Question

From the fact that a definite integral is a limit of a sum, we can say that ${\int }_0^2\left(x^2\right)dx$ = .

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

Answer 1

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

By definition we know,
${\int }_a^{b}f\left(x\right)dx$ = lim $n\to \infty \frac{b-a}{n}\left[f\right(a)+f(a+h)+f(a+2h)...f(a+(n-1)h\left)\right]$
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,
${\int }_0^2\left(x^2\right)dx$ = ${lim}_{n\to \infty }\frac{2}{n}\left[f\right(0)+f(0+\frac{2}{n})+f(0+\frac{4}{n})...f(0+(n-1)\frac{2}{n}\left)\right]$
$={lim}_{n\to \infty }\frac{2}{n}[0+\frac{2^2}{n^2}+\frac{4^2}{n^2}+.....\frac{\left(2\right(n-1){)}^2}{n^2}]$
$={lim}_{n\to \infty }\frac{2}{n}\left(\frac{2^2+4^2+6^2+...\left(2\right(n-1){)}^2}{n^2}\right)$
$={lim}_{n\to \infty }\frac{2}{n}\left(\frac{2^2(1^2+2^2+3^2...(n-1{)}^2)}{n^2}\right)$
$={lim}_{n\to \infty }\frac{2}{n}.\frac{4}{n^2}.\frac{(n-1)\left(n\right)\left(2\right(n-1)+1)}{6}$
$={lim}_{n\to \infty }\frac{8}{6}(1-\frac{1}{n})(2-\frac{1}{n})$
$=\frac{8}{6}.2$ (on applying limits)
$=\frac{8}{3}$