Asymptotic error formula for the trapezoidal rule

[kalish1]

I need to use the asymptotic error formula for the trapezoidal rule to estimate the number n of subdivisions to evaluate $\int_{0}^{2}e^{-x^2}dx$ to the accuracy $\epsilon=10^{-10}$. I also need to find the approximate integral in this case. I would like to know if my attempt is correct. Thanks in advance for any help.

**My attempt:** $E_n^T(f)\approx -h^2/12[f'(b)-f'(a)]. f(x)=e^{-x^2}, f'(x)=-2xe^{-x^2}.$

So $E_n^T(f)\approx h^2/(4e^4)$ and since $h=1/n$, we have to find an n that satisfies the inequality $\frac{1}{4e^4n^2}\leq 10^{-10}$. We obtain $n \approx 6767$. The approximate integral is ??

 

[Ackbach]

Hmm. I don't quite get what you get. I have
\begin{align*}
E_{N}(f)& \approx - \frac{(b-a)^{2}}{12 N^{2}}[f'(b)-f'(a)] \\
&= \frac{2^{2}}{12 N^{2}}[4e^{-4}-0] \\
&= \frac{4}{3 N^{2}e^{4}}.
\end{align*}
Hence, we need
$$ \frac{4}{3 N^{2}e^{4}} \le 10^{-10},$$
or
$$N \ge \frac{2 \cdot 10^{5}}{ \sqrt{3} \, e^{2}} \approx 15627.2,$$
so let $N=15628$. You must set up a trapezoidal scheme with this many sub-intervals, and evaluate. You're going to have:
$$A \approx \frac{2-0}{2 \cdot 15628} \sum_{j=0}^{15627}[f(x_{j})+f(x_{j+1})]
=\frac{1}{15628} \sum_{j=0}^{15627}[e^{-(2j/15628)^{2}}+e^{-(2(j+1)/15628)^{2}}]$$
$$=\frac{1}{15628} \sum_{j=0}^{15627}[e^{-(j/7814)^{2}}+e^{-((j+1)/7814)^{2}}].$$

 

[kalish1]

Great! CCC. (Crystal clear clarification.)