Approximating Definite Integrals — Question 6

Question 6 – Solution

Step 1: Write the trapezoidal error bound. $$|E_T|\le\frac{K(b-a)^3}{12n^2}.$$ Here $K=e$ and $b-a=1$, so $$|E_T|\le\frac{e}{12n^2}.$$ Step 2: Require the bound to be at most $10^{-4}$. $$\begin{align*} \frac{e}{12n^2}&\le10^{-4},\\ e&\le0.0012n^2,\\ n^2&\ge\frac{e}{0.0012}\approx2265.23,\\ n&\ge\sqrt{2265.23}\approx47.594. \end{align*}$$ Step 3: Choose the smallest integer. $$\boxed{n=48}$$