Answer
$$0.358$$
Work Step by Step
Given $$\int_{1}^{4} \frac{d x}{x^{3}+1}$$
Since $\Delta x=\dfrac{b-a}{n}=\dfrac{3}{6}=0.5$, 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_{6}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+2f(x_{3})+\cdots+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 0.358
\end{align*}
