THIS POWER POINT PRESENTATION IS CREATED AND WRITTEN BY

DR. JULIA ARNOLDUSING THE 5TH EDITION OF S. T. TAN’S TEXT

APPLIED CALCULUS

Evaluating Definite Integrals

Rule 1: a

a

dxxf 0)(

Rule 2: b

a

a

b

dxxfdxxf )()(

Rule 3: c, a constant b

a

b

a

dxxfcdxxcf )()(

Rule 4: b

a

b

a

b

a

dxxgdxxfxgxf )()()]()([

Rule 5: bcadxxfdxxfdxxfb

c

c

a

b

a

,)()()(

Properties of the Definite Integral

Example 1: Evaluatedx

x

x

1

03

2

1

2ln31

02ln31

1ln31

2ln31

10ln31

11ln31

1ln31

ln31

131

31

10

3

1

02

1

0

21

03

2

xu

duux

du

u

xdx

x

x x

x

x

x

Solution:

dxx

du

dxxdu

xu

2

2

3

3

3

1

First of all evaluating the integral above is different (sometimes from finding the area under the curve).

Consider these next two problems:

dxx )1(2

0

2

2. Find the area under the curve bounded by the curve, the x axis and the vertical lines x = 0 and x = 2

1. Evaluate dxx )1(2

0

2

First of all evaluating the integral above is different (sometimes from finding the area under the curve).

Consider these next two problems:

dxx )1(2

0

2

2. Find the area under the curve bounded by the curve, the x axis and the vertical lines x = 0 and x = 2

1. Evaluate dxx )1(2

0

2 Solution

32

36

38

0232

3)1()()1(

3

20

32

0

2

0

22

0

2

xx

dxdxxdxx

dxx )1(2

0

2

2. Find the area under the curve bounded by the curve, the x axis and the vertical lines x = 0 and x = 2

First let’s look at the graph.

Part of the area lies below the x-axis

When x = 1, f(1)= 0

(1,0)Let’s find the area below the x axis by integrating from 0 to 1.

dxx )1(2

0

2

2. Find the area under the curve bounded by the curve, the x axis and the vertical lines x = 0 and x = 2

(1,0)

32

0131

31 1

0

31

0

2

xx

dxx

As you can see from the result, areas below the x-axis are negative, although area itself is a positive quantity. Thus the area below the x- axis is 2/3.

dxx )1(2

0

2

2. Find the area under the curve bounded by the curve, the x axis and the vertical lines x = 0 and x = 2

34

131

36

38

131

238

31 2

1

32

1

2

xx

dxx

Now let’s find the area from 1 to 2.

(1,0)

The area is then 4/3 + 2/3 = 6/3or 2.

Average Value of a Function

Suppose f is integrable on [a,b]. Then the average value of f over [a,b] is

b

a

dxxfab

)(1

Find the average value of over the interval [0,4] xxf )(

34

68

461

0461

32

41

041

3

23

40

234

0

xdxx

Average Value of a Function

Find the average value of over the interval [0,4] xxf )(

34

68

461

0461

32

41

041

3

23

40

234

0

xdxx

Y= 4/3

Now that you’ve seen some examples, see if you can answer the following questions.

The area under the curve is the same as the definite integral?True or False (Run your mouse over the answer you think is correct

True

False

Now that you’ve seen some examples, see if you can answer the following questions.

The area under the curve is the same as the definite integral?True or False (Run your mouse over the answer you think is correct

True As seen in an earlier example. Go back if you don’t recall it.The definite integral was equal to 2/3, while the area under the curve was 6/3

To find out the area under a curve, you shouldA. View the graphB. Separate the areas below the x axis from the areas

above the x-axisC. Both of the aboveD. None of the above

A B C D

To find out the area under a curve, you shouldA. View the graphB. Separate the areas below the x axis from the areas

above the x-axisC. Both of the aboveD. None of the above

A B C

You should probably do both A, and B

Evaluate the definite integral and find the area under the curve for:

dxxx 2

0

21

34

38

4032

42

0

2

341

34

342

0

232

0

2

xx

dxxxdxxx

0316

316

032

62

0

2

36321

45

4532

0

22

0

2

xxdx

xx

xdxxx

Which of the following is the definite integral?

Evaluate the definite integral and find the area under the curve for:

dxxx 2

0

21

34

38

4032

42

0

2

341

34

342

0

232

0

2

xx

dxxxdxxx

Which of the following is the definite integral?

Remember you must multiply out polynomials before integrating them.

Evaluate the definite integral and find the area under the curve for:

dxxx 2

0

21

-1 1 2 3 4 5

-2

-1

1

2

3

4

Using the graph below the area under the curve is which of the following?

dxxxdxxx

dxxx

2

1

231

0

23

2

0

21

dxxxdxxx

dxxx

2

1

230

1

23

2

0

21

Evaluate the definite integral and find the area under the curve for:

dxxx 2

0

21

-1 1 2 3 4 5

-2

-1

1

2

3

4

Using the graph below the area under the curve is which of the following?

dxxxdxxx

dxxx

2

1

230

1

23

2

0

21

If you don’t change anything, you will get the definite integral value, which is what you would get for the first answer.Notice the limits of integration have been reversed in the first integral.

This causes the integral value which is negative from 0 to 1 to become positive for 1 to 0.

23

1218

1217

121

1217

121

1232

1248

121

38

416

31

41

32

42

1

2

34

121

31

41

01

0

34

1

343434

3434

2

1

230

1

23

2

0

2

Add

xx

xx

dxxxdxxx

dxxx

Now let’s compute the integrals:

The area under the curve from [0,2] is 3/2.While the definite integral from [0,2] is 4/3.

In a real life application of this material, one can find average yearly sales, concentration of a drug in the bloodstream, flow of blood in an artery, depreciation using the double declining-balance method.However, we will leave it to the business professional , or biologist to apply the material.Go to the homework.