Answer
$$T_{6}=1.17 ,\ \ \ \ M_{6}= 1.206$$
Work Step by Step
Given $$\int_{0}^{\pi/2}\sqrt{\sin x}d x,\ \ \ N=6 $$
Since $\Delta x=\dfrac{b-a}{N}=\dfrac{\pi/2}{6}=\frac{\pi}{12} $
Therefore, the sub intervals consist of
$$[0,\pi/12],[\pi/12,2\pi/12],[2\pi/12,3\pi/12],[3\pi/12,4\pi/12],[4\pi/12,5\pi/12],[5\pi/12,\pi/2]$$
The midpoints of these sub intervals are
$$\{\frac{\pi}{24}, \frac{3 \pi}{24}, \ldots, \frac{11 \pi}{24}\}$$
Hence
\begin{align*}
M_{n}&= \sum_{i=1}^{n}f(m_i)\Delta x\\
M_{6}&= \left[ f(m_1)+ f(m_2)+ ..............+f(m_5)\right]\Delta x\\
&=\frac{\pi}{12}\left[ f( \pi/24)+ f(3\pi/24)+\cdots + f(11\pi/24)\right]\\
&\approx 1.206
\end{align*}
Now find $T_6$
\begin{align*}
T_{n}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+.....+2f(x_{n-1})+f(x_n)\right]\Delta x\\
T_{6}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+2f(x_{3})+2f(x_4)+2f(x_5)+f(x_6)\right]\Delta x\\
&=\frac{\pi}{24}\left[f(0)+2f(\pi/12)+2f(\pi/6)+\cdots +2f(5\pi/12)+f(\pi/2)\right] \\
&\approx 1.17
\end{align*}
