ≤ x ≤ 3 with n = 5 is shown below. The length of each subinterval is

.4.05

13

x

3

x

x1 x2 x3 x4 x5

1 1.4 1.8 2.2 2.6



The corresponding Riemann sum is

CONTINUECONTINUEDD

xfxfxfxfxf 6.22.28.14.11

xfffff 6.22.28.14.11

12.154.06.22.28.14.11 33333

So, we estimate the area to be 15.12 (square units).


Approximating Area Using Left Endpoints

CONTINUECONTINUEDD

0

5

10

15

20

25

30

1 1.4 1.8 2.2 2.6 3 3.4



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

The velocity of a car (in feet per second) is recorded from the speedometer every 10 seconds, beginning 5 seconds after the car starts to move. See Table 2. Use a Riemann sum to estimate the distance the car travels during the first 60 seconds. (Note: Each velocity is given at the middle of a 10-second interval. The first interval extends from 0 to 10, and so on.)

Since measurements of the car’s velocity were taken every ten seconds, we will use . Now, upon seeing the graph of the car’s velocity, we can construct a Riemann sum to estimate how far the car traveled.

10x



5, 20

15, 44

25, 3235, 39

45, 65

55, 80

0

10

20

30

40

50

60

70

80

90

0 10 20 30 40 50 60

Time

Vel

oci

ty

CONTINUECONTINUEDD



Therefore, we estimate that the distance the car traveled is 2800 feet.

CONTINUECONTINUEDD

tvtvtvtvtvtv 55453525155

tvvvvvv 55453525155

10806539324420

2800

§ 6.3

Definite Integrals and the Fundamental Theorem


Calculating Definite Integrals

The Fundamental Theorem of Calculus

Area Under a Curve as an Antiderivative

Section Outline



Δx = (b – a)/n, x1, x2, …., xn are selected points from a partition [a, b].

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0 0.2 0.4 0.6 0.8 1 1.2



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Calculate the following integral.

1

05.0 dxx

The figure shows the graph of the function f (x) = x + 0.5. Since f (x) is nonnegative for 0 ≤ x ≤ 1, the definite integral of f (x) equals the area of the shaded region in the figure below.

10.5

1



The region consists of a rectangle and a triangle. By geometry,

5.05.01heightwidthrectangle of area

CONTINUECONTINUEDD

5.0112

1heightwidth

2

1 triangleof area

Thus the area under the graph is 0.5 + 0.5 = 1, and hence

.15.01

0 dxx



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Calculate the following integral.

1

1xdx

The figure shows the graph of the function f (x) = x on the interval -1 ≤ x ≤ 1. The area of the triangle above the x-axis is 0.5 and the area of the triangle below the x-axis is 0.5. Therefore, from geometry we find that

.05.05.01

1 xdx



-1.5

-1

-0.5

0

0.5

1

1.5

-1.5 -1 -0.5 0 0.5 1 1.5

CONTINUECONTINUEDD



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Use the Fundamental Theorem of Calculus to calculate the following integral.

1

0

5.031 13 dxex x

An antiderivative of 3x1/3 – 1 – e0.5x is . Therefore, by the fundamental theorem,

xexxxF 5.034 24

9

01131

0

5.031 FFdxex x

05.03415.034 200

4

9211

4

9ee

05.0 200

4

9211

4

9ee

047.02047.2



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

(Heat Diffusion) Some food is placed in a freezer. After t hours the temperature of the food is dropping at the rate of r(t) degrees Fahrenheit per hour, where .3/412 2 ttr

(a) Compute the area under the graph of y = r(t) over the interval 0 ≤ t ≤ 2.(b) What does the area in part (a) represent?

(a) To compute the area under the graph of y = r(t) over the interval 0 ≤ t ≤ 2, we evaluate the following.

20

2

0

2 )3/(4123/412 t

tttdtt

)30/(4012)32/(4212

533.243

4

5

116



(b) Since the area under a graph can represent the amount of change in a quantity, the area in part (a) represents the amount of change in the temperature between hour t = 0 and hour t = 2. That change is 24.533 degrees Fahrenheit.

CONTINUECONTINUEDD


Area Under a Curve as an Antiderivative

§ 6.4

Areas in the xy-Plane

Properties of Definite Integrals

Area Between Two Curves

Finding the Area Between Two Curves

Section Outline


Properties of Definite Integrals


Area Between Two Curves



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the area of the region between y = x2 – 3x and the x-axis (y = 0) from x = 0 to x = 4.

