Question

Find ${\int }_0^2(x^2+1)dx$ as the limit of a sum :

• (A) $\frac{4}{3}$
• (B) $\frac{14}{3}$
• (C) $\frac{14}{5}$
• (D) None of these

Answer 1

The Correct Option is C

Explanation:
We know that:${\int }_a^{b}f(x)dx=(b-a)\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}(f(a)+f(a+h)+\dots +f(a+(n-1)h))$Putting $a=0,b=2,h=\frac{b-a}{n}=\frac{2-0}{n}=\frac{2}{n}$in ${\int }_0^2{x}^2+1dx$$I=(2-0)\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}(f(0)+f(n)+f(2n)+\dots +fn-1)h$$f(0)=1$$f(h)=h^2+1$$=\left(\frac{4}{n^2}\right)+1$$f((n-1)h)=(n-1{)}^2\times \frac{4}{n^2}+1$$\therefore I=2\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}((1+1+\dots n\text{ times })+(0+\frac{4}{n^2}+\frac{16}{n^2}+\dots +\frac{(n-1{)}^2}{n^2}))$$=2\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}(n+\frac{4}{n}\frac{(n-1)n(2n-1)}{6})$$=2\underset{n\to \mathrm{\infty }}{lim}(1+\frac{2}{3}(1-\frac{1}{n})(2-\frac{1}{n}))$$=2\times (1+\frac{4}{3})$$=\frac{14}{3}$Hence, the correct option is (C).