Norm Prokup

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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Computing Definite Integrals using Substitution - Problem 2

Norm Prokup
Norm Prokup

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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When finding the definite integral of a function, you may need to use substitution in order to first find the indefinite integral. Recall that for the method of substitution, replace an expression of x with w. Find its derivative to find dw. Then, integrate as usual with respect to w.

After integrating, replace the w's with the appropriate expression of x. Then, plug in the limits of integration and subtract (if taking the integral from a to b, and the result of the integral is F(x), the result will be F(b) - F(a)).

Let’s try another definite integral using the method of substitution. Here we are asked to solve the integral from 0 to 3 of 800x over x² plus 16 the quantity squared, using the method of substitution. This looks like a really terrible integral. It’s actually quite easier if you use substitution.

I’m going to substitute for the x² plus 16 in the denominator. So w equals x² plus 16. And then my dw is going to be 2xdx. Now I get 800x so that’s planning. That’s going to be 400 times this. I still need to substitute for the limits of the integration. So let's see how the substitution works out. If x equals 0, what is w going to be? It’s going to be 0² plus 16, 16. If x equals 3, 3² is 9, 9 plus 16 is 25. So w equals 25.

And so my new integral becomes the integral from 16 to 25 of 800x. That’s 400 times 2xdx. So I have 400dw on top, over and this is just w². Let me write this in a slightly different way. Integral from 16 to 25, 400w to the -2.

Because I want to use the power rule for writing anti-derivatives on this guy. The antiderivative of w to the -2 is w to the -1 over -1. You add 1 to the exponent. So I’ll need that when I use the fundamental theorem of calculus. -400w to the -1 and I’ll evaluate that from 16 to 25. So -400 divided by 25, that’s what I get when I plug in 25 in here. When I plug the 16 in, I have to subtract that -400 over 16. These don’t look so great but they actually come out very nice. This is 400 over 25, that’s actually 16 and 400 over 16 is actually 25. So this is -16, minus minus, plus 25 and that’s 9.

So the whole integral turns out to just to be 9. Some people really resist the idea of switching over the limits integration to the new variable, but it’s a good hard to get into it. Actually makes for a very clean answer. I find that it's more difficult not to switch the limits of integration. So try to get over that initial feeling of reluctance to switch the limits of integration.

Looks like it’s complicated but it actually makes the integration much easier.

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