*Warning: This article is NOT for high school students. I assume you'll know some more advanced math syntax that you'd normally pick up in university. I also assume you're familiar with FTC (or FToC), or that you've been exposed to it, at the very least. I might do a more in depth version later, but not right now.*

I know, I know, the FTC tells us that we can integrate continuous functions and that the derivative of an integral gives you back your original function.

That's pretty much the least sexy way to think about it ever. Guys and girls, remember that math is the most sexy thing you can do for your partner, so it's important that you get it right.

Here's the full unsexy version of the FTC:

Suppose that \(f\) is continuous on \( I=\{x:a\leq x \leq b\}\) and that \(F\) is defined by \[F(x)=\int_a^xf(t)dt.\] Then \(F\) is continuous on \(I\) and \(F’=f(x)\) for each \(x \in I_1=\{x:a<x<b\}\).

Let's take a moment to understand what they are saying (I'll keep it short) and then we'll dive into why it makes SO much sense geometrically.

Do you know the geometrical significance of \(F(x)\)? If you don't, take a look at how it's defined. It's just the area under the \(f\) curve, starting at \(t = a\) and ending at \(t = x\). Nothing new here.

What we claim next is where things get interesting. We claim that the derivative, \(F’\) , gives us back our original function \(f\).

Actually I kinda lied to you when I said that was the interesting part. I mean, I suppose that was the interesting part of the FTC when you first saw it. By now, however, most students are quite comfortable with the relationship between derivatives and anti-derivatives. Everyone and their dog knows that they "undo" each other.

As it turns out, there is a super intuitive way to think about the FTC. And this my friends, is why we are gathered here today.

The first thing I want you to see is that if \(F'(x)=f(x)\), that means that \[\frac{dF}{dx}=f(x) \]

or in other words:

Every super small change in the \(x\) variable (we use \(dx\) to represent this change) will produce a change of \(f(x)\) in the \(F\) variable.

Re-read that sentence until you get it get it get it get it. Another way to see it is that for every increment of \(dx\) in the \(x\) variable, we get an increment of \(f(x)\) in the \(F\) variable (which is like the \(y\) variable here).

The next thing I want you to remember is that our function \(F\) represents the area under the \(f\) curve up to whatever \(x\) value you specify.

With that said, we can now re-state the FTC in a super sweet way. \[\frac{dF}{dx}=f(x) \]

means that:

every super small change in the \(x\) variable (\(dx\)) will produce a change of \(f(x)\) in the area under the \(f\) curve.

OK, now that's pimpin sweet. THAT, is something I can try to wrap my puny brain around and visualize! Awwwww yeah!

So let's do that. We're so close!

Shel, shut up and show us a pimpin sweet picture!

Shut up people! I'll show you a damn picture! And I'll make it pimpin sweet, as commanded!

Pic 1: Notice that when \(x=2\), the added area, or change in area, is a box of height approximately equal to \(f(2)\)

Now, everything should be coming together for you with regards to FTC, and life in general. OF COURSE it makes sense now. Of COURSE the change in area at some \(x\) value must be the area of the little red \(f(x)dx\) sliver that you add on to the end of it.

Thanks for reading you weirdos! I hope this brought some intuition to the FTC, something that has eluded me for some time!

Now go and think about math! :P Until next time!