Antiderivative Intro

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We discuss the antiderivative at a conceptual level before we dive into the mechanics.

We’ve spent a lot of time now developing the theory of derivatives, and then looking at a number of applications. But there is an obvious question we should consider; can we reverse this process? Even if we could, would that be useful?

Video Lecture

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Let’s consider the position function $s(t) = -4.9t^2 + 20t + 2$ . We know that, using derivatives, we can get a function for velocity, $v(t)=-9.8t + 20$ , and a function for acceleration, $a(t) = -9.8$ . But what if we started with the acceleration function?

If all we know is that the acceleration is $a(t) = -9.8$ , then we could try and deduce what velocity function that acceleration came from. Some consideration might lead one to realize that we get constant functions like $a(t)=-9.8$ after taking the derivative of a line. After all, the acceleration being a constant $-9.8$ is telling us that the slope of the velocity graph should be $-9.8$ for every point - thus it is a line with a slope of $-9.8$ , i.e. $v(t) = -9.8t$ .

1 : But wait, this doesn’t quite match our previous velocity function, which was $v(t) = -9.8t + 20$ . Why not?

Because we did something wrong when figuring out the velocity function from the acceleration function. Because there is a flaw in our logic, not every constant function comes from a line. Because the derivative of a constant is zero - which makes it difficult to know if there was a constant in the velocity function when all you have is the acceleration function. Uh... something, something ... math reasons?

Indeed, since taking a derivative of a constant results in zero, there is no way to tell what the constant in the velocity function was before a derivative converted it to the acceleration function. The difference between the $v(t)$ we deduced from $a(t)$ , and the real $v(t)$ we started with, is that one of them had a constant of $20$ and one of them had a constant of $0$ .

This is a demonstration of one of the flaws in trying to reverse the process of differentiation. Indeed, although we can kind of reverse the process of differentiation, we can’t quite do it perfectly. Some information is lost when we take a derivative, like in this example the value of the constant added to $v(t)$ was lost when we took a derivative to get $a(t)$ . This lost information is often referred to as initial conditions and we will see more about this (including why it has that name), and how to handle this, in future segments.

When we reverse the process of differentiation to get a function whose derivative is the original function, we call this new function an antiderivative. Explicitly:

Antiderivative Fix $f(x)$ as some function. Then, if $F(x)$ is a function such that $F'(x) = f(x)$ , then we call $F(x)$ an antiderivative of $f(x)$ .

So we’ve seen that the process of taking a derivative can be reversed, but not fully. The process of taking a derivative is a destructive process - obliterating some initial condition information. As a result, reversing this process inevitably runs into the problem that any answer is then missing that initial condition information... unless we can get it from somewhere else.