Let f(x) = $x^{2}-x$ then $\int\limits_{1}^{4}f(x)dx=\int\limits_{1}^{4}(x^{2}-x)dx$ $a = 1, b = 4, nh = 3$ Now, f (a) = f (1) = 1 – 1 = 0 $f(a+h)=f(1+h)$ $=(1+h)^{2}-(1+h)$ $=h^{2}+h$ $f(a+2h)=f(1+2h)$ $=(1+2h)^{2}-(1+2h)$ $=4h^{2}+2h$ $f(a+3h)=f(1+3h)$ $=(1+3h)^{2}-(1+3h)$ $=9h^{2}+3h$ ........................... $f(a+(n-1)h)=f(1+(n-1)h)$ $=(1+(n-1)h)^{2}-(1+(n-1)h)$ $=(n-1)^{2}h^{2}+(n-1)h$ By definition$=\lim\limits_{h\to 0}\left[h^{2}\frac{(n-1)n(2n-1)}{6}+\frac{h(n-1)n}{2}\right]$ $=\lim\limits_{h\to 0}\left[\frac{(3-h)3(6-h)}{6}+\frac{(3-h)3}{2}\right]$ $=\frac{3(3)(6)}{6}+\frac{3(3)}{2}=\frac{27}{2}$