 ###### 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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# Simple Substitutions - 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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For an exponential function, use the substitution on the function the exponent is raised to. In other words, to find the integral of ef(x), make w = f(x). Then, use the substitution method as usual: find dw by taking the derivative of w with respect to x, substitute the x's in the integrand with the expression for w, and integrate as usual. Replace the w's with the expression for x.

Let’s take a look at another problem. I want to perform this antidifferentiation and check my answer by differentiating. I want to try something first. Recall that we have this formula; the integral of e to the xdx is e to the x plus c. Why won’t it work to apply this formula directly? Why can’t I just write equals e to the 2 minus 5x plus c?

Let’s check that by differentiating. Does it work? Remember you can always check your antiderivative, your answer, by differentiating it. You should get what you started with inside the integral. When I do that, I have to differentiate e to the 2 minus 5x, and that’s e to the 2 minus 5x times the derivative of the inside part. The derivative of this is -5. Now this is not what I started with. It’s -5 times what I started with. So this didn’t work.

We have to use the method of substitution instead. Let me recopy this integral up here. E to the 2 minus 5xdx. Now with the method of substitution you want to look inside and here we have a linear function, 2 minus 5x. So that’s what I’m going to substitute for. I also need to calculate the dw, because I need to change this dx into dw. So dw is going to be -5dx.

Now you’ll notice after I switch the 2 minus 5x to w, I have the integral of to the w. The dx, I don’t have a -5 here so I need to actually get rid of that -5. I’ll multiply both sides by -1/5 and now I just have a dx which I can replace completely by -1/5dw. This becomes times -1/5dw.

The -1/5 can pull out in front, and then this is just straight forward e to the w plus c. -1/5 e to the w plus c. Now w was 2 minus 5x so this is -1/5 e to the 2 minus 5x plus c.

Let’s check this answer by differentiating it. Always a good habit to check if you’re not sure. Derivative with respect to x of -1/5e to the 2 minus 5x plus c is. I’ll have the -1/5 still. I have e to the 2 minus 5x, but I have to multiply by the derivative of 2 minus 5x which is -5. Of course the derivative of the plus c is just 0.

You can see that the -1/5 and the -5 are going to cancel and give me e to the 2 minus 5x which is perfect. That’s exactly what I started with in my integral and so that confirms that this is the correct answer; -1/5e to the 2 minus 5x plus c.