#### opus

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**1. The problem statement, all variables and given/known data**

Determine the series that is equal to the integral ##\int_0^1 x^2\cos(x^3)dx##

**2. Relevant equations**

**3. The attempt at a solution**

So I didn't really know what I was doing but I did end up with the correct solution.

What I did was to find a Taylor Series for the integrand, this turned out to be ##\sum_{n=0}^\infty (-1)n\frac{x^{6n+2}}{(2n)!}##

Now I have ##\int_0^1\sum_{n=0}^\infty (-1)^n\frac{x^{6n+2}}{(2n)!}dx##

After some research, I found that for power series (for which a Taylor series is) you can put the integral inside the sum (if this is incorrect, please let me know).

Now I have ##\sum_{n=0}^\infty (-1)^n\frac{1}{(2n)!}\int_0^1 x^{6n+2}dx##

This looks like a mess, but not sure how I should have grouped this the proper way. That is, should I have put everything after ##\sum## in brackets?

Eventually, I end up evaluating and I get ##\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!(6n+3)}##

Now normally, a series would never equal an integral to my understanding because series are for positive integers. But this is a power series, which involves x. So is it true that this is why the series can equal the integral in this case? Because x can take on ANY real positive values now?

I guess thinking about a series like a "bar graph" kind of thing isn't helping me much in trying to understand this.