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Approximate $\int_0^1 e^{-x^2} d x$ to two decimal places.
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The second derivative of $f=e^{-x^2}$ is $\left(4 x^2-2\right) e^{-x^2}$, and it is not hard to see that on $[0,1],\left|\left(4 x^2-2\right) e^{-x^2}\right| \leq 2$. We begin by estimating the number of subintervals we are likely to need. To get two decimal places of accuracy, we will certainly need $E(\Delta x)<0.005$ or
\begin{aligned} \frac{1}{12}(2) \frac{1}{n^2} &<0.005 \\ \frac{1}{6}(200) &<n^2 \\ 5.77 \approx \sqrt{\frac{100}{3}} &<n \end{aligned}
With $n=6$, the error estimate is thus $1 / 6^3<0.0047$. We compute the trapezoid approximation for six intervals:
$\left(\frac{f(0)}{2}+f(1 / 6)+f(2 / 6)+\cdots+f(5 / 6)+\frac{f(1)}{2}\right) \frac{1}{6} \approx 0.74512 .$
So the true value of the integral is between $0.74512-0.0047=0.74042$ and $0.74512+$ $0.0047=0.74982$. Unfortunately, the first rounds to $0.74$ and the second rounds to $0.75$, so we can't be sure of the correct value in the second decimal place; we need to pick a larger $n$. As it turns out, we need to go to $n=12$ to get two bounds that both round to the same value, which turns out to be $0.75$. For comparison, using 12 rectangles to approximate the area gives $0.7727$, which is considerably less accurate than the approximation using six trapezoids.

In practice it generally pays to start by requiring better than the maximum possible error; for example, we might have initially required $E(\Delta x)<0.001$, or
\begin{aligned} \frac{1}{12}(2) \frac{1}{n^2} &<0.001 \\ \frac{1}{6}(1000) &<n^2 \\ 12.91 \approx \sqrt{\frac{500}{3}} &<n \end{aligned}
Had we immediately tried $n=13$ this would have given us the desired answer.
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