Evaluate: ∫14(x2−x)dx as limit of sums.Answer: 227Solutionf(x)=x2−x, a=1,b=4,h=n3,nh=3, with f(1+rh)=(1+rh)2−(1+rh)=r2h2+rh.∫14f(x)dx=h→0limhr=0∑n−1(r2h2+rh)=h→0lim[h36(n−1)n(2n−1)+h22(n−1)n]As n→∞ with nh=3:=3(nh)3+2(nh)2=327+29=9+29=227CBSE Class 12 Board · 2010 · Subjective