5.7 Antiderivatives
We know how to solve the derivative problem: given a function, find its derivative. But many problems
in mathematics and its applications require us to sole the inverse of the derivative problem: given a
function f , find a function F whose derivative is f . If such a function F exists, it is called an
antiderivative of f .
Definition. A function F is called an antiderivative of f on an interval I if F '( x)  f ( x) for all x in
I .
For example, x 2 is an antiderivative of 2x . We discover that x 2  1 is an antiderivative of 2x , so the
antiderivative of a function is not unique. In fact, x 2  c ( c is a constant) is an antiderivative of 2x too.
In general, we have the following.
Theorem. If F is an antiderivative of f on an interval I , then the most general antiderivative of f
on I is
F ( x)  c
where c is a constant.
Properties: Given F ( x ) is an antiderivative of f ( x ) , and G( x ) is an antiderivative of g ( x ) , then
F ( x)  G( x) is an antiderivative of f ( x)  g ( x) , and cF ( x ) is an antiderivative of cf ( x ) .
Table of the antiderivatives.
function
x
n
(n  1)
1
x
ex
sin x
cos x
sec2 x
csc2 x
sec x tan x
csc x cot x
1
1  x2
1
1  x2
antiderivative
x n 1
c
n 1
ln x  c
ex  c
 cos x  c
sin x  c
tan x  c
 cot x  c
sec x  c
 csc x  c
sin 1 x  c (  cos1 x  c )
tan 1 x  c
( cot 1 x  c)
Example 1. Find all functions f such that f '( x)  sin x  2 x 2  x 4 x 3  4(1  x 2 ) 1
1
3


'
Solution. Notice that (  cos x ) '  sin x,  x 3   x 2 , (tan 1 x ) ' 
1
,
1  x2
'
x x x
4
3
13/4
x
7/4


 1 1 74 
x   x 7/4

7
1

4


'
 4 11 
or  x 4   x 7/4
 11 
2 3 4 114
4
Thus, we have f ( x )   cos x  x  x 
c
3
11
1  x2
Example 2. Find the general antiderivative r(t )  2 i  3t j
Solution. The general antiderivative is
3
3
2t i  t 2 j  c  2t i  t 2 j  c1 i  c2 j
2
2
Example 3. Find f such that f ''( x )  x 2 , f (1)  1 , f '(1)  0
Solution. f '( x ) 
f ( x) 
1 3
x c
3
1 4
x  cx  c1
12
Since f (1)  1 and f '(1)  0 , we have
1
1
f '(1)  0  (1)3  c  0  c  
3
3
1
1
5
f (1)  1  (1)4  c(1)  c1  1   c  c1  1  c1 
12
12
4
Thus, f ( x ) 
1 4 1
5
x  x
12
3
4
Example 3. Given that the graph of f passes through the point (1, 2) and that the slope of its tangent
line at ( x, f ( x )) is x 2  1. Find f ( x ) and f (0)
Solution. f '( x )  x 2  1
f ( x) 
1 3
x xc
3
The graph passes through (1, 2) which means f (1)  2
1 3
(1)  1  c  2
3
8
c
3
1
8
f ( x)  x3  x 
3
3
8
f (0) 
3
Example 4. A particle moves in a straight line and has acceleration given by a(t )  4t  1 . Its initial
velocity is v(0  2 cm/s and its initial displacement is s(0  5. Find its position function s(t ).
Solution. v '(t )  a(t )
v '(t )  4t  1
v ( t )  2t 2  t  c
v (0)  2
2(0) 2  0  c  2
c2
v ( t )  2t 2  t  2
s '(t )  v (t )
s '(t )  2t 2  t  2
2 3 1 2
t  t  2t  c1
3
2
s(0)  5
s(t ) 
c1  5
s(t ) 
2 3 1 2
t  t  2t  5
3
2