To find the antiderivative of a power of x, there is a rule you can follow: increase the power by 1, and divide by the new power. For example, the antiderivative of x is x2/2 + c, and the antiderivative of x4 is x5/5 + c. Remember to add the constant c at the end of the antiderivative.
More generally, if f'(x)=xn, for some number n, its antiderivative is f(x)=xn+1/(n+1).
Let’s try a harder problem. I want to find the anti-derivatives of first; little f(x) equals x to the 4th and then little f(x) equals x to the n any power function. Starting with x to the 4th. To find the anti-derivative of little f(x) equals x to the 4th. I need to find a function whose derivative is x to the 4th.
Remembering what derivatives do to power functions, they reduce the power by 1. So for example, the derivative with respect to x, of x to the 5th, I’ll get an x to the 4th out of this, because powers are reduced by 1. I’ll get 5x to the 4th. But that’s to exactly what I want. I want just x to the 4th. So what can I do to the original function to make the derivative come out just x to the 4th? I could divide it by 5. I can use 1/5x to the 5th. And my derivative will be 1/5 times 5x to the 4th. The 5 and the 1/5 cancel and I get x to the 4th.
So that tells me that one of the anti-derivatives for this function, is 1/5 x to the 5th. And I have a feeling that tells me once I have one anti-derivative of x to the 4th, I have all of them, just by adding a plus c. That gives me all the rest of them. Certainly, when c is 0, I get 1/5x to the 5th that is an anti-derivative. But c could be any other constant as well.
Let’s take a look at the second part; Anti-derivatives of f(x) equals x to the n. The next thing about this problem is once we solve it, we will have a formula for the anti-derivatives of any power function x to the n.
Well let’s use the same kind of ideas before. First, let’s recall we need to differentiate x to the n. What happens to the power? Comes out in front and goes down by 1. So what we want to end up with on the right side here, is x to then to n, not x to the n minus 1. So I have to start with one higher power. X to the n plus 1. Now when I differentiate this by the same rule, the n plus 1 comes out in front. X to the n, this goes down by 1. This is good, n plus 1 is just a constant and I have an x to the n like I need.
I can do a trick like I did before. I can make this constant come out 1 if I start with a function that has 1 over that constant in front of it. So 1 over n plus 1 times x to the n plus 1. Let’s see how that works. First of all this constant multiple is not touched by the derivative. We have 1 over n plus 1 times, and the derivative of x to the n plus 1 as we just saw, n plus 1 x to the n. These cancel and you get x to the n. Exactly what we need.
So this tells me that one anti-derivative of x to the n is one over n plus 1 x to the n plus 1. And of course, we can get all the rest of them by adding a constant. So this is a nice formula you can use any time you are anti-differentiating a power function. X to the n. The anti-derivatives of x to the n are 1 over n plus 1, x to the n plus 1, plus c.