# Thread: Asymptotic error formula for the trapezoidal rule

1. ## Asymptotic error formula for the trapezoidal rule

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 ??

2. ## Re: Asymptotic error formula for the trapezoidal rule

Rewriting this with the "tex" boxes:
Originally Posted by abscissa
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 $h=\frac{1}{n}$, 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 ??