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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

so we're told that f of X is equal to the infinite series we're going from N equals one to infinity of n plus 1 over 4 to the n plus 1 times X to the N and what we want to figure out is what is the definite integral from 0 to 1 of this f of X and like always if you feel inspired I encourage you to feel inspired pause the video and see if you could work through this on your own or at any time while I'm working through it pause it and try to keep on going all right 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 of X is this is this series so I could write the sum from N equals 1 to infinity of n plus 1 over 4 to the n plus 1 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 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 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 of X plus h of X and I just kept going on and on and on DX well this is the same thing as the sum of the integrals as the integral from 0 to 1 of G of X G of X DX plus the integral from 0 to 1 H of 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 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 this is going to be equal to the sum from N equals 1 to infinity of the integral the definite integral of each of these terms so I'm going to write it like this so of the integral from 0 to 1 of n plus 1 over 4 to the n plus 1 power times X to the N then it is DX so once again now we're taking the sum of each of these terms so let's evaluate 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 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 we're going to divide by n plus one so we have this original n plus 1 over 4 to the n plus 1 and 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 the incremented exponent this just comes out of hi I often call it the inverse pot or the the anti power rule or reversing the power rule so it's X to the n plus 1 over n plus 1 I just took the antiderivative now we're going to go from 0 to 1 for each of these terms and before we do that we can simplify we have an N plus 1 we have an N plus 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 equals 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 1 to the n plus 1 over 4 to the n plus 1 actually yeah why don't I write it that way 1 over 1 to the n plus 1 over 4 to the n plus 1 minus 0 to the n plus 1 over 4 to the n plus 1 so we're not going to even have to write that I could write 0 to the n plus 1 over 4 to the n plus 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 equals 1 to infinity we almost are going to get to our drum roll of 1/4 to the n plus 1 now you might immediately recognize this this is an finit geometric series what is the first term here well the first term first first term is well when n is equal to when n is equal to 1 the first term here is 1/4 to the second power did I do that right yeah when n is equal to 1 it's going to be so this is going to be 1/4 to the 2nd power which is equal to 1 over 16 so that's our first term and then our common ratio common common ratio here well that's going to be well we're going to ever we're going to keep multiplying by 1/4 so our common ratio here is 1/4 and so for an infinite geometric series this is it's our common ratio the well is less than or it's it's its absolute value is less than 1 we know that this is going to converge and it's going to converge to the value our first term 1 over 16 divided by 1 minus the common ratio 1 minus 1/4 so this is 3/4 so it's equal to 1 over 16 times 4 over 3 so 1 for this is going to be equal to 1/12 and we're done and this seemed really daunting at first but we just said to realize okay in an integral of a sum even an infinite sum well that's going to 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 realize oh we just have a we have an infinite geometric series which we know how to find the sum of and we're done
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