Answer
$$M_6=63.28,\ \ T_6=64.68$$
Work Step by Step
Given $$\int_{1}^{4} x^{3} d x $$
Since $\Delta x=\dfrac{b-a}{n}=\dfrac{3}{6}=\frac{1}{2}$
Therefore, the sub intervals consist of
$$[1,1.5],[1.5,2],[2,2.5],[2.5,3],[3,3.5], [3.5,4] $$
The midpoints of these sub intervals are
$$\{1.25,1.75,2.25,2.75, 3.25,3.75\}$$
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_6)\right]\Delta x\\
&=\frac{1}{2}\left[ f(1.25)+ f(1.75)+ f(2.25)+ f(2.75)+ f(3.25)+f(3.75)\right]\\
&\approx 63.28
\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_{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\\
&=\dfrac{1}{4}\left[f(1)+2f(1.5)+2f(2)+2f(2.5)+2f(3)+2f(3.5)+f(4)\right] \\
&\approx 64.68
\end{align*}
