chapter 6 integration section 5 the fundamental theorem of calculus
TRANSCRIPT
Chapter 6
Integration
Section 5
The Fundamental Theorem of Calculus
2
Objectives for Section 6.5 Fundamental Theorem of Calculus
■ The student will be able to evaluate definite integrals.
■ The student will be able to calculate the average value of a function using the definite integral.
3
Fundamental Theorem of Calculus
If f is a continuous function on the closed interval [a, b], and F is any antiderivative of f, then
)()()()( ba aFbFxFdxxf
b
a
4
By the fundamental theorem we can evaluate
easily and exactly. We simply calculate
Evaluating Definite Integrals
b
adxxf )(
)()( aFbF
5
Definite Integral Properties
[ f (x) g (x)] dx
a
b
f (x) dxa
b
g (x) dxa
b
f (x)dx
a
b
f (x)dxa
c
f (x)dxc
b
kf (x)dx
a
b
k f (x)dxa
b
f (x)dx a
b
f (x)dxb
a
f (x)dx 0a
a
6
Example 1
Make a drawing to confirm your answer.
105151535553
1
3
1
x
xdx
0 ≤ x ≤ 4
–1 ≤ y ≤ 6
7
Example 2
42
1
2
9
2
3
1
23
1
x
xdxx
Make a drawing to confirm your answer.
0 ≤ x ≤ 4
–1 ≤ y ≤ 4
8
Example 3
9093
3
0
33
0
2
x
xdxx
0 ≤ x ≤ 4
–2 ≤ y ≤ 10
9
Example 4
Let u = 2x, du = 2 dx?1
1
2 dxe x
1
2eu du
x 1
1
eu
2 x 1
1
e2x
2 x 1
1
e2
2
e 2
2
3.6268604
10
Example 5
0.69314718 2ln 1ln – 2ln
ln1 2
1
2
1
xxdx
x
11
Example 6
207.78515
1ln – )/2(e– 1/3– 3ln )/2(e 9
ln23
1
26
3
1
23
3
1
22
x
x
x
xex
dxx
ex
This is a combination of the previous three problems
12
Example 7
Let u = x3 + 4, du = 3x2 dx?4
5
0 3
2
dx
x
x
1
3
3x2
x3 4dx
0
5
1
3
1
udu
x0
5
ln u
3 x 0
5
ln(x3 4)
3 x 0
5
(ln 129)/3 Š (ln 4)/3
1.1578393
13
Example 7(revisited)
1.1578393 4)/3(ln – 129)/3(ln 3
ln 129
4
u
u
On the previous slide, we made the back substitution from u back to x. Instead, we could have just evaluated the definite integral in terms of u:
5
0
35
0 3
)4(ln
3
ln
xx
xu
14
Numerical Integration on a Graphing Calculator
Use some of the examples from previous slides:
2
1
1dx
x
5
0 3
2
4dx
x
x
Example 5:
Example 7:
0 ≤ x ≤ 3
–1 ≤ y ≤ 3
–1 ≤ x ≤ 6
–0.2 ≤ y ≤ 0.5
15
Example 8
From past records a management service determined that the rate of increase in maintenance cost for an apartment building (in dollars per year) is given by M ´(x) = 90x2 + 5,000, where M(x) is the total accumulated cost of maintenance for x years.
Write a definite integral that will give the total maintenance cost from the end of the second year to the end of the seventh year. Evaluate the integral.
16
Example 8
From past records a management service determined that the rate of increase in maintenance cost for an apartment building (in dollars per year) is given by M ´(x) = 90x2 + 5,000, where M(x) is the total accumulated cost of maintenance for x years.
$29,480 15,000– 810– 35,000 10,290
500030000,5907
3
37
3
2
xxxdxx
Write a definite integral that will give the total maintenance cost from the end of the second year to the end of the seventh year. Evaluate the integral.
Solution:
17
Using Definite Integrals for Average Values
The average value of a continuous function f over [a, b] is
dxxfab
b
a)(
1
Note this is the area under the curve divided by the width. Hence, the result is the average height or average value.
18
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.
b) Find the average value of the cost function over the interval [0, 1000].
c) Write a description of the difference between part a) and part b).
Example
19
a) Find the average cost per unit if 1000 dictionaries are produced
Solution: The average cost is
Example(continued)
1020000)(
)( xx
xCxC
30101000
20000)1000( C
20
Example(continued)
b) Find the average value of the cost function over the interval [0, 1000]
Solution:
25,000 5,000 20,000
)520000(1000
1
)1020000(1000
1)(
1
1000
0
2
1000
0
x
b
a
xx
dxxdxxfab
21
Example(continued)
c) Write a description of the difference between part a and part b
Solution: If you just do the set-up for printing, it costs $20,000. This is the cost for printing 0 dictionaries.
If you print 1,000 dictionaries, it costs $30,000. That is $30 per dictionary (part a).
If you print some random number of dictionaries (between 0 and 1000), on average it costs $25,000 (part b).
Those two numbers really have not much to do with one another.
22
Summary
We can find the average value of a function f by
We can evaluate a definite integral by the fundamental theorem of calculus:
)()()()( aFbFxFdxxf ba
b
a
dxxfab
b
a)(
1