The integral is approximated by the sum,
\(\displaystyle\int_0^2f(x)\,\mathrm dx\approx\sum_{n=0}^4\frac12\times\frac{f(x_n)+f(x_{n+1})}2=\frac14\sum_{n=0}^3(f(x_n)+f(x_{n+1}))\)
where \(f(x)=x^3\) and \(x_n=\dfrac12n\), giving you
\(\displaystyle\frac14\sum_{n=0}^3\bigg(\left(\frac n2\right)^3+\left(\frac{n+1}2\right)^3\bigg)\)
\(\displaystyle\frac1{32}\sum_{n=0}^3(n^3+(n+1)^3)\)
\(\displaystyle\frac1{32}\sum_{n=0}^3(2n^3+3n^2+3n+1)\)
Faulhaber's formulas make short work of computing the sum. You have
\(\displaystyle\sum_{n=0}^k1=k+1\)
\(\displaystyle\sum_{n=0}^kn=\frac{k(k+1)}2\)
\(\displaystyle\sum_{n=0}^kn^2=\frac{k(k+1)(2k+1)}6\)
\(\displaystyle\sum_{n=0}^kn^3=\frac{k^2(k+1)^2}4\)
which gives
\(\displaystyle\frac1{16}\sum_{n=0}^3n^3+\frac3{32}\sum_{n=0}^3n^2+\frac3{32}\sum_{n=0}^3n+\frac1{32}\sum_{n=0}^31\)
\(\displaystyle\frac{36}{16}+\frac{42}{32}+\frac{18}{32}+\frac4{32}\)
\(\implies\displaystyle\int_0^2x^3\,\mathrm dx\approx\frac{17}4=4.25\)
