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# Integrating power series

AP.CALC:
LIM‑8 (EU)
,
LIM‑8.D (LO)
,
LIM‑8.D.6 (EK)
,
LIM‑8.G (LO)
,
LIM‑8.G.1 (EK)

## Video transcript

- [Instructor] So we're told that f(x) is equal to the infinite series, we're going from n equals one to infinity of n plus one over four to the n plus one, times x to the n. And what we wanna figure out is, what is the definite integral from zero to one of this f(x)? And like always, if you feel inspired, and 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. Well, let's just rewrite this a little bit. This is going to be the same thing as the integral from zero to one. F(x) is this series, so I could write the sum from n equals one to infinity of n plus one over four to the n plus one, times x to the n. And now what I'm about to do might be the thing that might be new to some of you, but this is essentially, we're taking a definite integral of a sum of terms. 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 zero to one, and let's say I had a bunch of terms here. I could even call them functions. Let's say it was g(x) plus h(x), and I just kept going on and on and on, dx, well, this is the same thing as a sum of the integrals, as the integral from zero to one of g(x), g(x) dx plus the integral from zero to one h(x) dx, plus, and we go on and on and on forever. However many of these terms are. This comes straight out of our integration properties. 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 equals one to infinity of the definite integral of each of these terms. So I'm gonna write it like this. So of the integral from zero to one of n plus one over four to the n plus oneth power, times x to the n, and then it is dx. Once again, now we're taking the sum of each of these terms. Let's evaluate this business right over here. That is going to, I'll just keep writing it out. This is going to be equal to the sum from n equals one to infinity, and then the stuff that I just underlined in orange, this is going to be, let's see, we take the antiderivative here. We are going to get to x to the n plus one, and then we're gonna divide by n plus one. So we have this original n plus one over four to the n plus one, and that's just a constant when we think in terms of x, for any one of these terms, and then here we'd wanna increment the exponent, and then divide by that incremented exponent. This just comes out of, I often call it the inverse, or the anti-power rule, or reversing the power rule. So it's x to the n plus one over n plus one. I just took the antiderivative, and we're gonna go from zero to one for each of these terms. Before we do that, we can simplify. We have an n plus one, we have an n plus one, and so we can rewrite all of this. This is going to be the same thing, we're gonna take the sum from n equals one to infinity, and this is going to be, what we have in here, when x is equal to one, it is one, we could write one to the n plus one over four to the n plus one. Actually, yeah, why don't I write it that way. One to the n plus one over four to the n plus one, minus zero to the n plus one over four to the n plus one, so we're not gonna even have to write that. I could write zero to the n plus one over four to the n plus one, but this is clearly just zero. 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 equals one to infinity. And we almost are gonna get to our drumroll of 1/4 to the n plus one. Now you might immediately recognize this. This is an infinite geometric series. What is the first term here? Well, the first term is, well, when n is equal to one, the first term here is 1/4 to the second power. Did I do that right? Yeah. When n is equal to one, it's going to be, so this is going to be 1/4 to the second power, which is equal to 1/16, so that's our first term. And then our common ratio here, well that's gonna be, well, we're just gonna keep multiplying by 1/4, so our common ratio here is 1/4. And so for an infinite geometric series, this is, since our common ratio, its absolute value is less than one, we know that this is going to converge, and it's gonna converge to the value, our first term, 1/16, divided by one minus the common ratio, one minus 1/4, so this is 3/4, so it's equal to 1/16 times 4/3. This is going to be equal to 1/12. And we're done. And this seemed really daunting at first, but we just have to realize, okay, an integral of a sum, even an infinite sum, well, that's gonna be the sum of these infinite integrals. We take the antiderivative 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.
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