Upon sketching the graphs we can see that the two graphs cross; and by setting x2 – 3x = 0, we find that they cross when x = 0 and when x = 3. Thus one graph does not always lie above the other from x = 0 to x = 4, so that we cannot directly apply our rule for finding the area between two curves. However, the difficulty is easily surmounted if we break the region into two parts, namely the area from x = 0 to x = 3 and the area from x = 3 to x = 4. For from x = 0 to x = 3, y = 0 is on top; and from x = 3 to x = 4, y = x2 – 3x is on top. Consequently,



CONTINUECONTINUEDD

dxxxxx 4

3

2 034 to3 from area

dxxx 4

3

2 3

.833.12

27924

3

64

23

3

4

3

23

xx

dxxxxx 3

0

2 303 to0 from area

dxxx 3

0

2 3

.5.4002

279

23

3

3

0

23

xx

Thus the total area is 4.5 + 1.833 = 6.333.



CONTINUECONTINUEDD

-4

-2

0

2

4

6

8

10

12

-2 -1 0 1 2 3 4 5

y = x2 – 3x

y = 0



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Write down a definite integral or sum of definite integrals that gives the area of the shaded portion of the figure.

Since the two shaded regions are (1) disjoint and (2) have different functions on top, we will need a separate integral for each. Therefore



Therefore, to represent all the shaded regions, we have

dxxfdxxfxx 2

1

2

102 to1 from area

CONTINUECONTINUEDD

dxxfdxxfxx 4

3

4

304 to3 from area

.4

3

2

1 dxxfdxxf



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Two rockets are fired simultaneously straight up into the air. Their velocities (in meters per second) are v1(t) and v2(t), respectively, and v1(t) ≥ v2(t) for t ≥ 0. Let A denote the area of the region between the graphs of y = v1(t) and y = v2(t) for 0 ≤ t ≤ 10. What physical interpretation may be given to the value of A?

Since v1(t) ≥ v2(t) for t ≥ 0, this suggests that the first rocket is always traveling at least as fast as the second rocket. Therefore, we have

dttvtvA 10

0 21

dttvdttv 10

0 2

10

0 1

10

02

10

01 tsts



10

021 tsts

CONTINUECONTINUEDD

But again, since v1(t) ≥ v2(t) for t ≥ 0, we know that . So,

this implies that . This means that the position of the first rocket is always at least as high (up in the air) as that of the second rocket. That is, the first rocket is always higher up than the second rocket (or at the same height).

010

0 21 dttvtv

010

021 tsts

§ 6.5

Applications of the Definite Integral

Average Value of a Function Over an Interval

Consumers’ Surplus

Future Value of an Income Stream

Volume of a Solid of Revolution

Section Outline



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Determine the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1.

Using (2) with a = -1 and b = 1, the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1 is equal to

.111

1 1

1

dxx

An antiderivative of 1 – x is . Therefore,2

2xx

.1

2

3

2

1

2

1

2

11

2

11

2

1

22

11

2

1 221

1

21

1

xxdxx

So, the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1 is 1.



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

(Average Temperature) During a certain 12-hour period the temperature at time

t (measured in hours from the start of the period) was degrees. What was the average temperature during that period?

2

3

1447 tt

The average temperature during the 12-hour period from t = 0 to t = 12 is 12

0

3212

0

2

9247

12

1

3

1447

012

1

t

ttdttt

9

002047

9

121221247

12

1 32

32

.degrees 55066012

1



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the consumers’ surplus for the following demand curve at the given sales level x.

Since 20 units are sold, the price must be

20;10

3 xx

p

.12310

203 B

Therefore, the consumers’ surplus is

dxx

dxx

20

0

20

0 1021

103

20

0

2

202

xx



That is, the consumers’ surplus is $20.

20

002

20

20202

22CONTINUECONTINUE

DD

.20020



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

(Future Value) Suppose that money is deposited daily into a savings account at an annual rate of $2000. If the account pays 6% interest compounded continuously, approximately how much will be in the account at the end of 2 years?

Divide the time interval from 0 to 2 years into daily subintervals. Each

subinterval is then of duration years. Let t1, t2, ..., tn be points chosen

from these subintervals. Since we deposit money at an annual rate of $2000, the amount deposited during one of the subintervals is dollars. If this amount is deposited at time ti, the dollars will earn interest for the remaining 2 – ti years. The total amount resulting from this one deposit at time ti is then

365

1t

t2000t2000

.2000 206.0 itte



Add the effects of the deposits at times t1, t2, ..., tn to arrive at the total balance in the account:

.200020002000 206.0206.0206.0 21 teteteA nttt

CONTINUECONTINUEDD

This is a Riemann sum for the function on the interval 0 ≤ t ≤ 2. Since is very small when compared with the interval, the total amount in the account, A, is approximately

tetf 206.02000t

.4250106.0

2000

06.0

20002000 12.0

2

0

206.02

0

206.0

eedte tt

That is, the approximate balance in the account at the end of 2 years is $4250.



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the volume of a solid of revolution generated by revolving about the x-axis the region under the following curve.

Here g(x) = x2, and

2 to1 from 2 xxxy

.5

31132

512

55volume 55

2

1

52

1

42

1

22 x

dxxdxx

chapter 6 the definite integral. antidifferentiation areas and riemann sums definite integrals...

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