Here it is:[br][b][br]The Fundamental Theorem of Calculus:[/b] If you need to calculate the integral of some function [code]f(x)[/code] from[code] x=a[/code] to [code]x=b[/code] (either because your life depends on it, it has unfortunately appeared on a standardized test you're taking, or for really any other reason), and if you are able to find a different function[code] F(x)[/code] such that its derivative [code]F'(x)[/code] is equal to [code]f(x)[/code], then you can use [code]F(x)[/code] to calculate the integral in question. Specifically:[br][br][math]\int_a^bf\left(x\right)dx=F\left(b\right)-F\left(a\right)[/math]

I remember seeing the Fundamental Theorem of Calculus ("FTC") the first time when I was in high school, and I have to be honest: I really didn't think it was true. It simply doesn't seem like it should be possible that derivatives should have anything to do with integrals. And even if they do, it doesn't seem possible the relationship could be this incredibly straightforward. It's so straightforward, it seems impossible to me. But I was wrong. The FTC is true.[br][br]Let's take a minute and be sure we really understand what it means. So let's suppose we want to integrate something simple like [code]f(x)=2x+2[/code] from [code]x=1[/code] to [code]x=3[/code]. In other words, suppose we are trying to calculate [br][br][math]\int_1^32x+2dx[/math][br][br]Obviously, we could just ask Geogebra to do this for us, but let's try to use the FTC before we do that. [br][br]The FTC says if we can find another function [code]F(x)[/code] whose derivative, [code]F'(x)[/code], is equal to [code]f(x)[/code], then we can use [code]F(x)[/code] to calculate the integral. The word used to talk about the function[code] F(x)[/code] is the [b]antiderivative of[code] f(x)[/code][/b]. This is a good name. Since the [i]derivative[/i] of [code]F(x)[/code] is [code]f(x)[/code], it makes sense to say that the [i]antiderivative[/i] of[code] f(x)[/code] is [code]F(x)[/code]. I know capitalization seems like an exceedingly subtle way to distinguish between two [i]different[/i] functions, but this convention actually works out pretty well once you become accustomed to it.[br][br]For this particular puzzle, [code]F(x)=x^2+2x[/code] fits the bill. Indeed,[code] F'(x)=2x+2[/code] , and this is exactly equal to [code]f(x)[/code], so [code][/code][b][code]F(x)[/code] is an antiderivative of [code]f(x)[/code][/b]. Use the [url=https://www.geogebra.org/m/x39ys4d7#material/p8jdmayj]Monkey Rules[/url] to check. If you skipped the Monkey Rules, that's fine too. You'll need to use Geogebra to check that [code]F(x)[/code] has derivative equal to [code]f(x)[/code]. Either way, we'll return to how to find antiderivatives later. For now, we're going to use them without worrying about where they come from. All we'll need to do is check that they work by calculating a derivative.[br][br]Let's get back to the FTC. In the previous paragraph we found that the function [code]F(x)=x^2+2x[/code] has derivative equal to [code]f(x)=2x+2[/code]. The FTC says we can use [code]F(x)[/code] to calculate the integral of [code]f(x)[/code] as follows:[br][br][math]\int_1^32x+2dx=F\left(3\right)-F\left(1\right)=\left(3^2+2\cdot3\right)-\left(1^2+2\cdot1\right)=\left(15\right)-\left(3\right)=12[/math][br][br]I don't know about you, but that seems too easy to be true. No rectangles. No samples. How could that be?[br][br]Let's check it in Geogebra. Type [code]integral(2x+2,1,3)[/code] into the input bar.

[url=https://www.youtube.com/watch?v=gXBJRz_33Y8]You did it![/url] It doesn't seem like it should be possible, but it worked!

Whenever you're ready, move on to the next lesson for an activity that further illustrates what's going on the under the hood of the FTC.