Evaluate ∫02(x2+x+1)dx as limit of a sumAnswer: 320Solutionf(x)=x2+x+1, a=0,b=2,h=n2,nh=2, with f(rh)=r2h2+rh+1.∫02f(x)dx=h→0limhr=0∑n−1(r2h2+rh+1)=h→0lim[h36(n−1)n(2n−1)+h22(n−1)n+nh]As n→∞ with nh=2:=3(nh)3+2(nh)2+nh=38+2+2=320CBSE Class 12 Board · 2006 · Subjective