Section 4.1 – Antiderivatives
and Indefinite Integration
Reversing Differentiation
We have seen how to use derivatives to solve various
contextual problems. For instance, if the position of a
particle is known, then both the velocity and acceleration
can be calculated by taking a derivative:
s  t   3t 2  5t  7
s '  t   v  t   6t  5
v ' t   a t   6
Position Function
The derivative of the
Position Function is the
Velocity Function
The derivative of the
Velocity Function is the
Acceleration Function
But what if ONLY the acceleration of a particle is known? It
would be useful to determine its velocity or its position at
a particular time. For this case, a derivative is given and
the problem is that of finding the corresponding function.
a  t   32
Acceleration
Function
v ' t   a t 
s ''  t   a  t 
What function has a
derivative of 32?
What function has a
second derivative of 32?
Antiderivative
A function F is called an antiderivative of a
given function f on an interval I if:
F ' x  f  x
for all x in I.
Example:
F  x   1x is an antiderivative of f  x    x12
because F '  x    x12
The Uniqueness of Antiderivatives
Suppose f  x   3x 2 , find an antiderivative of f. That is,
2
find a function F(x) such that F '  x   3x .
F  x  x
3
Using the Power Rule in
Reverse
Is this the only function whose derivative is 3x2?
H  x  x  5
K  x   x  11
M  x  x  
H '  x   3x 2
K '  x   3x 2
M '  x   3x 2
3
3
3
There are infinite functions whose derivative is 3x2
whose general form is:
C is a constant real
G  x  x  C
3
number (parameter)
Antiderivatives of the Same Function Differ
by a Constant
If F is an antiderivative of the continuous
function f, then any other antiderivative, G,
of f must have the form:
G  x  F  x  C
In other words, two antiderivatives of the
same function differ by a constant.
Differential Equation
A differential equation is any equation that
contains derivatives. If a question asks you to
“solve a differential equation,” you need to find
the original equation (most answers will be in
the form y=).
Example:
The following is a differential
equation because it contains the
derivative of G:
The general solution
to the differential
equation is:
G ' x  2x
G  x   x2  C
Example 1
Find the general antiderivative for the given function.
f  x   x5
Using the opposite of
the Power Rule, a first
guess might be:
If:
F  x   16 x6
General
Solution:
F  x   x6
Then:
Divide this result by
6 to get x5
But:
F '  x   6 x5
F '  x   16  6 x5  x 5
F  x   16 x6  C
Example 2
Find the general antiderivative for the given function.
f  x   sin x
Using the opposite of the
Trigonometric Derivatives, a
first guess might be:
If:
F  x   cos x
F  x    cos x
General
Solution:
Then:
Multiply this result
by -1 to get sinx
But:
F '  x    sin x
F '  x     sin x  sin x
F  x    cos x  C
Example 3
Find the general antiderivative for the given function.
f  x   5 x3
Using the opposite of
the Power Rule, a first
guess might be:
If:
F  x   5x4
F  x   54 x 4
General
Solution:
Then:
Divide this result by
4 to get 5x3
But:
F '  x   20 x3
F '  x   54  4 x3  5x3
F  x   54 x4  C
Example 4
Find the general antiderivative for the given function.
f  x 
1
x

1
x1 2
x
1 2
Rewrite if necessary
Using the opposite of
the Power Rule, a first
guess might be:
If:
F  x   x1 2
F  x   2 x1 2
General
Solution:
Then:
But:
Multiply this result
by 2 to get x-1/2
F '  x   12 x1 2
F '  x   2  12 x1 2  x 1 2
F  x  2 x  C
Example 5
Find the general antiderivative for the given function.
f  x  

2
3
cos 2 x

9
cos 2 2 x
 9sec 2x
2
Multiply this result by
1/2 to get 9sec22x
Rewrite if necessary
Using the opposite
trigonometry F
derivatives:
If:
F  x   92 tan 2 x
 x   9 tan 2 x
Then:
General
Solution:
But:
F '  x   18sec 2 2 x
F '  x   92  2  sec 2 2 x  9sec 2 2x
F  x   92 tan 2 x  C
Antiderivative Notation
The notation
Integral

