Answer
$$T_4=2.75,\ \ M_4= 2.625 $$
Work Step by Step
Given $$\int_{0}^{2} x^{2} d x $$
We know that
$\Delta x=\dfrac{b-a}{n}=\dfrac{2}{4}=\frac{1}{2}$
Therefore, the sub intervals consist of
$$[0,0.5],[0.5,1],[1,1.5],[1.5,2] $$
The midpoints of these sub intervals are
$$\{0.25,0.75,1.25,1.75\}$$
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\\
&=\frac{1}{2}\left[ f(0.25)+ f(0.75)+ f(1.25)+ f(1.75)+ f(9)\right]\\
&\approx 2.625
\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}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+2f(x_{3})+f(x_4)\right]\Delta x\\
&=\dfrac{1}{4}\left[f(0)+2f(0.5)+2f(1)+2f(1.5)+f(2)\right] \\
&\approx 2.75
\end{align*}
