MATH 136
Introduction to Antiderivatives
Given a function f (x) , we wish to find another function F(x) such that F ′(x) = f (x) .
The new function F(x) is called the antiderivative, or indefinite integral of f (x) . We
denote the operation with the “integral” symbol ∫. In order to express that we want the
antiderivative of f (x) , with respect to the variable x , we write ∫ f (x) dx .
Since the derivative of F(x) + C will still be f (x) for any constant C , we always
include to term + C with a general antiderivative. Here are the basic antiderivatives
that we should commit to memory:
∫ f (x) dx
F(x) + C
∫ a dx
ax + C
∫ a x dx
a x2
+C
2
a x n +1
+C
n+1
n
∫ a x dx
(Power Rule for n ≠ −1)
∫ sin x dx
– cos x + C
∫ cos x dx
sin x + C
∫ sec
2
x dx
∫ sec x tan x dx
tan x + C
sec x + C
1
∫ x dx
∫e
x
dx
ln x + C
x
e +C
∫ f (m x + b) dx
1
F(m x + b) + C
m
∫ f ′(x) dx
f (x) + C
The Power Rule
a x n +1
n
+ C . That is, to integrate x simply raise
n+1
the power by 1 and then divide by the resulting exponent.
The general Power Rule is ∫ a x n dx =
Note: When dividing by a fraction, you may “invert and multiply.”
Example 1. Evaluate the antiderivatives.
3
(i) ∫ 12x dx
(ii) ∫ (8x
4/ 3
 6
12 
(iii) ∫  1/ 3 − 4  dx
x
x 
− 3 x ) dx
Solution. (i) Apply the Power Rule directly:
∫ 12x
3
dx =
12 4
x + C = 3x 4 + C .
4
3
(The derivative of the function F(x) = 3x 4 + C is the original function 12x .)
(ii)
∫ (8x
4/ 3
− 3 x ) dx = 8 ×
3 7/ 3
2
24 7/ 3
x
− 3 × x 3/ 2 + C =
x
− 2x 3/ 2 + C .
7
3
7
4/ 3
For x , we raise the power by 1 to 7/3, and then divide by 7/3 which is the same
1/ 2
as multiplying by 3/7. Likewise, for x
we raise the power to 3/2 and then divide by
3/2, which is the same as multiplying by 2/3.
(iii) We first re-write the exponents, then integrate with the Power Rule:
 6
3 2/ 3
1 −3
12 
−1/ 3
− 12 ×
x + C = 9x 2 / 3 + 4x −3 + C .
− 12 x −4 )dx = 6 × x
1/ 3 − 4  dx = ∫ (6 x
2
−3
x
x
∫ 
(Check that the derivative of the result is equal to the original function.)
Example 2. Evaluate the antiderivatives.
(a)
∫ sin(4x) dx
(b)
∫
10 − 4x dx
(c)
8
∫ (6 + 3x)5 dx
Solution. If we can integrate f (x) to obtain an antiderivative F(x) , then we can replace
x by the linear function mx + b and still integrate. That is, if ∫ f (x) dx = F(x) + C , then
1
∫ f (m x + b) dx = m F(m x + b) + C . We additionally divide by the slope m to “undo” the
Chain Rule.
1
(a) Since ∫ sin x dx = – cos x + C , then ∫ sin(4x) dx = − cos(4x) + C .
4
1
The derivative of the function F(x) = − cos(4x) + C is in fact sin(4x) .
4
(b) Since
∫
x dx =
∫x
1/ 2
dx =
integrate:
∫
(c)
2 3/ 2
+ C , then we can replace x by 10 − 4x and still
x
3
1
 12
3/ 2
3/2
10 − 4x dx = ∫ (10 − 4x)1/ 2 dx =
+ C = − (10 − 4x) + C .
 −4  3 (10 − 4x)
6
8
∫ (6 + 3x)5 dx = ∫ 8(6 + 3x)
= −
−5
2
1
1
−4
−4
dx = 8 ×  ×
(6
+
3x)
+
C
=
−
(6
+
3x)
+C
 3 −4
3
2
+C.
3(6 + 3x)4
Example 3. Evaluate the antiderivatives:
(a) ∫
1 −3x
e
dx
4
(b) ∫
10
dx
5 − 2x
1
m x+b
Because ∫ e x dx = e x + C , we have ∫ e
dx = e m x+ b + C .
m
1 −3x
1
1 −3x
1 −3x
∫ 4 e dx = 4 × (−3) e + C = − 12 e + C .
Solution.
(b)
(a)
Because
10
1
1
1
∫ x dx = ln x + C , we have ∫ m x + b dx = m ln m x + b + C .
Thus,
Thus,
10
∫ 5 − 2x dx = −2 × ln 5 − 2 x + C = − 5ln 5 − 2 x + C .
Solving for the Constant C
If we know the value of f (x) at some specific x , then we can solve for the generic
constant C when evaluating
∫ f (x) dx .
Example 4. Find f (x) if f ′(x) = 4 cos(2x) and f (π / 4) = –6.
Solution. Since we are given the derivative f ′(x) , we can say that f (x) = ∫ f ′(x) dx .
(That is, the antiderivative of the derivative of f is the function f .) Thus,
f (x) =
4
∫ 4 cos(2x) dx = 2 sin(2x) + C = 2 sin(2x ) + C .
Thus, f (x) = 2 sin(2x ) + C . But f (π / 4) = –6; thus
–6 = f (π / 4) = 2 sin(2 × π / 4) + C = 2 sin(π / 2) + C = 2 + C .
So C = –8 and f (x) = 2 sin(2x ) − 8 . (Check that the conditions of the problem are
satisfied.)
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Example 5. Find f (t) if f ′′(t) = 4 , f ′(−2) = –4, and f (2) = 3/2.
t
Solution. First f ′(t) =
f ′(−2) = –4; so
∫ f ′′(t) dt = ∫ 12t
−4
dt = −4t −3 + C . So f ′(t) = −4t −3 + C . But then
–4 = f ′(−2) = −4(−2) −3 + C =
−3
Thus, C = –9/2 and f ′(t) = −4t −
Now f (t) =
∫ f ′(t ) dt
1
+C.
2
9
.
2

−3 9 
−2 9
−2 9
= ∫  −4t −  dt = 2t − t + D . So, f (t) = 2t − t + D .

2
2
2
But we know f (2) = 3/2; thus, 3/2 = 2(2)
2 9
that D = 10. Hence, f (t) = 2 − t + 10 .
2
t
−2
−
9
(2) + D = 1/2 – 9 + D , which implies
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