Question

Find $\underset{0}{\overset{2}{\int }}{x}^{2}dx$ as the limit of a sum

• A. $4$
• B. $8$
• C. $\frac{4}{3}$
• D. $\frac{8}{3}$

Answer 1

The correct option is D $\frac{8}{3}$

By definition we know,
$\underset{b}{\overset{a}{\int }}f\left(x\right)dx=\underset{n\to \infty }{lim}\frac{b-a}{n}\left[f\right(a)+f(a+h)+f(a+2h)...f(a+(n-1)h\left)\right]$
Where $h=\frac{b-a}{n}$
Here, $a=0;b=2;f\left(x\right)=x^2;h=\frac{2-0}{n}=\frac{2}{n}$
Therefore,
$\underset{0}{\overset{2}{\int }}\left(x^2\right)dx=\underset{n\to \infty }{lim}\frac{2}{n}\left[f\right(0)+f(0+\frac{2}{n})+f(0+\frac{4}{n})...f(0+(n-1)\frac{2}{n}\left)\right]$
$=\underset{n\to \infty }{lim}\frac{2}{n}[0+\frac{2^2}{n^2}+\frac{4^2}{n^2}+.....\frac{2(n-1{)}^2}{n^2}]$
$=\underset{n\to \infty }{lim}\frac{2}{n}\frac{2^2+4^2+6^2+...2(n-1{)}^2}{n^2}$
=$\underset{n\to \infty }{lim}\frac{2}{n}\frac{2^2(1^2+2^2+3^2...(n-1{)}^2)}{n^2}$
$=\underset{n\to \infty }{lim}\frac{2}{n}.\frac{4}{n^2}.\frac{(n-1)\left(n\right)\left(2\right(n-1)+1)}{6}$
$=\underset{n\to \infty }{lim}\frac{8}{6}(1-\frac{1}{n})(2-\frac{1}{n})$
$=\frac{8}{6}.2$ (on applying limits)
$=\frac{8}{3}$