1
Section 4.4: The Fundamental Theorem of Calculus
Practice HW from Larson Textbook (not to hand in)
p. 257 # 5-35 odd, 43-47 odd, 73-77 odd
Definite Integral
The definite integral is an integral of the form
b
 f ( x) dx .
a
This integral is read as the integral from a to b of f ( x ) dx . The numbers a and b are said
to be the limits of integration. For our problems, a < b.
Definite Integrals are evaluated using The Fundamental Theorem of Calculus.
Fundamental Theorem of Calculus
Let f (x) be a continuous function for a  x  b and F (x) be an antiderivative of f (x) .
Then
b
 f ( x) dx  F ( x) a  F (b)  F (a)
b
a
2
Example 1: Evaluate  2 x dx .
1
Solution:
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2
3
Example 2: Evaluate  (3x 2  x  2) dx .
1
Solution:
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Additional Integration Formulas
1
1.

2
 1  x 2 dx 
1 x2
1
dx 
3
1
2
Example 3: Evaluate
 (e
0
2x

1
1 x
2
) dx .
Solution:
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4
3
Example 4: Evaluate  ( x 3  1) 2 dx .
1
Solution:
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5
e
Example 5: Evaluate

4  u2
1
u3
du .
Solution: We rewrite and evaluate the integral as follows:
e

1
4  u2
u3
 4 u2 
du    3  3  du

u 
1u
e
Rewrite fraction property (a  b) / c  a / c  b / c
e
1

   4u 3   du
u
1
Simplify
u e
 u 2

 4
 ln u 
 2


 u 1
2
 (
u2
2
 (
 (

e2
2
e2
 3
u e
 ln u )
u 1
 ln e )  (
2
(1) 2
 1)  (2  0)
e2
2
Integrate
 ln 1 )
Note ln | e | ln e  1, ln | 1 | ln 1  0
1 2
2
e2
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6

Example 6: Evaluate  (3  sin x) dx .
0
Solution:
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Recall that if f ( x)  0 for a  x  b . Then
b
Definite Integral:
 f ( x) dx 
a
Area Between f ( x) and the
x axis for a  x  b
7
Example 7: Find the area under the graph of y  x 2  1 on [0, 2].
Solution:
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8
2
Example 7: Evaluate
2
x
 e dx .
1
Solution: On this one, there is no anti-derivative formula to integrate the function e  x .
Thus, the Fundamental Theorem of Calculus cannot be applied here. However, this
integral can be approximated using left hand (lower) sums, right hand (upper) sums, or
with the midpoint rule (see Example 4 in the Section 4.2/4.3 notes). It can be shown with
Maple that
2
2
x
 e dx  1.628905524
2
1
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3
Example 8: Evaluate
 2 x  1 dx
0
Solution: Note that 2x  1  0 when x 
1
1
. Thus, when x  , 2x  1  0 and when
2
2
1
, 2x  1  0 . Since the absolute value of a quantity is always positive, we can say by
2
the definition of the absolute value that
x
 (2 x  1) when x  1 / 2
| 2 x  1 | 
 2 x  1 when x  1 / 2
Thus, we can say that
3
1/ 2
3
0
0
1/ 2
1/ 2
3
 2 x  1 dx    (2 x  1) dx   (2 x  1) dx

 (2 x  1) dx   (2 x  1) dx

0
  x2  x

(Distribut e negative in first integral)
1/ 2
x 1 / 2
x 0
 ( x 2  x)
3
x 1 / 2
 ( (1 / 2) 2  1 / 2)  0  ((3) 2  3)  ((1 / 2) 2  1 / 2)
 1 / 4  1 / 2  9  3  1 / 4  1 / 2
13
  6.5
2
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9
Note: For a function f (x) that is both positive and negative over an interval, the total area is
the area enclosed by the positive part of the curve minus the negative part of the curve.
Example 9: Consider the function f ( x)   x 3  4 x 2  3x over the interval 0  x  3 . The
graph of the function over this interval is given by
Find the total area enclosed between the function and the x axis.
Solution:
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10
Average Value of a Function
The average value of a function measures the average value (average y value) for a
function over a closed interval. Its formula is given as follows.
Formula for Average Value of a Function
If a function f is integrable on a closed interval [a, b], then
b
Average Value of f
1

f ( x) dx
on interval [a, b]
b  a a

Example 10: Find the average value of the function f ( x)  cos x on the interval [0, ] .
2
Solution: For this problem, a  0 and b 
Average Value of f
on interval [0,  / 2]

