TITLE: Approximating $\int_0^1 \cos(x^2)dx$ with power series
QUESTION [5 upvotes]: I would like to calculate $\int_0^1 \cos(x^2)dx$ with an error smaller than $10^{-6}$ (this error should be proven.).
I have a strategy, but I am not quite sure if this is a valid one. Here is what I did:
I know that the series representation of the cosine is:$$\cos(x) = \sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k)!}$$
Since the integral boundaries I want to calculate only range from $0$ to $1$, I can use the fact that $a_k :=\frac{x^{2k}}{(2k)!}$ is a falling sequence for $x \in [0,1]$. Then I would use the estimation from the Leibnitz criterion, that for positive falling $a_k$ with 
$$s_n := \sum_{k=1}^n (-1)^k a_k$$
the inequality $|s - s_n|\le a_{n+1}$ holds, whereas $s$ denotes the limit of $s_k$.
Then I could say that for $n=9$:
$$|\cos(x)-s_9| \le \frac{1}{(2\cdot 10)!}$$ 
Then I would calculate $\int_0^1 \sum_{k=0}^9 (-1)^k \frac{x^{4k}}{(2k)!}dx$
The estimate I get seems to be correct, however I am not sure whether or not my prove above is sufficient. Any help would be greatly appreciated!

REPLY [2 votes]: You need to change the order of what what you're doing.  First integrate, then bound the series:
$$\int_0^1 \cos(x^2) \; dx = \int_0^1 \sum_{k=0}^{\infty} \frac{(-1)^k x^{4k}}{(2k)!} \; dx =\sum_{k=0}^{\infty} \frac{(-1)^k}{(4k+1)(2k)!}. $$
So the error in $s_n$ is bounded by $$\frac{1}{(4(n+1)+1)(2(n+1))!}.$$ 
And $n=4$ suffices.