 ###### 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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# Indefinite Integrals - Problem 3

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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One useful property of indefinite integrals is the constant multiple rule. This rule means that you can pull constants out of the integral, which can simplify the problem. For example, the integral of 2x + 4 is the same as the 2 multiplied by the integral of x + 2. However, it is important that only constants—not variables—are pulled out of the integral.

Another property is the power rule of antiderivatives (remember that to take the integral of something is to antidifferentiate). The power rule states that the antiderivative of a xn is xn+1/n + c, where c is some constant.

There is no product or quotient rule for antiderivatives, so to solve the integral of a product, you must multiply or divide the two functions.

Let's do some much harder examples. I want to find some antiderivatives. In this example I'm not going to check my antidifferentiation by differentiation. Remember you can always do that, if you're unsure of your answer. It's a good habit to get into. Let me come back to it later when I feel like our work is a little bit more prone to error. Let's skip it this time.

So here, I want to antidifferentiate x³ times (2x² minus 3). Now remember there is no product rule for antidifferentiation. You can't just antidifferentiate each of these guys, and multiply their products. You have to multiply through.

So let's do that. If I multiply this through, I'm going to get x³ times 2x², that's 2x to the 5th. Remember you add the exponents. So 2x to the 5th minus 3x³. Much better. Now this I can antidifferentiate just by separating it out, and getting the constants out.

So first, let's separate it into a sum. I have 2x to the 5th dx plus the integral of -3x³ dx. So what I've just done is I've used the sum rule. You have the integral of the sum it equals the sum of the integrals. Now I'm going to use the constant multiple rule to pull the 2, and the -3 out of each of these integrals respectively. 2 times the integral of x to the 5th dx minus 3 times the integral of x³.

Now I have each of these integrals in a form where I can apply this property; x to the n. The integral is 1 over n plus 1, x to the n plus 1. So here where my n is 5, n plus 1 is going to be 6. So it'll be 2 times 1/6 x to the 6th minus 3 times 1/4x to the 4th. The n here is 3, so n plus 1 will be 4. Then I tuck in my plus c at the end.

So all that's left to do is simplify my fractions. 2 over 6 is the same as 1/3, 3/4 x to the fourth plus c. We're not checking this time. So hopefully that's the right answer. 1/3x to the sixth minus 3/4 x to the fourth plus c. These are the antiderivatives of this function.

Let's do another one. Now here we have a quotient of two functions. Just like there is no product rule for antidifferentiation, there's no quotient rule either. So you really have to combine these, so you get like a sum or difference of powers of x.

So that's the first thing I'm going to do. I'm going to write each of these as just powers of x. I have 22x³ minus 4 over the cube root of x is the same as x to the 1/3dx. So this equals; now what I can do, is, I can essentially distribute this x to the 1/3 over each of these terms. Think of it this way. Think of it as 22x³ over x to the 1/3 minus 4 over x to the 1/3.

So let me simplify each of these so I have a single power of x. I've got 22x³ over x to the 1/3 is x to the 3 minus 1/3. 3 is 9/3. 9/3 minus 1/3 is 8/3, minus 4. Then 4 over x to the 1/3 is the same as x to the -1/3. So this is a much nicer form for integration purposes. I have each of these written as a power of x. Now I can separate the sum.

So that's one term. The other one is notice, I said sum even though there's a negative sign here. I can pull that negative inside. This is the same as adding -4x to the -1/3. So -4x to the -1/3. Then I can pull the constants outside.

Pretty soon I'll be doing these two steps in a single step. Once you get used to all the properties, but for right now I'm separating them. So I want to pull this -4 out. Integral of -1/3 to the x.

So now I have each integral as just the integral of some power of x. So remember, x to the 8/3, I have to add 1 to the exponent, that becomes 11 over 3. So x to the 11 over 3. I have to divide by 11 over 3 which is the same as multiplying by the reciprocal; 3/11, minus 4. I have to do the same thing with x to the -1/3. I add 1. -1/3 plus 1 is 2/3. So I get x to the 2/3. I have to divide by 2/3 which is the same as multiplying by 3/2, and then I'll add my c at the end.

Some cancellations happens. The 22 and the 11 cancel leaving a 2. 2 times 3 is 6. 6x to the 11/3, minus, here I have some cancellation. The 4 and the 2 cancel leaving a 2. 2 times 3 again is 6 x to the 2/3, and then plus c. These are my antiderivatives; 6x to the 11/3 minus 6x to the 2/3 plus c.