Question

Evaluate: ${\int }_2^3{x}^{2}dx$ as a limit of sum.

Answer 1

$I=\stackrel{3}{\underset{2}{\int }}{x}^{2}dx$

$=(3-2)\underset{n\to \infty }{lim}\frac{1}{n}\left(\right(2{)}^2+(2+h{)}^2+(2+2h{)}^2+(2+(n-1\left)h{)}^2\right)$

$=\underset{n\to \infty }{lim}\left(n\right(2^2)+h^2(1+2^2+3^2+...(n-1{)}^2)+\left(2\right)\left(2\right)\left(h\right)(1+2+3+.....(n-1\left)\right)$

$=\underset{n\to \infty }{lim}\frac{1}{n}(4n+h^2(n-1\left)\right(n\left)\right(2n-1)+4h\frac{(n-1)\left(n\right)}{2})$

we know $h=\frac{3-2}{n}=\frac{1}{n}$

$\therefore$

$\underset{n\to \infty }{lim}(4+\frac{(1-\frac{1}{n})(2-\frac{1}{n})}{6n^2}+\frac{2(n-1)}{n})$

$\underset{n\to \infty }{lim}(4+\frac{(1-\frac{1}{n})(2-\frac{1}{n})}{6}+2(1-\frac{1}{n}))$

$=4+\frac{1\times 2}{6}+2$

$=6+\frac{1}{3}=\frac{19}{3}$