Integrating power series | Series | AP Calculus BC | Khan Academy

So we're told that ( f(x) ) is equal to the infinite series we're going from ( n = 1 ) to infinity of ( \frac{n + 1}{4^{n + 1}} x^n ). What we want to figure out is what is the definite integral from 0 to 1 of this ( f(x) ). And like always, if you feel inspired, I encourage you to feel inspired, pause the video, and see if you can work through this on your own. Or at any time while I'm working through it, pause it and try to keep on going.

Alright, well, let's just rewrite this a little bit. This is going to be the same thing as the integral from 0 to 1. ( f(x) ) is this series, so I could write the sum from ( n = 1 ) to infinity of ( \frac{n + 1}{4^{n + 1}} x^n ). Now what I'm about to do might be the thing that might be new to some of you, but this is essentially taking a definite integral of a sum of terms, and that's the same thing as taking the sum of a bunch of definite integrals. Let me make that clear.

So if I had a, let's say this is a definite integral from 0 to 1, and let's say I had a bunch of terms here, I could even call them functions. Let's say it was ( g(x) + h(x) ), and I just kept going on and on and on ( dx ). Well, this is the same thing as the sum of the integrals: the integral from 0 to 1 of ( g(x) , dx + ) the integral from 0 to 1 of ( h(x) , dx + ) and we go on and on and on forever, however many of these terms are. This comes straight out of our integration properties.

So we can do the exact same thing here, although we'll just do it with the sigma notation. This is going to be equal to the sum from ( n = 1 ) to infinity of the integral, the definite integral of each of these terms. So I'm going to write it like this: the integral from 0 to 1 of ( \frac{n + 1}{4^{n + 1}} x^n , dx ).

So once again, now we're taking the sum of each of these terms. So, let's evaluate this business right over here. So that is going to, I'll just keep writing it out, this is going to be equal to the sum from ( n = 1 ) to infinity, and then the stuff that I just underlined in orange. This is going to be, let's see, we take the anti-derivative here, we are going to get ( \frac{x^{n + 1}}{n + 1} ).

So we have this original ( \frac{n + 1}{4^{n + 1}} ), and that's just a constant when we think in terms of ( x ) for any one of these terms. And then here, we'd want to increment the exponent and then divide by that incremented exponent. This just comes out of, I often call it the inverse pi or the anti-power rule or reversing the power rule. So it's ( \frac{x^{n + 1}}{n + 1} ). I just took the anti-derivative, and we're going to go from 0 to 1 for each of these terms.

Before we do that, we can simplify. We have an ( n + 1 ), we have an ( n + 1 ), and so we can rewrite all of this. This is going to be the same thing. We're going to take the sum from ( n = 1 ) to infinity, and this is going to be what we have in here when ( x ) is equal to 1. It is 1. We could write ( \frac{1^{n + 1}}{4^{n + 1}} ) actually. Why don't I write it that way?

( \frac{1}{1^{n + 1}} 4^{n + 1}} - 0^{n + 1} over 4^{n + 1} ). So we're not going to even have to write that. I could write ( \frac{0^{n + 1}}{4^{n + 1}} ), but this is clearly just 0.

And then this, and this is starting to get nice and simple. Now this is going to be the same thing. This is equal to the sum from ( n = 1 ) to infinity of ( \frac{1}{4^{n + 1}} ). Now you might immediately recognize this; this is an infinite geometric series. What is the first term here?

Well, the first term, first first term is, well, when ( n ) is equal to 1, the first term here is ( \frac{1}{4^2} ). Did I do that right? Yeah. When ( n ) is equal to 1, it's going to be so this is going to be ( \frac{1}{4^2} ), which is equal to ( \frac{1}{16} ). So that's our first term.

And then our common ratio here, well that's going to be, well we're going to keep multiplying by ( \frac{1}{4} ). So our common ratio here is ( \frac{1}{4} ). And so for an infinite geometric series, since our common ratio, well, is less than or it's, its absolute value is less than one, we know that this is going to converge, and it's going to converge to the value. Our first term ( \frac{1}{16} ) divided by one minus the common ratio ( 1 - \frac{1}{4} ).

So this is ( \frac{3}{4} ). So it's equal to ( \frac{1}{16} \times \frac{4}{3} ). So ( \frac{1}{12} ). And we're done! This seemed really daunting at first, but we just had to realize, okay, an integral of a sum, even an infinite sum, well that's going to be the sum of these infinite integrals. We take the anti-derivative of these infinite integrals which we were able to do, which is kind of a cool thing—one of the powers of symbolic mathematics. And then we realized, oh, we just have an infinite geometric series which we know how to find the sum of, and we're done.