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Problem 31 Hard Difficulty

Use a power series to approximate the definite integral to six decimal places.
$ \int^{0.2}_0 x \ln (1 + x^2) dx $

Answer

$\approx 0.000395$

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Video Transcript

So what we have here is the integral from 0 to 0 to x times the natural log of one plus X squared DX, um, and the Power Series. What we have is the summation. This is the natural log of one plus X equals the summation of an equals one to infinity of negative one to the end, minus one of X over and corrects to the end over. And if we replace X with X squared, what we end up getting is the summation from n equals one to infinity of negative one to the N minus one of times X to the two n over end. So then when we multiply by X, what we get here when we multiply by X that's going to give us. Um, since we're doing that, it's going to end up giving us again the summation. So we can just copy that portion right here. The only thing that's gonna be changing is what's in the top. So as a result, we can duplicate this and this is just going to have a plus one right here. Then we can integrate this, um, integrating the summation. What? This is gonna look like is just now this from 0 to 0 to, um, and recall that will be integrate. Now, this is going to be, um, two and plus two. And and this is going to be X to the two n plus two. Um, Then what we end up getting is that this is going to equal summation from zero to infinity. We're sorry and equals one to infinity of negative one to the N minus one of 0.2 to end plus two over two and plus two times n minus zero. If we look at how and how what happens as an increases, we see that the term is, uh, very small. So we'll use the alternating series estimation therapy. So we have is that s minus two absolute value. That is less than ankle to be three, which is less than zero point 000 000107 So knowing this, we just need to add all the terms before B three, and what we end up getting as a result is that this is going to end up giving us approximately 0.0 395