Indefinite Integrals
Chapter intro text here…
8.1 Antiderivatives
In the previous chapters we learned how to find marginal cost given a cost function, marginal
revenue given a revenue function, and marginal profit given a profit function. Is it possible to do
the reverse process? Can we find a profit function p( x) if we are given the marginal profit
function, p '( x) ? Lets take a simple function like f '( x)  2 x . We know from experience that
f '( x)  2 x is the derivative of the function f ( x)  x2 , therefore we call f ( x)  x2 the
antiderivative of f '( x)  2 x
Example 8.1 Given
g ( x)  4x3  3x2  4x  1,
f ( x)  x4  x3  2x2  x  2
Solution 
show
that
its
antiderivative
is
To show f ( x ) is the antiderivative of g ( x) we need to take the derivative of
f ( x)  x  x  2x2  x  2 .
4
3
d
d
f ( x)   x 4  x 3  2 x 2  x  2   4 x 3  3x 2  4 x  1  g ( x ) ,
dx
dx
Since
d
f ( x)  g ( x) , f ( x ) is the antiderivative of g ( x) .
dx

Recall that the derivative of a function tells us about the slope of a line tangent to the graph of the
function at some value x = a.
The graphs of a few functions of the form
4
3
2
f ( x)  x  x  2x  x  C are shown in figure 8.1 below.
Figure 8.1 Graphs of f ( x)  x4  x3  2x2  x  1 and g ( x)  x4  x3  2x2  x  3
Note to authors: This is saved
as chpt8_fig1.eps also.
Since the shape of each curve is the same and the curves are just translations of one another, we
would predict that the slope of the tangent lines at all values of x = a must be the same for f ( x )
and g ( x ) . Figure 8.2 below has the tangent line drawn for both curves at x = –1.
Figure
8.2
Graphs of
4
3
2
g ( x)  x  x  2x  x  3 at x = –1.
tangent
lines
to
f ( x)  x4  x3  2x2  x  1
and
Since the two tangent lines at x = –1 are parallel, f '(1)  g '(1) . Thus, it seems safe to
conclude that f '(a)  g '(a) for all values of x = a. We can also show this algebraically and see
that f '( x)  4x3  3x2  4x  1 and g '( x)  4x3  3x2  4x  1 . So, which function, f ( x ) or
g ( x ) is the antiderivative of 4 x3  3x 2  4 x  1 ? The answer is that both f ( x ) and g ( x ) are
antiderivatives of 4 x3  3x 2  4 x  1 .
As a matter of fact, any function of the
form f ( x)  x4  x3  2x2  x  C , where C is some constant, is an antiderivative of
Thus, the most general antiderivative of h( x)  4 x3  3x2  4 x  1
4 x3  3x 2  4 x  1 .
is H ( x)  x4  x3  2 x2  x  C .
To actually find an antiderivative we need to use the following rule.
Power Rule for Antiderivatives
If n  1 then the most general antiderivative of x n is
1 n1
x  C , where C is an arbitrary constant.
n 1
Example 8.2
Find the most general antiderivative of f '( x)  x2 .
Solution 
Using the power rule for antiderivatives we obtain
f ( x) 
1 21 1 3
x  x C .
2 1
3

This process of antidifferentiation is also know as integration. Integration uses the symbol

(called the integral sign) to denote that the antiderivative of a function is to be taken. When the
integral sign is used we always find the most general antiderivative of the function that
immediately follows the integral sign. The arbitrary constant C that comes from finding the most
general antiderivative is called the constant of integration.
x
3
Example 8.3
Find
Solution 
We need to find the most general antiderivative of x using the power rule.
2
dx
3
2
x
3
2
1
3
x ( 2 1)  C
( 2  1)
1
3
2

x( 2  2 )  C
3
2
( 2  2)
1 5
 x 2 C
dx 
3
5
2
2 5
 x 2 C
5
Notice that the integral sign and the dx are no longer written after the power rule is applied. This
is because the operation of integration has been performed. You can think of this in the same
manner you do the plus sign when you figure 2 + 2 = 4. What happened to the plus sign on the
right side of the equation? It was no longer written because you performed the operation of
addition.

