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Computing Definite Integrals using Substitution - Problem 1

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Teacher/Instructor 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.

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. For ln(x), recall that its derivative is 1/x. 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 use the method of substitution to solve another definite integral. Here I’m asked to solve the integral from 1 to e of natural log x over x. First thing I want to do, is rewrite this integral in another form. From 1 to e, I’ll pull the natural log x in front. And this is the same as natural log x times 1 over x dx.

When we substitute, you'll see why I did this. I’m going to substitute for natural log x. Because in here I don’t actually see there is so much as a composite function, but I see that natural log has its derivative also in the integral. So I’m going to substitute for that. And the derivative is 1 over x dx. A perfect substitution. The w will take care of natural log, and the dw will take care of all the rest. But let’s take care of the limits of integration as well.

So when x is 1 what’s w? The natural log of 1 is 0. When x is e, we have a natural log of e. A natural log of its own base, that’s going to be 1. So the new limits are 0 and 1. This becomes w and this becomes dw. So the integral of w is ½w². So I’ll take this from 0 to 1. First at 1, ½ of 1² is ½ of 1, minus ½ of 0² which is just 0. The answer is just a half. That’s it.

So it’s really easy if you switch the limits over, to just go ahead and evaluate the integral with respect to w. And then plug in the limits.

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