1

2

. Thus, the average value is given by
2
 /2
0
 cos x dx
0
1
x  / 2
 1 sin x x 0

2



2

2

(sin(  / 2)  sin( 0))
(1  0)
2

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11
The Second Fundamental Theorem of Calculus
If f is continuous on an open interval I containing a, then for every x in the interval
x

d 
  f (t ) dt   f ( x)
dx  a

Example 11: Using the Second Fundamental Theorem of Calculus to find F (x) if
x
F ( x)  
t2
2
1 t 1
dt
Solution:
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Recall that the chain rule of differentiation was used to differentiate a composition of two
functions. The formula for the chain rule was given by:
Chain Rule of Differentiation
d
( f ( g ( x ))  f ( g ( x )) g ( x )
dx
Suppose we let F  f ( g ( x )) and let u  g (x ) . Then F  f (u ) and hence
du
dF
 f (u ) and
 g ( x )
du
dx
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For F  f ( g ( x )) and u  g (x ) we can use these ideas to rewrite the chain rule as follows:
F ( x ) 
dF
d
du dF du

( f ( g ( x ))  f ( g ( x )) g ( x )  f (u )

dx dx
dx du dx
This gives another way to write the chain rule, which we summarize as follows:
Chain Rule: Alternative Form
If we want to differentiate the composition F ( x )  f ( g ( x )) , we set u  g (x ) and compute
the following:
F ( x ) 
dF dF du


dx du dx
For example, suppose we want to differentiate
F ( x)  ( x 2  7)10
Then by setting u  x 2  7 , we have
F  u10
Hence,
du
dF
 10u 9 and
 2x .
du
dx
Thus, for F ( x)  ( x 2  7)10
F ( x ) 
dF dF du


 10u 9 ( 2 x )  10( x 2  7) 9 ( 2)  20 x ( x 2  7) 9 .
dx du dx
This way of expressing the chain rule can be useful when using the Second Fundamental
Theorem of Calculus. Suppose
g( x)
F ( x) 
 f (t ) dt
a
Then setting u  g (x ) gives
13
u
F   f (t ) dt
a
Then
u
 du
dF du d 
du

F ( x ) 


f
(
t
)
dt
 f (u ) 
 f ( g ( x ))  g ( x )

 dx
du dx du  a
dx

Example 12: Using the Second Fundamental Theorem of Calculus to find F (x) if
x3
F ( x) 
 sin t
dt
1
Solution:
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Example 13: Using the Second Fundamental Theorem of Calculus to find F (x) if
ln x
F ( x) 
e
t
dt
x
Solution: We start by rewriting the integral as follows:
0
F ( x )   e t dt 
ln x
x
e
t
dt
0
Next, we can use the property of integration where
0
x
x
0
t
t
 e dt    e dt
Thus, the integral becomes
x
F ( x )    e dt 
t
0
ln x
e
t
dt
0
Hence,
x
 d  ln x t 
d 
t
F ( x ) 
 e dt     e dt 
 dx 

dx  0

 0

The first term in the sum we can use the Second Fundamental Theorem directly and write:
x
x

d 
d  t 
t

 e dt  
e dt  e x



dx  0
dx

0

For the second term in the sum
ln x
d 
dx 
e
0
t

du 1
 ). Then
dt  , we write u  ln x (note

dx x

u
t
t
 e dt   e dt and we can write
0
ln x
0
(continued on next page)
15
d 
dx 
Hence,
d 
dx 
u

d  t  du
u 1
ln x 1

 e dt   du   e dt   dx  e x  e x 
0

0

ln x
t
1
x
x

1
Note e ln x  x

 x

t
  1. Using the fact we calculated above that d   e t dt   e x ,
e
dt



dx  0
0


ln x
we have
F ( x ) 
x
 d  ln x

d 
  e t dt     e t dt   e x  1
 dx 

dx  0

 0

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