Answer
$$ M_4=5.333,\ \ \ T_4=5.146 $$
Work Step by Step
Given $$\int_{0}^{4} \sqrt{x} d x $$
Since $\Delta x=\dfrac{b-a}{n}=\dfrac{4}{4}=1$
Therefore, the sub intervals consist of
$$[0,1],[1,2],[2,3],[3,4] $$
The midpoints of these sub intervals are
$$\{0.5,1.5,2.5,3.5\}$$
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\\
&=\left[ f(0.5)+ f(1.5)+ f(2.5)+ f(3.5)\right]\\
&\approx 5.33333
\end{align*}
To find $T_4$
\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_{4}&= \frac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+2f(x_{3})+f(x_4)\right]\Delta x\\
&=\dfrac{1}{2}\left[f(0)+2f(1)+2f(2)+2f(3)+f(4)\right] \\
&\approx 5.1462
\end{align*}
