Indefinite Integrals - Problem 1
Let's solve some indefinite integrals. We're going to need two properties first. This property; the constant multiple rule.
If you're integrating the constant times some function, you can pull the constant out of the integrals, that's really helpful. We use that in these examples. The second is the power rule of anti-derivatives. When you're integrating x to the n, the anti-derivatives are 1 over n plus 1, x to the n plus 1 plus c. So we use both of those.
Let's take a look at this problem. It says perform the antidifferentiation. That's good because it reminds us that solving indefinite integrals means antidifferentiating; coming up with the antiderivatives of this function. Check by differentiating your answer. This is something students don't do enough of. You should always check your answer when you're not sure about what you come up with. So I want to try to get you in the habit by doing that myself.
So first of all, let's take the opportunity to use the constant multiple rule on this integral. 3 is a constant, so I can pull it out. That makes the integral look like this; x to the 4th dx. So this is 3 times the integral of x to the 4th dx. This integral can be solved using this formula. So n is 4, so I'm going to have 1 over 4 plus 1; 1 over 5. So it's 3 times 1 over 5 x to the 5, plus, let's call it c1.
Now I'm going to distribute the 3 over these terms. I get 3/5x to the 5th plus 3 times c1. Usually what you do in the end, is you just convert this to a new constant, and call it c. 3 times c1 is just an arbitrary constant, so you can just call it c. Think of it as renaming the constant. The renaming of 3 times c1, just c. So that's my set of antiderivatives. 3/5x to the 5th plus c.
Now I want to check by differentiating. So I'm going to differentiate this, and hopefully, I'll get this back again. So let me do that here. The derivative with respect to x of 3/5 x to the fifth plus c. So I get 3/5 time, and the derivative of x to the 5th is 5x to the fourth. The derivative of the plus c part is just 0. Now 3/5 times 5 is 3. So I get 3x to the 4th. That is the function I began with. So this is correct. That checks.
Let's take a look at another example. At the integral of 12 over x to the 5th. Now I want to rewrite this in a form that will allow me to use the power rule. When you have a power of x in the denominator, you can still use the power rule. This is the same as 12 times x to the -5 dx. The power rule will apply to this function.
First, I want to take the 12 out using the constant rule. So 12 times the integral of x to the -5 dx. The power rule says, that's it's going to be x to the -5. I always add 1. Maybe I should write this up. X to the -5 plus 1 over -5 plus 1, plus, I'll call it c1.
Now -5 plus 1 is -4. So this is 12 times x to the -4 over -14 plus c1. Then I can distribute the 12. So the 12 divided by -4 is going to give me -3x to the -4 plus 12c1. Then I can rename this just c. I could also write this as -3 over x to the 4th. So that's my set of antiderivatives; -3 over x to the 4th plus c.
Now I should check this by differentiation. I'm going to start with it in this form, since when I differentiate, I also use the power rule for derivatives. So the derivative of -3x to the -4 plus c is -3 times the derivative of x to the -4. That's -4x to the, remember the power rule is down by, so from -4 to -5. Then, the derivative of plus c, 0. -3 times -4 is 12. So I get 12x to the -5. Let's just check back. That is what I started with, 12x to the -5. So this is correct. That means that this is my correct answer. These are my antiderivatives of 12 over x to the 5th.