Calculus 1-Antiderivatives, guaranteed 100% Pass

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Antiderivatives


Def. A function 𝐹 is an antiderivative of 𝑓 on an interval 𝐼 provided
𝐹 ′ (𝑥 ) = 𝑓(𝑥), for all 𝑥 in 𝐼.


Antiderivatives will become very important when we talk about integration.

If 𝐹(𝑥) and 𝐺(𝑥) are both antiderivatives of 𝑓(𝑥), i.e. 𝐹 ′ (𝑥 ) = 𝑓(𝑥) and
𝐺 ′ (𝑥 ) = 𝑓(𝑥), Then 𝐺 ′ (𝑥 ) = 𝐹′(𝑥). As we saw from the Mean Value
Theorem, this means that 𝐺 (𝑥 ) = 𝐹 (𝑥 ) + 𝐶, where 𝐶 is a constant.
So an antiderivative of a function is not unique, but any antiderivative of a
function 𝑓 differs from any other antiderivative of 𝑓 by a constant.


So far we have developed a number of formulas for derivatives of functions. To
find an antiderivative of a function 𝑓(𝑥), we need to go “backwards”. That is we
need to find a function 𝐹(𝑥) such that 𝐹’(𝑥) = 𝑓(𝑥).




Ex. Find all antiderivatives of 𝑓(𝑥) = 1.


This one isn’t so bad. We have to find a function 𝐹(𝑥) such that 𝐹’(𝑥) = 1. If
we take 𝐹(𝑥) = 𝑥 , that will work. So 𝐹(𝑥) = 𝑥 is an antiderivative of
𝑓(𝑥) = 1. If we want all antiderivatives of 𝑓(𝑥) = 1, we take 𝐹(𝑥) = 𝑥 + 𝐶.

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Ex. Find all antiderivatives of 𝑓(𝑥) = 𝑥.


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So we need a function 𝐹(𝑥) such that 𝐹’(𝑥) = 𝑥 . If we take 𝐹 (𝑥 ) = 𝑥 2 we
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will have an antiderivative of 𝑓(𝑥) = 𝑥 . To get all antiderivatives we take
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𝐹 (𝑥 ) = 𝑥 2 + 𝐶.
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Ex. Find all antiderivatives of ℎ(𝑥 ) = 𝑥 𝑛 , where 𝑛 is a positive integer.


We know when we take a derivative of 𝑥 𝑛 , 𝑛 a positive integer, we get
𝑑
(𝑥 𝑛 ) = 𝑛𝑥 𝑛−1 .
𝑑𝑥

So if we want to go “backwards” let’s try taking as an antiderivative of 𝑥 𝑛 ,
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𝐻 (𝑥 ) = 𝑥 𝑛+1 .
𝑛+1

Notice that if 𝑛 is a positive integer (or even a positive rational number) that

𝐻′(𝑥 ) = 𝑥 𝑛 .
So all of the antiderivatives of ℎ(𝑥 ) = 𝑥 𝑛 , 𝑛 a positive integer, are given by
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𝐻 (𝑥 ) = 𝑥 𝑛+1 + 𝐶.
𝑛+1

In particular, when 𝑛 = 3, all of the antiderivatives of 𝑔(𝑥 ) = 𝑥 3 are given by
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𝐺 (𝑥 ) = 𝑥 4 + 𝐶.
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When 𝑛 = 2, all of the antiderivatives of 𝑓 (𝑥 ) = 𝑥 2 are given by
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𝐹 (𝑥 ) = 𝑥 3 + 𝐶.
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