Examples

Simpson’s Rule on the Gaussian distribution

The function $e^{x^2}$ is very important for probability and statistics, but it cannot be integrated analytically.

Apply Simpson’s Rule to approximate the integral:

$${\int }_0^1{e}^{x^2}dx$$

with $\mathrm{\Delta }x=0.1$ and $n=10$. What error bound is guaranteed for this approximation?

Solution

(1) We need a table of values of $x_i$ and $y_i=f(x_i)$:

$x_i:$	$0.0$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$	$0.7$	$0.8$	$0.9$	$1.0$
$f(x_i):$	$e^{{0.0}^2}$	$e^{{0.1}^2}$	$e^{{0.2}^2}$	$e^{{0.3}^2}$	$e^{{0.4}^2}$	$e^{{0.5}^2}$	$e^{{0.6}^2}$	$e^{{0.7}^2}$	$e^{{0.8}^2}$	$e^{{0.9}^2}$	$e^{{1.0}^2}$
$\approx$	$1.000$	$1.010$	$1.041$	$1.094$	$1.174$	$1.284$	$1.433$	$1.632$	$1.896$	$2.248$	$2.718$

These can be plugged into the Simpson Rule formula to obtain our desired approximation:

$$\begin{array}{c}{S}_{10}=\frac{1}{3}\cdot 0.1\cdot (1.000+4\cdot 1.010+2\cdot 1.041+4\cdot 1.094+\cdots +2\cdot 1.896+4\cdot 2.248+2.718)\\ \\ \approx 1.463\end{array}$$

To find the error bound we need to find the smallest number we can manage for $K_4$.

Take four derivatives and simplify:

$$f^{(4)}(x)=(12+48x^2+16x^4)e^{x^2}$$

On the interval $x\in [\mathrm{0,1}]$, this function is maximized at $x=1$. Use that for the optimal $K_4$:

$$f^{(4)}(1.000)=206.589$$

Finally we plug this into the error bound formula:

$$\frac{K_4{(b-a)}^5}{180n^4}=\frac{206.589\cdot {1.000}^5}{180\cdot {10}^4}\approx 0.0001$$ $$\gg \gg \mathrm{Error}(S_{10})\le 0.0001$$