The following rules will help us find antiderivatives of more complicated functions.
Let k and C be constants, then
1)  k dx  kx  C
Example 8.4
2)
 k f ( x) dx  k  f (x) dx
3)
 [ f ( x)  g ( x)] dx   f ( x) dx   g ( x) dx
Find the indefinite integral.
a)
 1
2x 2 
b)   2 
 dx
3 
 3x
3
5
  2 x  7  dx
Solution 
a) Using the rules from above and the power rule we get
 2x
5
 7  dx   2 x 5 dx   7 dx
 2 x5 dx   7 dx
 1 51 
 2
x   7x  C
 5 1

1
 x6  7 x  C
3
Differentiating the answer will demonstrate that the answer checks.
b)
 1
2x 2 
1
2x 2

dx

dx



  3x 2 3   3x 2  3 dx
1
2 3
  x 2 dx   x 2 dx
3
3

1 1
3
 2 1
 
x 21   
x ( 2 1)   C
3  2  1
 3   3 2  1

1
2
2
5


   x 1    x 2   C
3
3 5 
3
3
5
1 4x 2
 
C
3x 15

The last two antiderivatives we need to investigate at are those that involve the natural logarithm
and the exponential function with base e. Recall that
d
1
d x
ln x  and
e  ex .
dx
x
dx
This leads us to the following rules.
1)
2)  e x dx  e x  C
1
1
 x dx   x dx  ln | x | C
Example 8.5
Find the indefinite integral.
3
  x  2e
x

 5  dx

Solution 
3
  x  2e
x
1

 5  dx  3 dx  2 e x dx  5 dx
x

 3ln | x |  2 e x  5 x  C

Example 8.6
Find f(x) given f '( x)  4 x  3  e x if f (0)  9 .
Solution 
First find
  4x  3
dx 
4 2
x  3x  e x  C  2 x 2  3x  e x  C
2
Since f (0)  9 , we get
9  2(0)2  3(0)  e0  C
,
9  1  C
10  C
so f ( x)  2 x2  3x  e x  10 .

8.2 Integration using substitution
For differentiation we had special rules like chain rule, product rule, and quotient rule. Since
integration is the inverse process of differentiation, it seems reasonable to conclude that special
rules exist for integration. One special rule we will study for integration is called substitution,
commonly called u-substitution. Integration by substitution allows us to recover functions that
required the chain rule to take its derivative.
For example, to find the derivative of f ( x)  ( x 2  1)3 , we use the chain rule and set u  x 2  1 ,
so u '  2x . Thus, when given f ( x)  ( x 2  1)3  u 3
f '( x)  3u 2  u '
 3( x 2  1) 2  (2 x)
 6 x( x 2  1) 2
Thus,

d 2
( x  1)3  6 x( x 2  1) 2 or equivocally 6 x( x 2  1) 2 dx  ( x 2  1)3  C .
dx
But, how can we recover f ( x ) given f '( x)  6 x( x 2  1)2 ? We must use substitution. We begin
by letting u  x 2  1 and then take the derivative of both sides with respect to x
du
 2x
dx
and solve the above equation for dx
du
 dx .
2x
Now substitute u for x2  1 and
du
for dx then simplify.
2x

6 x( x 2  1) 2 dx 

6x u2
du
2x
  3u 2 du
 3 u 2 du
We can use the general power rule for antiderivatives on the last integral
1 
3  u3   C  u3  C
3 
Finally we must substitute x 2  1 for u and we have recovered f ( x ) from above.
( x 2  1) 3  C .
It can, however, be difficult to determine what expression to make u. The table below shows
different types of problems that involve substitution. In each case let u = f(x).

i)  f '( x)  ( f ( x)) n  dx
Example 8.7
a)

ii)
  f '( x)  e
f ( x)
 dx
iii)
  f '( x) 
n
f ( x)  dx
iv)

f '( x )
dx
f ( x)
Find
x 2 x 3  1 dx
b)

4 xe 2 x
2
1
dx
c)
 3x 2  2 
  x3  2 x  dx
Solution 
a) This is a type iii in the table above. Since f ( x)  x3  1 we let u  x3  1 . Now differentiate
this equation with respect to x and then solve for dx.
du
 3x 2
dx
 du  3x 2 dx 
du
 dx
3x 2
Now substitute u for x3  1 and
du
for dx.
3x 2
x
2
x3  1 dx   x 2 u
du
3x 2
After simplifying the integral will be completely in terms of u.