Indefinite
Integral
Constant of
Integration
f  x  dx  F  x   C
Variable of Integration
Means that F is an antiderivative of f. It is called
the indefinite integral of f and satisfies the
condition that F '  x   f  x  for all x in the
domain of f.
New Notation with old Examples
Find each of the following indefinite integrals.
a.
 x dx
b.
 sin x dx
c.
3
5 4
5
x
dx

4 x C

d.

5
1
x
 x C
1
6
6
  cos x  C
dx  2 x  C
Basic Integration Rules
Let f and g be functions and x a variable; a, b, and c be constant; and C is an
arbitrary constant.
Constant Multiple
 cf  x  dx  c  f  x  dx
Sum Rule
  f  x   g  x  dx   f  x  dx   g  x  dx
Difference Rule
  f  x   g  x  dx   f  x  dx   g  x  dx
Constant Rule
(zero)
 0 dx  C
Basic Integration Rules
Let f and g be functions and x a variable; a, b, and c be constant; and C is an
arbitrary constant.
Constant Rule
(non-zero)
Power Rule
Trigonometric Rule
Trigonometric Rule
 a dx  ax  C
n
x
 dx 
1
n 1
x n 1  C
 sin x dx   cos x  C
 cos x dx  sin x  C
Basic Integration Rules
Let f and g be functions and x a variable; a, b, and c be constant; and C is an
arbitrary constant.
Trigonometric Rule
Trigonometric Rule
Trigonometric Rule
Trigonometric Rule
 sec
2
x dx  tan x  C
 sec x tan x dx  sec x  C
 csc x cot x dx   csc x  C
 csc x dx   cot x  C
2
Example 1
Evaluate   x  3 x  7  dx
5
2
  x5 dx   3x 2 dx   7 dx
Sum and Difference
Rules
  x5 dx  3 x 2 dx   7 dx
Constant Multiple

1
5 1
x
5 1
 3
1
2 1
x
2 1
 16 x 6  x 3  7 x  C
 7x  C
Power and Constant
Rules
Simplify
Example 2
Evaluate
 5

x  4sin x dx
   5 x1 2  4sin x  dx
Rewrite
  5 x1 2 dx   4sin x dx
Sum Rule
 5 x1 2 dx  4  sin x dx
Constant Multiple Rule
 5
1
1 21
1 2 1
x
 4    cos x   C
 103 x 3 2  4 cos x  C
Power and Trig
Rules
Simplify
Example 3
The graph of a certain function F has slope 4 x  5 at
each point (x,y) and contains the point (1,2). Find the
function F.
3
Integrate:
3
3
4
x

5
dx

4
x
  dx   5 dx

 4  x 3 dx   5 dx
 4  311 x31  5 x  C
 x4  5x  C
Use the Initial
Condition to
find C:
2  1  5 1  C
Difference
Rule
Constant Multiple Rule
Power and Constant Rules
Simplify
4
2  4  C
6C
F  x   x4  5x  6
Example 4
A particle moves along a coordinate axis in such a way that
3
its acceleration is modeled by a  t   2t for time t>0. If
the particle is at s=5 when t=1 and has velocity v=-2 at
this time, where is it when t=4?
Integrate the acceleration to find velocity:
v  t     2t 3  dt  2   t 3  dt  2  311 t 31  C  t 2  C
Use the Initial Condition to find C for velocity:
2   1  C
2
C  1
v  t   t 2  1
Integrate the Velocity to find position:
s  t     t 2  1 dt    t 2 dt   1 dt   211 t 21  1t  C  t 1  t  C
Use the Initial Condition to find C for position:
5  1  1  C
1
Answer the
Question:
C 5
1
s  4    4   4  5  1.25
s  t   t 1  t  5