Answer
$\int_{0}^{\pi}x~sin^2~x~dx \approx 2.4674$
Work Step by Step
$\Delta x = \frac{b-a}{n} = \frac{\pi-0}{4} = \frac{\pi}{4}$
We can find the midpoints of the four subintervals:
$x_1 = \frac{\pi}{8}$
$x_2 = \frac{3\pi}{8}$
$x_3 = \frac{5\pi}{8}$
$x_4 = \frac{7\pi}{8}$
$\int_{0}^{\pi}x~sin^2~x~dx \approx \sum_{i=1}^{4} f(x_i)~\Delta x$
$\int_{0}^{\pi}x~sin^2~x~dx \approx \frac{\pi}{4}\cdot (\frac{\pi}{8}~sin^2~\frac{\pi}{8}+\frac{3\pi}{8}~sin^2~\frac{3\pi}{8}+\frac{5\pi}{8}~sin^2~\frac{5\pi}{8}+\frac{7\pi}{8}~sin^2~\frac{7\pi}{8})$
$\int_{0}^{\pi}x~sin^2~x~dx \approx \frac{\pi}{4}\cdot (3.14159)$
$\int_{0}^{\pi}x~sin^2~x~dx \approx 2.4674$