1
u du
3
1
1
 u  2 du

3
1  2 32 
  u C
3 3 
2 3
 u 2 C
9

Now we must substitute x3  1 for u to obtain the final answer.
3
2 3
2
x  1  C

9
b) This is a type ii in the table above. Since f ( x)  2 x2  1 we let u  2 x 2  1 and then
differentiate both sides of the equation with respect to x
du
 4 x  du  4 x dx 
dx
Now substitute u for 2 x2  1 ,
 4 xe
du
 dx .
4x
du
for dx, and the new limits of integration.
4x
2 x 2 1
dx   4 x e u
2
du
  e u du  e u  C  e 2 x 1  C
4x
c) This is a type iv in the table above, so let u  x 3  2 x , then


du
 3x 2  2  du  3x 2  2 dx 
dx
du
 dx
3x 2  2
Making all substitutions we get the following integral.
3x 2  2
du
du
3
 u  3x 2  2   u  ln u  C  ln x  2 x  C

8.3 Applications of Antiderivatives
Example 8.8 The rate of change in sales of bicycles at Ted’s bicycle shop for the year 2001 is
given by s( x)   x  7.5 , where x represents the month number in 2001 (i.e. x = 1 is January, x =
2 is February, … x = 12 is December). If the shop sold 28 bicycles in the month of February, find
the shop’s total sales function, TS ( x) .
Solution 
Since we are given the rate of change of sales, we can find the total sales
function, TS ( x) , by finding  s( x) dx .
TS ( x)   s ( x) dx   (  x  7.5) dx
   x dx  7.5 dx
1
  x 2  7.5 x  C
2
We also know that TS (2)  28 because the shop sold a total of 28 bicycles in the month of
February. Thus,
1
28   (2) 2  7.5(2)  C
2
28  13  C
15  C
1
So the sales function is TS ( x)   x 2  7.5 x  15 .
2

A
company’s
Profit
function,
in
dollars,
is
given
by
P( x)  0.3x  56.16 x  638 where x is the number of items sold. In addition, the company
has a marginal revenue function of MR( x)  0.6 x  76.16 . Determine the total cost for the
company to produce 10 items.
Example 8.9
2
Solution  Recall that Profit = Revenue – Cost, so P ( x)  R ( x)  C ( x) . Solving this
equation for C ( x ) we get C ( x)  R ( x)  P ( x) . Thus we need a revenue function and profit
function to obtain the cost function. We are given P( x)  0.3x 2  56.16 x  638 and we can
find the revenue function by integrating MR( x)  0.6 x  76.16 .

 0.6 x  76.16 dx

R( x)  MR( x) dx
 0.3x 2  76.16 x  C
Since no revenue is generated when zero items are sold, R (0)  0 . Thus
0  0.3(0) 2  76.16(0)  C
0C
So R( x)  0.3x 2  76.16 x . Now we can find the cost function C ( x ) .
C ( x)  R( x)  P( x)
 0.3x 2  76.16 x   0.3 x 2  56.16 x  638 
 0.3x 2  76.16 x  0.3 x 2  56.16 x  638
 20 x  638
To find the cost to produce 10 items we need to find C (10)  20(10)  638  838 . Thus, it costs
$838 to produce 10 of the company’s items.
Sample Quiz
Question
8.1
f ( x)  5 x 4  x 2  2 x  7
Show
that
5
3
2
1
F ( x)  x  3 x  x  7 x  10 .
Question 8.2 Find the most general antiderivative of g ( x) 
 dx
Question 8.3 Find

Question 8.4 Find
  4t  32  t  dt
Question 8.5 Find
 16 x
Question 8.6 Find
3
x 2  4e x 
1
t

7
x2
2
3

4 x 2  1 dx
e x 

 dx
 x
1
5
is
the
antiderivative
x 4  x3  4 x 2  3 .
of
Question 8.7 Find

 6 x2  x 
 3
 dx
2
 4x  x  2 
Question 8.8 Sonbyrne Sunglass Company found that during the month of March its rate of
change of sales could be modeled by s( x)  0.004 x3  0.213x 2  3x  8 , where x represents
the day of March. If the company knows that 17 pairs of sunglasses were sold on March 30, find
the total number of sunglasses the company sold on March 20.
Question 8.9 Larry's custom made furniture shop sells twin size bed frames for $250 each. If the
shop's marginal cost is represented by mc( x)  535.6
x  2 and he knows that it costs $1476.50 to
produce 1 bed, find the shop's profit, to the nearest dollar, for producing 20 beds. (Assume the
revenue function is linear.)
Question 8.10 Welch Construction has found that the value of the company tractor is decreasing
2
at a rate given by v(t )  390te0.012t , where t represents the number of years after purchase. If
the company purchased the tractor for $17650, what is the value of the tractor 3 years after
purchase?