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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Complicated Indefinite Integrals - 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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Some integrals require a knowledge of specific antiderivatives. For example, recall that the derivative of ln(x) is 1/x. So, the antiderivative of 1/x is ln|x| + c. This means that when 1/x or x-1 is in an integral, remember that you cannot use the power rule on it, and that its antiderivative is ln|x| + c. Then, you can antidifferentiate as you normally would.

Another antiderivative to remember is that the antiderivative of ex is ex + c.

We’re working on some tougher indefinite integral problems. Before I get started with this example, I want to remind you of some integration formulas. First, the power rule for antiderivatives. The integral of x to the n dx is 1 over n plus 1, x to the n plus 1 plus c, if n is not equal to -1. If n happens to equal -1, you have the integral of x to the -1, you have the special case, the natural log of the absolute value of x plus c. So we’ll see that come into play in this example. And then this other formula; the integral of e to the x is e to the x plus c.

Let’s do the example. It says perform the antidifferentiation. Our first problem has three terms. So the first thing I would do is separate the three terms and pull out the constants. So 32 times the integral of x to the 5.4 dx, plus 12 times the integral of e to the x dx plus the integral of -16 dx. Here I use the first version of power rule. I have a power of x other than -1, so I add 1 to the exponent, 6.4 and I divide by 6.4. So this will be 32 x to the 6.4 divided by 6.4 plus, and then the integral of e to the x is e to the x plus c. This is 12 e to the x, I’ll write the plus xc at the end. And then the integral of the constant is that constant times x. so -16x plus c.

Let me just simplify this a little bit. The key to simplifying this, is recognizing that, both of these numbers are multiples of 3.2. That’s 3.2 times 10, 3.2 times 2. I’m just going to do it at the side, 32 over 6.4 is 3.2 times 10, 3.2 times 2. 3.2's cancel you get 10 over 2 which is 5. This is just 5, x to the 6.4 plus 12e to the x minus 16x plus c.

Let’s take a look at another example. I have the integral of e to the x over 5, minus 1/5, minus 4 over 5x. Now first of all, let’s observe that e to the x over 5 is the same as 1/5e to the x. And let’s also observe that minus 4/5x, that’s -4/5 times 1/x, which is the same as -4/5 times x to the -1, so we get to use that special formula. Let's do it.

I’m going to separate these terms and pull out the constants. I’ll have 1/5e to the x, dx and I’m going to integrate -1/5 just a constant. And then I’ll pull the -4/5 out in front. I have -4/5, integral of x to the -1 dx. So three terms, first the integral of e to the x is e to the x. The integral of a constant is that constant times x, minus 4/5. And the integral of x to the -1, is natural log of the absolute value of x, and then I throw a plus c in there.

And that’s my answer. 1/5e to the x minus 1/5x minus 4/5 natural log of the absolute value of x plus c. Don’t forget this absolute value, it’s really important, the antiderivative of x to the -1.

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