Section 11
Antiderivatives
Our mind is capable of passing
beyond the dividing line we
have drawn for it. Beyond the
pairs of opposites of which the
world consists, other, new
insights begin.
Herman Hesse
OR many functions F ( x ), we can find the instantaneous rate of change
F 0 ( x ). Suppose we start off knowing F 0 ( x ), how fast a function is
F
changing. We may want to recreate the original function F ( x ). (We will
later discuss finding the original function in more complicated situations,
but to begin with, we’ll assume that we know the derivative.) We begin
with some terminology.
Definition
Given a function f ( x ), the function F ( x ) is an antiderivative of f ( x ) if
F 0 ( x ) = f ( x ).
Saying that F ( x ) is an antiderivative of f ( x ) is exactly the same thing
as saying that f ( x ) is the derivative of F ( x ); they are different ways of
saying
F 0 ( x ) = f ( x ).
Example. Since F ( x ) = 2x4 + 3x satisfies F 0 ( x ) = 8x3 + 3, we can say
that F ( x ) is an antiderivative of f ( x ) = 8x3 + 3.
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1
Remark. Antiderivatives are not unique; a function will have more than
one antiderivative (it will have an infinite number of them, in fact). For
example, the function G ( x ) = 2x4 + 3x + 5 is also an antiderivative of the
function f ( x ) = 8x3 + 3 from the previous example.
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Any time we have a rule for taking a derivative, we automatically
have a rule for finding an antiderivative. For example, since we know
that the derivative of F ( x ) = 12 x2 is f ( x ) = x, we immediately know that
an antiderivative of f ( x ) = x is F ( x ) = 12 x2 . Similarly, an antiderivative
of f ( x ) = x2 is F ( x ) = 13 x3 . We have, in fact, the following:
Proposition
An antiderivative of f ( x ) = x n (for n 6= −1) is
F(x) =
1
x n +1 .
n+1
Warning!!
Note that the above formula doesn’t work for n = −1; in fact, the
1
n+1 is undefined for n = −1. An antiderivative of f ( x ) =
formula n+
1x
1
−1 will be brought up later.
x = x
Example. An antiderivative of f ( x ) = x7 is F ( x ) = x7+1 /(7 + 1) =
x8 /8.
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Some special cases that you might want to remember separately are:
• An antiderivative of f ( x ) = x is F ( x ) = 12 x2 .
• An antiderivative of a constant, f ( x ) = c, is F ( x ) = cx.
Just as with derivatives, antiderivatives can be taken term-by-term,
and constant coefficients are left alone.
Example. An antiderivative of f ( x ) = 2x4 + 3x is F ( x ) = 2 · x5 /5 + 3 ·
x2 /2 = 52 x5 + 32 x2 .
2
2
Warning!!
Computing antiderivatives of polynomials has similarities to computing derivatives of polynomials, and it is easy to confuse the two. In
particular, if there is a constant when finding an antiderivative, the
constant will not vanish like it does when taking derivatives.
Next, recall
• If F ( x ) = sin( x ) then F 0 ( x ) = cos( x )
• If F ( x ) = cos( x ) then F 0 ( x ) = − sin( x )
• If F ( x ) = e x then F 0 ( x ) = e x
• If F ( x ) = ln( x ) then F 0 ( x ) = 1/x.
These give us the following antiderivatives.
Proposition
• An antiderivative of f ( x ) = sin( x ) is F ( x ) = − cos( x ).
• An antiderivative of f ( x ) = cos( x ) is F ( x ) = sin( x ).
• An antiderivative of f ( x ) = e x is F ( x ) = e x .
• An antiderivative of f ( x ) =
1
x
is F ( x ) = ln( x ).
Note that antiderivatives of sin and cos are different than the derivatives: if f ( x ) = sin( x ), the derivative is f 0 ( x ) = cos( x ) while an antiderivative is F ( x ) = − cos( x ). Also, we now know an antiderivative for
f ( x ) = x n for n = −1, namely F ( x ) = ln( x ).
As mentioned before, antiderivatives are not unique. The expressions
F ( x ) = x2 and G ( x ) = x2 + 2 are both antiderivatives of f ( x ) = 2x.
However, any two antiderivatives of the same function are related.
