## THE SECOND FUNDAMENTAL THEOREM OF CALCULUS

__This is perhaps the most important thing the Integral Sector of this website contains. So, be sure not to waste it.__

Well, we just learned about the First Fundamental Theorem of Calculus, and I remembered kicking a statement up there saying that the Second is the more useful one. This theorem will literally solve all definite integrals. Provided that you can find the indefinite integral of the function.

The second fundamental theorem of calculus goes like this:

Where F(x) is the anti-derivative, or indefinite integral of f(x). The proof of this theorem really lies with the first one. You can try on your own. I'll try to put it up in documents for download. But now, let's focus on what this theorem does. After all, in tests and in practice, it's the APPLIANCES of theorems that come into use, not the PROOFS. It's a whole different story for people who strive to be mathematicians, but they probably aren't reading this in the first place, so......

What does this say? First, it does not put any restrictions on the function lying inside the integral signs, nor on its lower bound and upper bound, a and b. This means this identity applies to ALL definite integrals. Secondly, we see this is a simple, simple expression, and it agrees with institution: the constant terms of the anti-derivatives cancel, making out a definite number for the definite integral. We can calculate the definite integral with a hand calculator.

Now, an example would be this: integrate y=x, 0<x<1. Of course, some folks would try to find the area using the area of triangles. And I admit for this problem it's simpler. But let's use the definite integral. So, the anti-derivative of y=x is y=0.5x^2+C, while C is a constant with any real value. Now, plug in x=1 and x=0 and minus the two expressions. (Note that C conveniently cancel.) You get what you want. The answer should agree with your triangle area.

After you convince yourself of the immaculately power of this theorem, seal it in your mind forever, next to the Chain Rule, Rules of Differentiation, and the next day for the dentist.

What does this say? First, it does not put any restrictions on the function lying inside the integral signs, nor on its lower bound and upper bound, a and b. This means this identity applies to ALL definite integrals. Secondly, we see this is a simple, simple expression, and it agrees with institution: the constant terms of the anti-derivatives cancel, making out a definite number for the definite integral. We can calculate the definite integral with a hand calculator.

Now, an example would be this: integrate y=x, 0<x<1. Of course, some folks would try to find the area using the area of triangles. And I admit for this problem it's simpler. But let's use the definite integral. So, the anti-derivative of y=x is y=0.5x^2+C, while C is a constant with any real value. Now, plug in x=1 and x=0 and minus the two expressions. (Note that C conveniently cancel.) You get what you want. The answer should agree with your triangle area.

After you convince yourself of the immaculately power of this theorem, seal it in your mind forever, next to the Chain Rule, Rules of Differentiation, and the next day for the dentist.