Use series to approximate the value of $\int_0^1{cos(x^2)dx}$ so that the error in your approximation is less than $\frac{1}{100}$.
My work:
$$f(x)-T_n(x)=R_n$$
$$R_n=|\frac{f^n (c) x^n}{n!}|$$
$f^n (c)$ is the nth derivative of $cos(x^2)$ and we choose c to be the value on the interval that returns the maximum value. $\frac{d^ncos(x^2)}{dx^n}$ is $(2x)^n \frac{+}{-}cos/sin(x^2)$. The x value on the interval 0 to 1 that returns the biggest value is x=1, and the trigonometric part of the term always returns 1 at that value. (I realize now this is likely a misuse of math).
So we define $f^n (c)$ as $2^n$ (that being the maximum value of the nth derivate on the interval 0 to 1. Now to find the Taylor series that returns a $R_n < \frac{1}{100}$ we just evaluate $\frac{2^n}{n!}$ at different n values. I found that at n=8, the remainder term was less than 0.01.
Now to figure out the rest of the problem, I create a Taylor series of $cos(x^2)$ to the 8th term and then integrate it and evaluate it at 1, right?
Can anyone point out the errors in math I am making. And further, a question I have is that if we are using a Taylor series centered at x=0 to evaluate an $\int_1^2{x}$ and the error bound on the interval from 0 to 2 is $\frac{+}{-} 0.02$ does that mean that using the Taylor series to find the difference in the integrand evaluated at 2 and 1 would have error bound $\frac{+}{-} 0.04$? I wonder this because you are using two terms with an error bound to evaluate a difference.
Considering the fact I likely completely messed up the solving process, it's honestly appropriate if you just discard my work to tell me how to do it right.
edit: I realize now that because this is an alternating series, using the last remaining term as the error bound is more efficient.