§12.5 The Fundamental Theorem of
Calculus
The student will learn about:
the definite integral,
the fundamental theorem of calculus,
and using definite integrals to find average
values.
1
Fundamental Theorem of Calculus
If f is a continuous function on the closed interval
[a, b] and F is any antiderivative of f, then

b
a
f ( x) dx  F( x) ba  F(b)  F(a)
2
Evaluating Definite Integrals
By the fundamental theorem we can evaluate

b
a
f ( x) dx
Easily and exactly. We simply calculate
F(b )  F(a )
3
Definite Integral Properties
REMINDER !
 a f (x) dx  0
a
 a f (x) dx    b f (x) dx
b
a
 a k  f (x) dx  k   a f (x) dx
b
b
 a [ f (x)  g (x) ] dx   a f (x) dx   a g (x) dx
b
b
b
 a f (x) dx   a f (x) dx   c f (x) dx
b
c
b
4
Example 1
3
1
 5 dx
 5 dx  5x
3
1
3
5 · 3 – 5 · 1 = 15 – 5 = 10

x1
Make a drawing to confirm your answer.
0x4
-1y6
5
Example 2
3
1
 x dx

3
1
2
x
x dx 
2
9 1

  4
x1
2 2
3
Make a drawing to confirm your answer.
0x4
-1y4
6
Example 3
3
1. 0
3
x
x dx 
3
2
3
x0
 9-0= 9
0x4
- 2  y  10
7
Example 4
1
2. 
1
e
2x
dx
Let u = 2x
du = 2 dx
u
1 1 u
e
1 e du 
2
2
1
x  1
1 1 2x
1 e 2 dx 
2

e
2x
2
1
x  1

e2 e 2

 3.6268604
2
2
8
Example 5
2
3. 
1
1
dx  ln x
x
2
x 1
= ln 2 – ln 1 = ln 2
= 0.69314718
9
Examples 6
3
4. 1
 2 2x
x  e 

3
2x
1
 dx
x
x
e

 ln x
3
2
3
x1
This is a combination
of the previous three
problems

= 9 + (e 6)/2 + ln 3 – 1/3 – (e2)/2 – ln 1
= 207.78515
10
Examples 7
2
1
3
x
x
Let u = + 4
5
5
dx

5. 
dx


0
0 3
3
2 dx
du
=
3x
3
x 4
x 4
2
x3
ln ( x  4) 5
ln u 5
1 51


0 du 
x0
x0
3
3
3 u
3
(ln 129)/3 – (ln 4)/3 = 1.1578393
11
Examples 7 REVISITED
5
5. 
0
x
2
x 4
3
dx 
Let u = x3 + 4
du = 3x2 dx
ln u 5
1 51
1 5 3x 2

du


dx

0 3
0
x0
3
3
u
3 x 4
At this point instead of substituting for u we can
replace the x value in terms of u. If x = 0 then u = 4 and
if x = 5, then u = 129.
ln u 129
 (ln 129)/3 – (ln 4)/3 = 1.1578393
u4
3
12
Numerical Integration on a Graphing Calculator
Use some of the examples from above.
3.  2
1
5
5. 0
1
dx
x
x
0x3
-1y3
2
x 4
3
dx
-1  x  6
- 0.2  y  0.5
13
Example 8
From past records a management services determined
that the rate of increase in maintenance cost for an
apartment building (in dollars per year) is given by
M ’ (x) = 90x 2 + 5,000 where 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 seventh year to
the end of the seventh year. Evaluate the integral.

