;)
Como no indicas n, lo haré para n=2
\
$$\begin{align}&\int_ \frac 1 2^1 x^3 dx\simeq \frac h 3 \Big[f(x_0)+4 f(x_1)+f(x_2) \Big]\\&\\&h= \frac{b-a} n=\frac{1-0.5} 2=\frac 1 4\\&\\&x_0= \frac 1 2\\&\\&x_1= x_0+h= \frac 1 2+ \frac 1 4= \frac 3 4\\&\\&x_2=x_1+h= \frac 3 4+ \frac 1 4=1\\&\\&\int_ \frac 1 2^1 x^3 dx\simeq \frac 1 {12} \Big[f(\frac 1 2)+4 f(\frac 3 4)+f(1) \Big]=\\&\\&\frac 1 {12} \Big[( \frac 1 2)^3+4( \frac 3 4)^3 +1^3 \Big]=\\&\\&\frac 1 {12} \Big[ \frac 1 8+ \frac {27}{16}+1 \Big]= \frac {15}{64}\\&\\&\end{align}$$
Saludos
;)
;)