Answer
$$25.976$$
Work Step by Step
Given $$\int_{2}^{4} \sqrt{6 t^{3}+1} d t $$
Since $\Delta x=\dfrac{b-a}{n}=\dfrac{2}{3} $, then \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_{3}&= \dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+2f(x_{3}) \right]\Delta x\\
&=\dfrac{1}{3}(\sqrt{6(2)^{3}+1}+2 \sqrt{6\left(\frac{8}{3}\right)^{3}+1}+2 \sqrt{6\left(\frac{10}{3}\right)^{3}+1}+\sqrt{6(4)^{3}+1})\\
&\approx 25.976
\end{align*}