Proposition
Let F ( x ) and G ( x ) both be antiderivatives of f ( x ). Then for some constant C,
G ( x ) = F ( x ) + C.
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Conversely, in the above situation, any expression of the form F ( x ) + C
will be an antiderivative of f ( x ). This means that as long as we know
one antiderivative of f ( x ), we know what they all look like. Since it is
often useful to deal with the entire class of antiderivatives, the following
definition is introduced.
Definition
The indefinite integral of a function f ( x ) is defined as
Z
f ( x )dx = F ( x ) + C,
where F ( x ) is any antiderivative of f ( x ) (so F 0 ( x ) = f ( x )) and C is an
arbitrary (unspecified) constant.
The arbitrary constant C is called the constant of integration,
and is
R
part of the indefinite integral. The reason for the notation and dx (as
well as the adjective
indefinite) will be discussed later. However, both the
R
integral sign and the dx are part of the notation.
Example. Let f ( x ) = 3x2 + 2x. Then
F1 ( x ) = x3 + x2
F2 ( x ) = x3 + x2 + 5
are antiderivatives, and the indefinite integral is
Z
f ( x )dx = x3 + x2 + C
In an indefinite integral, the constant C is called the constant of integration and is necessarily unspecified. If it has a value, then the expression
is a specific antiderivative rather than the indefinite integral.
Finding an indefinite integral is simply finding an antiderivative and
adding the constant of integration.
Example.
Z
1
1
1
(3x3 − 6x2 + 4x − 3)dx = 3 x4 − 6 x3 + 4 x2 − 3x + C
4
3
2
3
= x4 − 2x3 + 2x2 − 3x + C
4
4
Example. Find a function F ( x) satisfying F 0 ( x) = 4x3 − 3 and F (2) =
14.
We are told that F ( x ) is an antiderivative of 4x3 − 3. Since
Z
we know that
(4x3 − 3)dx = x4 − 3x + C,
F ( x ) = x4 − 3x + C
for some constant C. We are also given
F (2) = 14,
and from the formula for F ( x ), we also have
F (2) = 24 − 3 · 2 + C = 10 + C.
So we must have 14 = 10 + C, so C = 4. Finally,
F ( x ) = x4 − 3x + 4.
Example. A bike is traveling so that in t hours, it is traveling at f (t ) =
30 − t + 21 t 2 km/hr. It is currently 10 km away. Find a function giving
its position as a function of time.
Let F (t) be the position function. Then since the derivative of position is
velocity, we know
F 0 ( t ) = f ( t ),
and since it is currently 10 km away, we also know
F (0) = 10.
Since
Z
1
1
f (t)dt = 30t − t2 + t3 + C,
2
6
we know
1
1
F (t) = 30t − t2 + t3 + C
2
6
for some C. From this formula, we have
1
1
F (0) = 30 · 0 − 02 + 03 + C = C
2
6
5
and we are given
F (0) = 10,
so we get C = 10. In t hours, the bike’s position will be
1
1
F (t) = 30t − t2 + t3 + 10 km
2
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Example. Near the surface of the Earth, the acceleration due to gravity is (about a ) constant −32 feet/sec2 . A ball is thrown down at −50
feet/sec from the top of a 500 foot high tower. Let t = 0 seconds be the
time the ball is thrown.
(1) Find the velocity v(t ) as a function of t.
The derivative of velocity, v(t), is the acceleration. We are told that
v0 (t) = −32 feet/sec2 . Since
Z
(−32)dt = −32t + C
we know that v(t) = −32t + C for some constant C. We are told that
v(0) = 50 feet/sec, so
v(0) = −32(0) + C = −50,
so C = −50. That gives us a velocity function of
v(t) = −32t − 50 feet/sec.
(2) Find the height f (t ) as a function of t.
The derivative of the height, f 0 (t), is the velocity, so f 0 (t) = −32t − 50.
Since
Z
(−32t − 50)dt = −16t2 − 50t + C1
we know that f (t) = −16t2 − 50t + C1 for some constant C1 . Since
f (0) = −16(0)2 − 50(0) + C1 = 500,
we get C1 = 1000 and so
f (t) = −16t2 − 50t + 500 feet.
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