Use Simpson’s Rule with $n=4$ to approximate $\int_{\pi }^{2\pi }\dfrac{\sin x}{x}dx$. Express the answer rounded to three decimal places.

Solution We partition the interval $[\pi ,2\pi ]$ into the four subintervals $$\left[ \pi ,\dfrac{5\pi }{4}\right]\!\qquad \left[ \dfrac{5\pi }{4},\dfrac{ 3\pi }{2}\right]\!\qquad \left[ \dfrac{3\pi }{2},\dfrac{7\pi }{4}\right]\!\qquad \left[ \dfrac{7\pi }{4},2\pi \right]$$

each of width $\dfrac{\pi }{4}.$ The value of $f( x) =\dfrac{\sin x}{x}$ corresponding to each endpoint is $$\begin{eqnarray*} f(\pi) &=& 0\ \ \ f\!\left( \dfrac{5\pi }{4}\right) = - \dfrac{2\sqrt{2}}{5\pi}\ \ \ f\!\left( \dfrac{3\pi }{2} \right) =-\dfrac{2}{3\pi}\ \ \ f\!\left( \dfrac{7\pi }{4}\right) = -\dfrac{2\sqrt{2}}{7\pi}\quad f(2\pi) =0 \end{eqnarray*}$$

Then using Simpson’s Rule, we get $$\begin{eqnarray*} \int_{\pi }^{2\pi }\dfrac{\sin x}{x}\,dx &\approx &\dfrac{2\pi -\pi }{3\cdot 4} \left[ f(\pi )+4\, f\! \left( \dfrac{5\pi }{4}\right) +2 f\! \left( \dfrac{3\pi }{2}\right) +4 f\! \left( \dfrac{7\pi }{4}\right) +f( 2\pi ) \right] \nonumber \\[7pt] &= &\left( \dfrac{\pi }{12}\right) \left[ 0+4\left( -\dfrac{2\sqrt{2}}{ 5\pi }\right) +2\left( -\dfrac{2}{3\pi }\right) +4\left( -\dfrac{2\sqrt{2}}{ 7\pi }\right) +0\right] \approx -0.434 \end{eqnarray*}$$