Antiderivatives of Core Functions

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We develop rules to quickly determine antiderivatives of most of our core functions.

We know it is possible to reverse the derivative process, at least to a point, and to that end it is worth establishing the rules for the core functions whose derivative processes we’ve already managed to determine. In this segment we record those core function relationships.

Video Lecture

Text with Additional Details

So far we’ve developed derivative shortcut rules for the following core functions:

$\displaystyle \frac {d}{dx} x^n = nx^{n-1}$
$\displaystyle \frac {d}{dx} a^x = \ln (a)\cdot a^x$
$\displaystyle \frac {d}{dx} \log _b(x) = \frac {1}{x\ln (b)}$

To reverse the process, we “simply” need to go backward, using the equality as a reference. Let’s begin by looking at the polynomial rule.

Let’s say we want to find an antiderivative to $x^n$ , that is, we want a function $F(x)$ such that $F'(x) = x^n$ . We know we will need to increase the power by one, because the derivative of a polynomial lowers the power by one, so let’s start with a simple guess: $F(x) = x^{n+1}$ . Using this as a starting point, we can take the derivative and see how close we are.

1 : If $F(x) = x^{n+1}$ , then $F'(x) = \answer {(n+1)\cdot x^n}$

So calculating $F'(x)$ gives us that $F'(x) = (n+1)x^{(n+1)-1} = (n+1)x^n$ . This is close to what we wanted, but not quite right, because we have this extra “ $n+1$ ” coefficient. To counter this, we can modify our proposed $F(x)$ slightly by dividing by the extra $n+1$ term. Remember that this $n+1$ is actually a constant (since $n$ isn’t a variable but an arbitrary constant) which means the derivative process will ignore it. Thus our next guess for $F(x)$ would be $F(x) = \frac {1}{n+1}x^{n+1}$ . Then calculating $F'(x)$ we get

$F'(x) = (n+1)\frac {1}{n+1}x^{(n+1)-1} = \answer {x^n}$

And we have our antiderivative! Or, at least, we have one of the possible antiderivatives. Remember that the derivative of a constant is zero, so any added constant to our polynomial would also be a valid antiderivative, thus we encapsulate this information by including a “ $+ C$ ” term to denote that we can have any constant added on as well. We’ll discuss this more in a future segment, but for now, this gives us our rule for reversing the derivative process for polynomials - in particular if $f(x) = x^n$ and we want a $F(x)$ such that $F'(x) = f(x)$ , then $F(x) = \frac {1}{n+1}x^{n+1} + C$ .

There is an important exception here though: what about when $n=-1$ ? Our formula rapidly falls apart in this case, since the coefficient tries to divide by zero, and our power becomes zero so we end up with some terrible looking expression that can’t exist. But let’s backup and recall what it means when $n=-1$ . Essentially we are asking “what function has, as a derivative, $x^{-1}= \frac {1}{x}$ ?” And we actually already know this! Indeed, in this specific case (where $n=-1$ ) we have $F(x) = \ln (x) + C$ . Then calculating $F'(x) = \frac {1}{x}$ as desired.

2 : Which of the following is an antiderivative for the function $f(x) = 2x^3$ ?

$\frac {1}{2}x^4$ $6x$ $\frac {1}{4}x^4$ Something about $C$ ?

Next lets consider the exponential. We want a function $F(x)$ so that $F'(x) = a^x$ . Again, it helps to recall what the derivative does for exponentials as a starting point. Since the derivative of $a^x$ is itself, with an extra constant multiplied (in particular $\ln (a)$ ), lets start with $F(x) = a^x$ and see how close we are to the desired result. If we calculate $F'(x)$ for $F(x) = a^x$ we get $F'(x) = \ln (a) a^x$ , which is again almost what we want, but it has that extra factor of $\ln (a)$ . Again, since this is a constant (remember that $a$ here is a number, like $5$ or $e$ or $17$ , not a variable) we can go back and adjust $F(x)$ by dividing by that constant so that, when we take the derivative, the constant is canceled out. Moreover, remember that we still need our “ $+ C$ ” as well. So with this in mind, our new proposed $F(x)$ is $F(x) = \frac {1}{\ln (a)}a^x + C$ . Then calculating $F'(x)$ we get $F'(x) = \ln (a) \frac {1}{\ln (a)}a^x = a^x$ which is exactly what we wanted!

Finally, let’s consider the log function. Now we want $F(x)$ such that $F'(x) = \log _b(x)$ . But... it turns out this one is actually really difficult to find an antiderivative for. This might not be too surprising though because, remember, the derivative of $\log _b(x)$ doesn’t result in a log. Unlike polynomials and exponentials, when you take a derivative of a log function you get a totally different kind of function! Indeed, to actually find something whose derivative is log, we need techniques outside the scope of this course!

This goes to show that reversing the derivative process can actually be really difficult. Indeed, this is arguably the first time in your math career where there are problems that could be relatively easily stated, that nobody can actually solve. For example, it has been proven that there is no actual function whose derivative is $e^{x^2}$ . I’m not saying that it can’t be done in calculus one, or that it can’t be done by undergrads... I’m saying it cannot be done by anybody, even professional mathematicians! This is a wild new world we’re stepping into here, and it is easy to accidentally stumble into a real beast of a problem without realizing it. Moreover it requires some real cleverness to reverse some of these derivative processes, and we’re going to just scratch the surface during this semester.

We have reversed the derivative process for monomial type forms and exponentials. In particular we have that:

For $x^n$ for any $n\neq -1$ , then $F(x) = \frac {1}{n+1}x^{n+1} + C$ is the antiderivative, i.e. $F'(x) = x^n$ .
For $x^n$ when $n=-1$ (i.e. for $x^{-1} = \frac {1}{x}$ ), then $F(x) = \ln (x) + C$ is the antiderivative, i.e. $F'(x) = x^n$ .
For $a > 0$ , then $F(x) = \frac {1}{\ln (a)} a^x + C$ is the antiderivative, i.e. $F'(x) = a^x$ .
As a special case for the previous rule, when $a=e$ we get that $F(x) = e^x$ since then $F'(x) = e^x$ .

We also discovered that there are some elementary functions, like log, whose antiderivative are incredibly difficult to determine directly. Indeed, we even saw an example of a relatively basic looking function whose antiderivative simply cannot be determined directly, no matter how much mathematics you know!