Definition 6.2.1.
We use \(\int f(x)dx\) to denote the collection of *all* antiderivatives for \(f\)
Section 6.2 Antiderivatives
Due to the FTC, we now see that moving backwards through differentiation (finding antiderivatives) is important. It may not have occurred to you but while functions have only one derivative, they have more than one antiderivative. Whaaaa??? Sure, notice that \((x^2)'=2x\) and also \((x^2 + 1)'=2x\text{.}\) So, there is some sort of *collection* of antiderivatives for a given function. To denote this collection of antiderivatives, we adopt a new notation.
Unfortunately, this uses the notation we used earlier for integrals. That’s a historic fact and we have no power to change it. It differs in that the notation doesn’t use the endpoints of any definite interval (i.e. we are just using \(\int\) instead of \(\int_a^b\)). For that reason, some people add to the confusion by referring to the collection of all antiderivatives as the "indefinite integral". But, it is important to remember, this symbol denotes a collection of functions. It is not simply one function and it is certainly not an integral.
Perhaps the next most natural thing to wonder is, "How wild and weird does the full collection of antiderivatives for a given function get?" Well, the good news is ... not very weird.
Theorem 6.2.2.
If \(f\) is continuous on an interval and \(F\) is an antiderivative there, then \(\int f(x)dx = F + c\) where \(c\) is any constant
So, the collection of all antiderivatives of a given function can be found by just finding *one* antiderivative and then adding to it any constant you wish. That’s all of them. So, for example, we have
\begin{equation*}
\int 2xdx = x^2 + c
\end{equation*}
Since we have the FTC, it is clear that getting good at finding antiderivatives is critical. Even more, this theorem now points out we really just need to find one of them. So, how do we get good at finding one antiderivative for a given function? Well, it takes practice. That is generally taken up in a subsequent calculus course. But, let’s take a little bit of time to improve our skills at least a little bit. Remember, when you are looking at a function and are hoping to find where it came from via differentiation, you should be looking at the function for *clues* as to what differentiation properties may have been used to get you there. For example, consider something like
\begin{equation*}
\int 2xcos(x^2)dx
\end{equation*}
Where could this have come from? Well, it’s a product of two functions (\(2x\) and \(cos(x^2)\)). So, look back at differentiation properties and see if any of them result in a product of functions. You’ll quickly notice that the chain rule gives you a product. Specifically, \((f(g(x)))'=g'(x)f'(g(x))\text{.}\) So, it’s reasonable to expect that you got to \(2xcos(x^2)\) via the chain rule! If so, what functions are playing the roles of \(f\) and \(g\text{.}\) Well, the only *inside* function to a composition here is \(x^2\text{.}\) So that might be \(g\text{.}\) It’s inside \(cos\text{.}\) So, \(cos\) is apparently the \(f'\) which means that \(sin\) could be \(f\text{.}\) In all, it looks like we may have differentiated \(sin(x^2)\) (i.e.\(f(g(x))\) ). Let’s check!
\begin{equation*}
\begin{array}{lll}
(sin(x^2))' \amp = \amp (x^2)'sin'(x^2) \\
\amp = \amp 2xcos(x^2)
\end{array}
\end{equation*}
Bingo! We got it. We can now say that
\begin{equation*}
\int 2xcos(x^2)dx = sin(x^2) + c
\end{equation*}
