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### Course: AP®︎ Calculus BC (2017 edition) > Unit 12

Lesson 9: Power series intro# Integrating power series

Within its interval of convergence, the integral of a power series is the sum of integrals of individual terms: ∫Σf(x)dx=Σ∫f(x)dx. See how this is used to find the integral of a power series.

## Want to join the conversation?

- Apart from its just a problem. Can anyone tell, what is the integration of power series means or applications of differentiating/integrating a power series?(6 votes)
- It's also useful for Fourier series, and the Fourier series are useful to transform a function of time to a function of frequency and functions of frequency are useful in signal analysis and signal analysis....(11 votes)

- Why didn't we prove that the bounds 0 to 1 were within the interval of convergence?(7 votes)
- I guess you just have to assume, although if you do it out, it is in fact within the interval of convergence (-4,4).(6 votes)

- I just got confused at the end where he divided the first term 1/16 by 1 - 1/4.

How did he arrive to that conclusion?(3 votes)- 𝐺 = 𝑎 + 𝑎𝑟 + 𝑎𝑟^2 + 𝑎𝑟^3 + ... + 𝑎𝑟^(𝑛 − 1)

is a geometric sum with 𝑛 terms.

We realize that

𝐺 − 𝑟𝐺 = 𝑎 − 𝑎𝑟^𝑛 ⇔ 𝐺 = 𝑎(1 − 𝑟^𝑛)∕(1 − 𝑟)

With |𝑟| < 1 we get

lim(𝑛→∞) 𝐺 = 𝑎∕(1 − 𝑟)(4 votes)

- Why are integrate term by term in power series?(3 votes)
- Say we have an indefinite integral of a sum (a + b). In this case we can evaluate this integral as a sum of two integrals. In other words; integral of a+b equals itegral of a + integral of b.

Same reasoning can be used when thinking about integrating series: integrating whole series is the same as taking series of integrals.(3 votes)

- When can we exchange the series symbol with the integral symbol?(3 votes)
- We don't exchange the symbols, rather we are integrating the series. you can never just "exchange" symbols.(2 votes)

- Could you explain the very end of this? Why is the 1st term divided by (1-r) to find the actual value of the integral? And just to clarify this means that 1/12 is equal to the 3rd integral of f(x) from 0 to 1 right?(2 votes)
- The formula for the value of a geometric series is a/(1-r), where a is the first term and r is the common ratio. Check out Khan Academy's videos of this. And this indeed means 1/12 is equal to the integral of f(x) from 0 to 1.(1 vote)

- Can we integrate other types of series?(2 votes)
- Yes, as long as it can be written in Sigma notation.(1 vote)

- can this be used to approximate the integrals of functions that are impossible/really difficult to integrate?(by writing the function as a taylor series and then integrating the series)(2 votes)
- According to my understanding, a function is integrable if it is continuous, but is summation continuous? It seemed that the summation is discrete. Thank you1(1 vote)
- A function is integrable if it is continuous, but it doesn't have to be continuous; a function can be summed if it is defined at all indexes of summation (so yes, summation is discrete and does not require a continuous function).(3 votes)

- Is n being treated as a constant?(1 vote)
- Yep! n is the number of the term so it'll be a constant.(2 votes)

## 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.