Answer
$$\approx0.7450$$
Work Step by Step
Given $$\int_{\pi / 4}^{\pi / 2} \sqrt{\sin \theta} d \theta$$
We have: $\Delta x=\dfrac{b-a}{n}=\dfrac{\pi}{16}$
Therefore, the midpoints of these sub-intervals are
$$\frac{9 \pi}{32}, \frac{11 \pi}{32}, \frac{13 \pi}{32},\frac{15 \pi}{32} $$
Hence
\begin{align*}
M_{n}&= \sum_{i=1}^{n}f(m_i)\Delta x\\
M_{4}&= \left[ f(m_1)+ f(m_2)+ ..+f(m_4)\right]\Delta x\\
&=\Delta x\left(f\left(\frac{9 \pi}{32}\right)+f\left(\frac{11 \pi}{32}\right)+f\left(\frac{13 \pi}{32}\right)+f\left(\frac{15 \pi}{32}\right)\right)\\
&=\frac{\pi}{16}(\sqrt{\sin \frac{9 \pi}{32}}+\sqrt{\sin \frac{11 \pi}{32}}+\sqrt{\sin \frac{13 \pi}{32}}+\sqrt{\sin \frac{15 \pi}{32}})\\
&\approx 0.744978\approx0.7450
\end{align*}
