Estimate the error that results from using Simpson's Rule with $n =4$ to approximate $\int_{\pi }^{2\pi }\frac{\sin x}{x}dx$. (See Example 6.)

Solution To estimate the error, we need to find the absolute maximum value of $\vert f^{(4) }\vert$ on the interval $[\pi ,2\pi ]$. We begin by finding $f^{(4) }$: $$\begin{eqnarray*} f(x) &=&\dfrac{\sin x}{x} \\[5pt] f^{\prime} (x) &=&\frac{x\cos x-\sin x}{x^{2}}=\frac{\cos x}{x}-\frac{\sin x}{x^{2}} \\ && \\ f^{\prime \prime} (x) &=&\left[ \frac{-x\sin x-\cos x}{x^{2}} \right] -\left[ \frac{x^{2}\cos x-2x\sin x}{x^{4}}\right]\\[6pt] & = & \frac{2 \sin x}{x^{3}} -\frac{\sin x}{x}-\frac{2\cos x}{x^{2}} \\[15pt] f^{\prime \prime \prime} (x) &=&\left[ \frac{2x^{3}\cos x-6x^{2}\sin x}{x^{6}}\right] -\left[ \frac{x\cos x-\sin x}{x^{2}}\right] - \left[ \frac{-2x^{2}\sin x-4x\cos x}{x^{4}}\right] \\[6pt] &=&\frac{2\cos x}{x^{3}}-\frac{6\sin x}{x^{4}}-\frac{\cos x}{x}+\frac{ \sin x}{x^{2}}+\frac{2\sin x}{x^{2}}+\frac{4\cos x}{x^{3}}\\[6pt] &=&\frac{6 \cos x}{x^{3}}-\frac{\cos x}{x}+\frac{3 \sin x}{x^{4}} \\[6pt] && \\ f^{\,(4) }(x) &=&\left[ \frac{-6x^{3}\sin x-18x^{2}\cos x}{ x^{6}}\right] -\left[ \frac{-x\sin x-\cos x}{x^{2}}\right] \\[8pt] &&+\,\left[ \frac{3x^{2}\cos x-6x\sin x}{x^{4}}\right] -\left[ \frac{6x^{4}\cos x-24x^{3}\sin x}{x^{8}}\right] \\[8pt] &=&\frac{4 \cos x}{x^{2}}-\frac{24 \cos x}{x^{4}}+\frac{\sin x}{x}-\frac{ 12 \sin x}{x^{3}}+\frac{24 \sin x}{x^{5}} \end{eqnarray*}$$

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We use graphing technology to find the absolute maximum value of $\vert f^{(4)}\vert$. See Figure 20.

Figure 20 $y=\vert f^{(4)}\vert$ on the interval $[\pi, 2\pi]$.

Since on the interval $[\pi ,2\pi]$ the maximum value of $\vert f^{(4)}\vert \lt 0.176,$ we use $M=0.176$. Then the error that results from using Simpson's Rule to approximate $\int_{\pi }^{2\pi }\frac{\sin x}{x}dx$ is $$\hbox{Error} \leq \frac{M(b-a)^{5}}{180n^{4}}=\frac{0.176(2\pi -\pi )^{5}}{180\cdot 4^{4}}\approx 0.001$$

That is, $$\begin{eqnarray*} -0.434-0.001 & \leq& \int_{\pi }^{2\pi }\frac{\sin x}{x}dx \leq -0.434+0.001 \\[5pt] -0.435 & \leq& \int_{\pi }^{2\pi }\frac{\sin x}{x}dx \leq -0.433 \end{eqnarray*}$$