the fundamental theorem of calculus 4.4 - d-1

7/22/2019 The Fundamental Theorem of Calculus 4.4 - D-1

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Second Fundamental Theorem of Calculus

If f is continuous on [ ],a b , and ( ) ( )x

a

F x f t dt = , then

( ) ( )'F x f x= on [ ],a b .

Example 1: Evaluate 5

1

12x

d

dt dt

x.

Even though the upper limit of integration is a variable, we can use The Fundamental

Theorem of Calculus to evaluate this, since ( ) 512f t t= is a continuous function on any

interval. We have

65 6

11

12 12 2 26

|x x

tt dt x= = .

( )6 52 2 12d

dx x

x =

( ) 5 5

1

' 12 12x

F x t dt d

xx

d= =

The Second Fundamental Theorem of Calculus lets us go directly from

5 5

1

12 to 12x

xt dtd x

d


2/7

Example 2: If ( ) ( )2

2

cosx

F x t dt = , find ( )'F x .

Solution: In order to use the Second Fundamental Theorem of Calculus one of the

limits of integration has to be a single variable, not the variable squared.

Therefore, we are going to manipulate the problem to a form that lets us

use the theorem.

Let ( ) 2u x x= , and rewrite ( ) ( )2

2

cosx

F x t dt = as ( ) ( )( )

2

cos

u x

F x t dt = .

From the Chain Rule, ( ) ( )( ) ( )2' cos cos 2du

F x u x x xdx

= = .

You could also work the problem as follows:

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )( ) ( )

2 2

2

22

2 2

cos sin sin sin 2

' sin sin 2 2 cos

|x x

d

F x t dt t x

dF x x x

xx

= = =

= =


3/7

Example 3: If ( ) ( )4

2

5

3 2= + +x

F x x x dt , compute ( )'F x .

Solution: In order to use the Second Fundamental Theorem of Calculus one of the

limits of integration has to be a single variable. Therefore, we are going to

manipulate the problem in to a form that lets us use the theorem.

Let ( ) 4=u x x , and rewrite ( ) ( )4

2

5

3 2= + +x

F x x x dt as ( ) ( )( )

2

5

3 2u x

F x x x dt = + + .

From the Chain Rule,

( ) ( )( ) ( )( )( ) ( ) ( )( ) ( )

( ) ( )

22 4 4 3

8 4 3 11 7 3

' 3 2 3 2 4

3 2 4 4 12 8

= = + + = + +

= + + = + +

dF duF x u x u x x x

dux

du dx

x x x x

dx

x x

.

You could also work it as follows:

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

4 4

2 3 2

55

3 2 3 24 4 4

12 8 4

1 33 2 2

3 2

1 3 1 32 5 5 2 5

3 2 3 2

1 3 5352

3 2 6

| = + + = + +

= + + + +

= + +

x x

F x x x dt x x x

x x x

x x x

Now take the derivative with respect to x.

( ) 12 8 4 11 7 31 3 535

' 2 4 12 83 2 6

= + + = + +

dF x x x x x x x

dx


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Example 4: If ( )2

4

2

x

x

F x t dt = , compute ( )'F x .

Solution: In this case, both the upper and lower limit of integration are variables,

and we cannot apply The Second Fundamental Theorem of Calculus.

Therefore, we do what we usually do, rewrite the problem in an equivalent

form that will allow us to use the theorem.

( )2 2 20 2

4 4 4 4 4

2 2 0 0 0

x x x x

x x

F x t dt t dt t dt t dt t dt = = + = +

Lets now work this as two separate problems where ( )2

4

0

x

G x t dt = and ( )2

4

0

x

H x t dt= .

Let ( ) 2u x x= and rewrite ( )2

4

0

x

G x t dt = as ( )( )

4

0

u x

G x t dt = .

( )( )

( )( ) ( )4 44 4

0

' 2 2 32

u xd du du

G x t dt u x x xdu dx dx

= = = =

Let ( ) 2v x x= and rewrite ( )2

4

0

x

H x t dt= as ( )( )

4

0

v x

H x t dt= .

( )( )

( )( ) ( )444 2 9

0

' 2 2

v xd dv dv

H x t dt v x x x xdv dx dx

= = = =

( ) ( ) ( ) 4 9' ' ' 32 2= + = +F x G x H x x x .


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You could also work it this way:

( ) ( ) ( )

( )

2 2

5 54 5 2 10 5

22

10 5 9 4

1 1 1 1 322

5 5 5 5 5

1 32' 2 32

5 5

|= = = =

= =

x x

xx

F x t dt t x x x x

d

dx

xF x x x x

You might say, I like the second approach because it is less complicated. Sometimes the

second way is not possible. See the next example.

Example 5: For the function ( ) ( )2

3

4

ln 4x

F x t dt = + , find the equation of the tangent

line at 2x= .

Solution: The slope, ( )' 2F , can be calculated using The Second Fundamental

Theorem of Calculus.

( ) ( )( ) ( )

( ) ( ) ( )( ) ( )

32 6

6

' ln 4 2 2 ln 4

' 2 2 2 ln 2 4 4ln 68

F x x x x x

F

= + = +

= + =

The point of tangency is ( )( )2, 2F where ( ) ( )4

3

4

2 ln 4 0F t dt = + = since the upper and

lower limits of integration are the same.

The equation is ( )4ln 68 2y x= .


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Here is another example where the easiest way to solve it is to use The Second

Fundamental Theorem of Calculus.

Example 6: Compute ( )'F x if ( ) ( )3

3

2

cosx

F x t dt = .

Solution: Let ( ) 3u x x= and rewrite ( ) ( )3

3

2

cosx

F x t dt = as ( ) ( )( )

3

2

cos

u x

F x t dt = .

( ) ( )( )

( )( )( ) ( )( ) ( 333 3 2 2 9

2

' cos cos cos 3 3 cos

= = = = =

u x

dF du du duF x t dt u x x x

dx x

du dx d xdx x d

Proof of The Second Fundamental Theorem of Calculus:

If f is continuous on [ ],a b , and ( ) ( )x

a

F x f t dt = , then

( ) ( )'F x f x= on [ ],a b .

Using the definition of derivative, we have

( ) ( ) ( )

( ) ( )

( ) ( ) ( )

0 0

0 0

1' lim lim

1 1lim lim

+

+ +

+ = =

= + =

x h x

h h

a a

x h a x h

h ha x x

F x h F xF x f t dt f t dt

h h

f t dt f t dt f t dth h


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Look very carefully at the term ( )1

x h

x

f t dth

+

. It is the average value of the function

( )f t on the interval [ ],x x h+ (if 0h> ).

By the Mean Value Theorem, ( ) ( )1

x h

x

f t dt f ch

+

= for some c between andx x h+ .

Since c is between andx x h+ , we have c x as 0h .

Since f is continuous, it follows that ( ) ( ) ( ) ( )0 0

1' lim limx h

h hx

F x f t dt f c f xh

+

= = = .

the fundamental theorem of calculus 4.4 - d-1

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