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