11

The student will learn about:

§4.5 Application of Definite Integrals and Area Between Curves.

the average value of a function, the average value of a function, and finding the area between two curves.

2

Introduction• In this section we will use definite integrals for two

important purposes: finding average values of functions and finding areas between curves.

• Average values are used everywhere. Birth weights of babies are compared with average weights, and retirement benefits are determined by average income.

• Averages eliminate fluctuations, reducing a collection of numbers to a single “representative” number.

• Areas between curves are used to find quantities from trade deficits to lives saved by seat belts.

3

Average Value of a FunctionIntuitively, the average should represent a “leveling off” of the curve to a uniform height, the

horizontal line shown on the right.

This leveling should use the “hills” to fill in the “valleys,” maintaining the same total area under the curve. Therefore, the area under the horizontal line must equal the area under the curve.

Equating the two areas

44

Using Definite Integrals for Average Values

Average Value of a Continuous Function f over [a, b].

Note this is the area under the curve divided by the width. Hence, the result is the average height or average value.

b

a

1f (x) dx

b a

55

Application§6.5 # 70. The total cost (in dollars) of printing x dictionaries is C (x) = 20,000 + 10x

a. Find the average cost per unit if 1000 dictionaries are produced

Continued on next slide.

Note that the average cost is C(x)C(x) .

x

20000C(x) 10

x

C(1000) 3020000

101000

What does this mean?

66

Application - continuedThe total cost (in dollars) of printing x dictionaries is C (x) = 20,000 + 10x

b. Find the average value of the cost function over the interval [0, 1000]

Continued on next slide.

b

a

1f (x) dx

b a

1000

0

1(20000 10x) dx

1000

10002

x 0

1(20000x 5x )

1000

= 20,000 + 5,000 = 25,000

What does this mean?

77

Application - concludedThe total cost (in dollars) of printing x dictionaries is C (x) = 20,000 + 10x

c. Write a description of the difference between part a and part b

From part a the average cost is $30.00

From part b the average value of the cost is $25,000.

The average cost per dictionary of making 1,000 dictionaries is $30. The average total cost of making between 0 and 1,000 dictionaries is $25,000.

88

Study of Area ContinuedIn the previous section we studied the area between the graph and the x-axis. The graph was always above the axis but this is not always the case.

If the curve is below the x-axis the area is negative. We need only change the sign to positive to get the correct answer.

99

Study of Area ContinuedWith our calculator we can guarantee that the curve will be above the x-axis by graphing

y = abs [ f(x) ].

We will use this technique.

10

Area Between Curves

The area between two curves can be written as a single integral:

b

aabs f (x) g(x) dx

Show how to get abs function on the calculator.

1111

Study of Area ContinuedWe will now study the area between two curves f (x) and g (x).

We will use our knowledge of integrals to find the area under this new curve.

f (x)

g (x)We will define a new function abs [ (f (x)) – (g (x)) ] and that situation will graph as an entirely different third function.

1212

SUMMARY OF AREA PROBLEMS

1. Graph y = abs ( f (x) – g (x) ) in the interval of integration from a to b. In some cases you may need to use minimum to find the interval of integration.

That’s it!

1313

Find the area bounded by y = 4 – x 2; y = 0, 0 x 4.

Example 1 - By Calculator

4

0

2 )0()4( dxxabsArea

1414

Example 2 - By CalculatorFind the area bounded by y = 0.5 x 2 + 3; y = 2x, from x = 1 to 3.

3 2

1A abs [ (0.5x 3) (2x) ]dx

Form a new function abs [(f (x)) – (g (x))] .

1515

Example 3 - By CalculatorFind the area bounded by y = x 3 + 3x 2 - 1; y = x + 3 from x = -2 to 1.

1 3 21 2

A abs [ (x 3x 1) (x 3) ]dx

Form a new function abs [(f (x)) – (g (x))] .

1616

Example 4 - By CalculatorFind the area bounded by y = x 2 – 4 and y = 8 – 3x – 2x 2

1.56 2 2

2.56A abs [(x 4) (8 3x 2x )]dx 35.05

Find the x-intercepts [minimum]

x = - 2.56 and 1.56

Form a new function abs [(f (x)) – (g (x))] .

In this problem the limits of integration are not given.

1717

SUMMARY OF AREA PROBLEMS

1. Graph y = abs [(f (x)) – (g (x))] in the interval of integration from a to b. In some cases you may need to use minimum to find the interval of integration.

That’s it!

1818

Summary.

We found a methods to calculate the area between two curves.

We can find the average value of a function f by: b

a

1f (x) dx

b a

b

aabs f (x) g(x) dx

1919

ASSIGNMENT

§4.5 on my website.

7, 8, 9, 15, 16, 17.