Answer
$\int_{0}^{\pi/2} \dfrac{\sin x}{x} dx \approx 1.37$
Work Step by Step
Simpson's Rule states that
$S_{n}=\dfrac{1}{3}[y_0+4y_1+2y_2+..+4y_{N-3}+2y_{N-2}+4y_{N-1}+y_N]\Delta x$
Since, $\Delta t=\dfrac{\pi/2-0}{8}=\dfrac{\pi}{16} $
Thus, using Simpson's Rule, we have:
$S_{8}= \dfrac{1}{3}(\dfrac{\pi}{16} )[y_0+4y_1+2y_2+4y_3+2y_4+4y_5+2y_6+4y_7+y_8]\\=\dfrac{\pi}{48}[1+4[\dfrac{\sin (\pi/16)}{\pi/16}]+2[\dfrac{\sin (\pi/8)}{\pi/8}]+4[\dfrac{3\sin (\pi/16)}{3\pi/16}]+2[\dfrac{\sin (\pi/4)}{\pi/4}]+4[\dfrac{\sin (5\pi/16)}{5\pi/16}]+2[\dfrac{3\sin (\pi/8)}{3\pi/8}]+4[\dfrac{\sin (7\pi/16)}{7\pi/16}]+[\dfrac{\sin (\pi/2)}{\pi/2}] \approx 1.37$
Hence, $\int_{0}^{\pi/2} \dfrac{\sin x}{x} dx \approx 1.37$
