If we use the composite trapezoidal rule, then what is the least number of divisions $N$ for which the error of the integral $\int^1_0{e^{-x}}dx$ doesn't exceed $\frac{1}{12}\times10^{-2}$.My guess is 11 or 5. Kindly tell me which of the answer is correct?I obtained 11 as the answer by applying the formula $$\Big|\frac{(b-a)^3}{12N^2}\times{(e^{-x})^{\prime\prime}_{x=\varepsilon}}\Big| \text{ = } {\frac{10^{-2}}{12}}$$ where $\varepsilon \in [0,1]$ is chosen so that it maximizes the value of $e^{-\varepsilon}$ (which I believe occurs at $\varepsilon = 0$).Solving this equation, I get $N = 10$ (i.e. I must have atleast 11 equidistant divisions if I have to keep the error less than  $\frac{10^{-2}}{12}$).As far as 5 is concerned, I just used 5 equidistant intervals i.e $0,\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},1$ and I applied them in the trapezoidal rule. Now here's the problem:- When calculated the answer at $N=5$, I got a value greater that the actual value of the integral. Is it possible? If yes, why?Which of my answer is correct because I obtained 11 by a well-established formula and 5 was just an option I hit upon and am not really sure of 5's correctness.ThanksNote:- I am posting this question because everywhere else nobody is giving any reply at all. I don't know if this question belongs here. If it does, then kindly reply. If it doesn't,then feel free to erase or delete or whatever :)
Error calculation in trapezoidal rule
numerics;quadrature