Minimum number of subintervals n for the Composite Trapezoidal Rule

Question

I am trying to compute the minimum number of subintervals n for the Composite Trapezoidal Rule, in order for the approximation of the following integral to have 5 decimals correct. $$\int_0^2 \frac{1}{x+4}dx$$ So I am using: $$|E_n^T(f)|\leq\frac{(b-a)h^2}{12}M_2<ε$$ where $\mathbf{M_2}=\max_{x\in [a,b]} |f''(x)|=\frac{1}{32}$, $\mathbf{h}=\frac{b-a}{n}$, $\mathbf{ε}=\frac{1}{2}\cdot10^{-5}$. Solving for $n$ I get $n>64.5497...$, so I say that the minimum number of subintervals to achieve 5 decimals is 65. However, using matlab (or an online tool) I see that the Trapezoidal Rule needs only $n=48$ subintervals to achieve this accuracy. What am I doing wrong? Is something wrong with my calculations or is it possible that the number of subintervals needed in practice can actually be less than 65?

Answer 1

If you look at the formula $\frac2{(x+4)^3}$ for the second derivative, you can see that at $x=2$ it will have a somewhat smaller value than at $x=0$. From this lower bound of the derivative you can compute a lower bound for $n$, the truly minimal $n$ will be somewhere in-between.

You could also use the more exact error formula $$ \int_a^b f(x)dx=T(h)-\frac{h^2}{12}(f'(b)-f'(a))+\frac{h^4(b-a)}{720}f^{(4)}(\eta) $$ to get from the term of second degree in $h$ the estimate $$ n^2\approx\frac{(b-a)^2}{12}\frac{6^2-4^2}{6^2\cdot 4^2}\cdot 2\cdot10^5=\frac{10^6}{12\cdot 36} \implies n\approx 48.1125, $$ giving $n=49$ as the best number of segments. The fourth-order correction should not change this value.