7
3
90 x  5,000 dx  30 x 3 + 5,000x
2
|
7
x3
= 10,290 + 35,000 – 810 – 15,000
= $29,480
14
Using Definite Integrals for
Average Values
Average Value of a Continuous Function f over
[a, b].
1
b
 a f ( x) dx
ba
Note this is the area under the curve divided by
the width. Hence, the result is the average height
or average value.
15
Example
§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
C ( x)
Note that the average cost is C ( x) 
.
x
20000
C ( x) 
 10
x
20000
C (1000 ) 
 10  30
1000
Continued on next slide.
16
Example - continued
The 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]
1
1
b
f ( x) dx 

a
ba
1000

1
( 20000x  5x 2 )
1000
1000
1000
0
x0
(20000  10 x) dx

20,000 + 5,000 = 25,000
Continued on next slide.
17
Example - continued
The 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 us $30. The average total cost of making
between 0 and 1,000 dictionaries is $25,000.
18
Summary.
We can evaluate a definite integral by the
fundamental theorem of calculus:
 f (x) dx  F(x)  F(b)  F(a)
b
a
b
a
We can find the average value of a function f
by:
1
b
f ( x) dx

a
ba
19
Practice Problems
§6.5; 1, 3, 5, 9, 13,
17, 21, 23, 27, 31,
35, 39, 41, 45, 49,
53, 55, 61, 65, 69,
73, 81, 85.
20
§13.3 Integration by Parts
The student will learn to
integrate by parts.
21
Integration by Parts
Integration by parts is based on the product formula
for derivatives: d  f ( x) g ( x)   f ( x) g' ( x)  g ( x) f ' ( x)
dx
We rearrange the above equation to get:
d
 f ( x) g ( x)   g ( x) f ' ( x)
f ( x) g' ( x) 
dx
And integrating both sides yields:
d
 f ( x ) g' ( x )    f ( x) g ( x )    g ( x ) f ' ( x )
dx
Which reduces to:
 f (x) g' (x)  f (x) g (x)   g (x) f ' (x)
22
Integration by Parts Formula
If we let u = f (x) and v = g (x) in the preceding
formula the equation transforms into a more
convenient form:
Integration by Parts Formula
 u dv  u v   v du
The goal is to change the integral from  u dv to
 v du and have the second integral be easier than
the first.
23
Integration by Parts Formula
 u dv  u v   v du
This method is useful when the integral on the left
side is difficult and changing it into the integral on
the right side makes it easier.
Remember, we can easily check our results by
differentiating our answers to get the original
integrals.
Let’s look at an example.
24
Example
 u dv  u v   v du
Consider  x e dx
x
Our previous method of substitution does not work.
Examining the left side of the integration by parts
formula yields two possibilities.
Option 1
u
 x
dv
x
e dx
Option 2
u
dv
x
e
x dx

Let’s try option 1.
25
Example - continued
 u dv  u v   v du
x
x
e
dx

We have decided to let u = x and dv = e x dx, (Note: du
= dx and v = e x), yielding
 x e dx  x e   e dx
x
x
x
Which is easy to integrate
= x ex – ex + C
As mentioned this is easy to check by differentiating.
d
x e x – e x  C  x e x  e x  e x  x e x
dx
26
Selecting u and dv
 u dv  u v   v du
1. The product u dv must equal the original integrand.
2. It must be possible to integrate dv by one of our
known methods.
3. The new integral  v du should not be more
involved than the original integral  u dv .
4. For integrals involving x p e ax , try
u = x p and dv = e ax dx
5. For integrals involving x p ln x q , try
u = (ln x) q
and
dv = x p dx
27
Example 2
 u dv  u v   v du
3
x
 ln x dx
Let u = ln x
du = 1/x dx
and
dv = x 3 dx,
and
v = x 4/4, and the integral is
yielding
4
4
x
x
1
3
dx
 x ln x dx  ln x  
4
4 x
Which when integrated is:
x4
x4

ln x 
C
4
16
Check by differentiating.
d x
x
4x
 x 1
3
 C 
 x ln x 
 x 3 ln x
 ln x –
dx  4
16
16
 4 x
4
4
4
3
28
Summary.
We have developed integration by parts as a
useful technique for integrating more
complicated functions.
Integration by Parts Formula
 u dv  u v   v du
29
Practice Problems
§7.3; 1, 3, 5, 7, 11, 15,
19, 21, 23, 25, 29, 33,
37, 41, 43, 47, 49, 57,
63, 65.
30