Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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.
Unit
The Definite Integral