The fundamental theorem of Calculus is an important theorem relating antiderivatives and definite integrals in Calculus. The fundamental theorem of Calculus states that if a function f has an antiderivative F, then the definite integral of f from a to b is equal to F(b)-F(a). This theorem is useful for finding the net change, area, or average value of a function over a region.
I need to introduce a very important topic, "The Fundamental Theorem of Calculus." Here's the theorem right here. If f is a continuous function and capital F is an anti-derivative of little f then the definite integral from a to b of little f of x dx is capital F of b minus capital F of a. So again capital F is an anti-derivative of this inside function. This b is the same as this b, this a is the same as this a. So you can evaluate a definite integral exactly using an anti-derivative and just evaluating it and subtracting.
So let's see how that works out in an example. Says find the exact area under y=x squared plus 1 from x=0 to x=2. So the exact area equals the definite integral of this function from 0 to 2. That would be the integral from 0 to 2 of x squared plus 1 dx. So this is the integral I'm going to solve. Let's take it up here.
In this integral from 0 to 2, this is my little f of x. I need an anti-derivative for it and an anti-derivative would be capital f of x equals one third x cubed plus x. Now it's also true that one third x cubed plus x plus 1 is an anti-derivative of x squared plus 1. You can use any anti-derivative, it doesn't matter and that's why most people will choose to use the anti-derivative with a +0 here. So I need to evaluate this anti-derivative at 2 and then evaluate it at 0 and subtract. So this is going to equal capital f of 2 minus capital f of 0. Now capital f of 2 is one third of 2 cubed, one third of 2 cubed plus 2 minus capital f of 0 one third of 0 cubed plus 0. This is just going to be 0. One third of 2 cubed, 8 thirds plus 2-0. So this is going to be our be our answer. 2 is 6 thirds so this is 14 thirds or about 4 and 2 thirds. This is the exact value for the area under that curve and we got it using just a couple of calculations, the anti-derivative evaluated at 2 minus the anti-derivative evaluated at 0.
Now here's some helpful notation. When you're using the fundamental theorem of Calculus, you often want a place to put the anti-derivatives. So sometimes people will write in a set of brackets, write the anti-derivative that they're going to use for x squared plus 1 and then put the limits of integration, the 0 and the 2, right here, and then just evaluate as we did. So you'll see me using that notation in upcoming lessons.