BC 2
Euler Encore 3
Name
On the previous sheet, we found that if A(x) is the antiderivative of a function  such that

A(a) = 0, then A(b) =
b
a
  x  dx . This result leads us to:
A Fundamental Theorem of Calculus...
Let  be continuous on [a, b] and let F be any antiderivative of ƒ. Then
b
 x dx  F b  F a .
a

(1)
 
 
Suppose (x) = 3x2 + 2 and you wish to find

5
2
F(x)
  x  dx 
 
 3x  2 dx . Then
5
2
2
+ 2x + C (C an arbitrary constant) is the general antiderivative of . This gives
= x3
F  5  F  2  53  2  5  C   23  2  2  C   125  10  8  4  123

 

Note that the C's cancelled. For this reason, we usually can skip them entirely.
(2)

Evaluate
3
1
 5 x  1 dx .
In doing this, we introduce new notation:
 5x2

5
x

1
dx

 x



 2

1



(3)
3
3
 5  3 2
  5  12


 3  
  1   16
 2
  2

1 
Evaluate each of the following.
1 
2 
(a)
 3  2  dx
3 
x 

(b)


0
(c)
(d)
3

1
1
1 x
e2
1
dx
x

2
IMSA
sin  x  dx
2
dx
EE 3 p.1
Fall05
The (First) Fundamental Theorem of Calculus
Let ƒ be a continuous function defined in an open interval I containing a.
Then F  x  

x
a
  t  dt is a differentiable function for all x in I and

d  x
  t  dt     x  .

dx  a


(4)
 t
x
Find F  x  
4

 2t  3 dt .
3
Then differentiate to find
(5)
Then find
(6)
Find
d 

dx 
(7)
Find
d 

dx 
IMSA

Find F  x  
x
2
d 

dx 

dF d 
 
dx dx 
 t
x
4
3


 2t  3 dt  .

 sin  2t   cos  3t   dt .

x
 sin  2t   cos  3t   dt  .
2

 

sin t 2 dt  .
17


x

x
3
tan 1
 t  dt 
EE 3 p.2
Fall05
(8)
Find F  x  

x3
sec2  t  dt .
1
d 
Then find

dx 

x3
1

2
sec  t  dt  .


Was your answer to #8 surprising? At first glance, it does not seem to follow the First Fundamental
Theorem of Calculus. That is because the upper limit of the integral was x 3 and not x, which means
that little thing called the Chain Rule must be taken into consideration. So now the question is, how do
x
ex

d  e
10
we handle problems like
sin t dt  when we are unable to find the integral
sin t 10 dt ?



dx  1
1

The TI-89 can help us with the answer to this question.

(9)

 
Use the TI-89 to find each of the following:
x

d  e

a.
sin t10 dt  (Note the syntax is d(∫(sin(t^10),t,1,e^x),x)

dx  1



IMSA
 
 
b.
2
d  x cos t 

e
dt 

dx  


c.

d  ln x
1

dt 

dx  0
t5  t 2 1 
d
5

d  x

tan 1  ln t  dt 

dx  1





EE 3 p.3
Fall05
(10) Now it’s your turn. Put the TI-89 away and try these for yourself.
x

d  e

a.
cos t 8 dt 

dx  1


 

b.
5

d  2 x 12

ln  t  dt 

dx  e


c.
1

d  tan x
1

dt 
3 4
dx  2
t  t 3  2 



(11) Based on the results to #8, #9, and #10, complete the following statement (which is a generalized
form of the First Fundamental Theorem of Calculus).
d 

dx 

g x
a

  t  dt  

How does the Chain Rule come into play to produce this result?
Evaluate
IMSA
d 

dx 

g x
a

  t  dt  on the TI-89 to see if you get the same result.

EE 3 p.4
Fall05