Solution  (a) We partition the interval $[0,\pi]$ into four subintervals, each of width $\Delta x=\dfrac{\pi -0}{4}=\dfrac{\pi }{4}.$ $$\left[ 0,\dfrac{\pi }{4}\right] \qquad \left[ \dfrac{\pi }{4},\dfrac{ \pi }{2}\right]\hbox{} \qquad \left[ \dfrac{\pi }{2},\dfrac{3\pi }{4}\right] \hbox{}\qquad \left[\dfrac{3\pi }{4},\pi \right]$$

510

The values of $f( x) =\sin x$ corresponding to each endpoint are $$f(0) =0 \!\qquad f\!\left( \dfrac{\pi }{4} \right) =\dfrac{\sqrt{2}}{2}\!\qquad f\!\left( \dfrac{\pi }{2} \right) =1\!\qquad f\!\left( \dfrac{3\pi }{4}\right) = \dfrac{\sqrt{2}}{2}\!\qquad f (\pi) =0$$

See Figure 14. Now we use the Trapezoidal Rule: $$\begin{eqnarray*} \int_{0}^{\pi }\sin x\,dx &\approx &\dfrac{\pi -0}{2\cdot 4}\,\left[ \sin (0)+2\,\sin \left( \dfrac{\pi }{4}\right) +2\,\sin \left( \dfrac{\pi }{2} \right) +2\,\sin \left( \dfrac{3\pi }{4}\right) +\sin (\pi )\right] \\[6pt] &=&\dfrac{\pi }{8}\,\left[ 0+2\left( \dfrac{\sqrt{2}}{2}\right) +2\left( 1\right) +2\left( \dfrac{\sqrt{2}}{2}\right) +0\right] =\dfrac{\pi }{8} \big( 2+2\sqrt{2}\big) \\[6pt] &=& \dfrac{\pi }{4}\big( 1+\sqrt{2}\big) \approx 1.896 \end{eqnarray*}$$

Figure 14 $n=4$.

To approximate $\int_{0}^{\pi }\sin x\,dx$ using six subintervals, we partition $[0,\pi]$ into six subintervals, each of width $\Delta x=\dfrac{\pi -0}{6}=\dfrac{\pi }{6},$ namely, $$\left[ 0,\dfrac{\pi }{6}\right]\! \qquad \left[ \dfrac{\pi }{6},\dfrac{ \pi }{3}\right]\!\qquad \left[ \dfrac{\pi }{3},\dfrac{\pi }{2}\right]\! \qquad \left[ \dfrac{\pi }{2},\dfrac{2\pi }{3}\right]\!\qquad \left[\dfrac{2\pi }{3},\dfrac{5\pi }{6}\right]\! \qquad \left[ \dfrac{ 5\pi }{6},\pi \right]$$

See Figure 15. Then we use the Trapezoidal Rule. $$\begin{eqnarray*} \int_{0}^{\pi }\sin x\,dx &\approx & \dfrac{\pi }{2\cdot 6}\left[ 0+2\cdot \dfrac{1}{2}+2\cdot \dfrac{\sqrt{3}}{2}+2\cdot 1+2\cdot \dfrac{\sqrt{3}}{2} +2\cdot \dfrac{1}{2}+0\right] \\[5pt] &=&\dfrac{\pi }{12}(4+2\sqrt{3}) = \dfrac{\pi }{6}(2+\sqrt{3}) \approx 1.954 \end{eqnarray*}$$

Figure 15 $n=6.$

(b) The exact value of the integral is $$\int_{0}^{\pi }\sin x \, dx = \big[-\cos x\big] _{0}^{\pi }=-\cos \pi +\cos 0=1+1=2$$

The approximation using the Trapezoidal Rule with four subintervals underestimates the integral by $0.104.$ The approximation using six subintervals underestimates the integral by $0.046.$ Notice that the approximation using six subintervals is more accurate than the approximation using four subintervals.