Question

Evaluate the following integral by expressing as a limit of sum
${\int }_0^1{x}^{2}dx=\frac{A}{B}$

Find $A+B$

Answer 1

$weknow$
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx={lim}_{h\to 0}h[f\left(a\right)+f(a+h)+f(a+2h)+.......+f(a+(n-1)h)]$
$h=\frac{b-a}{n}$
$\underset{0}{\overset{1}{\int }}\left(x^2\right)dx={lim}_{h\to 0}h[f\left(0\right)+f(0+h)+f(0+2h)+.......+f(0+(n-1)h)].......\left(i\right)$
$h=\frac{1-0}{n}=\frac{1}{n}$
$f\left(0\right)={\left(0\right)}^2=0$
$f(0+h)={(0+h)}^2=h^2$
$f(0+2h)={(0+2h)}^2=4h^2$
$f(0+(n-1)h)={(0+(n-1)h)}^2={(n-1)}^2{h}^2$
$pluggingin\left(i\right)$
$\underset{1}{\overset{2}{\int }}(2x+5)dx={lim}_{h\to 0}h[0+\left(h^2\right)+\left(4h^2\right)+.....+\left({(n-1)}^2{h}^2\right)]$
$={lim}_{h\to 0}h\left[h^2(1+4+........{(n-1)}^2)\right]$
$={lim}_{h\to 0}h\left[h^2(1^2+2^2+........{(n-1)}^2)\right]$
$={lim}_{h\to 0}h\left[h^2\left(\frac{n(n-1)(2n-1)}{6}\right)\right]$
$={lim}_{n\to \propto }\frac{1}{n}\left[\frac{1}{n^2}\left(\frac{n(n-1)(2n-1)}{6}\right)\right]$
$={lim}_{n\to \propto }\frac{1}{n}\left[\frac{n^2}{n}\left(\frac{(1-\frac{1}{n})(2-\frac{1}{n})}{6}\right)\right]$
$={lim}_{n\to \propto }\frac{1}{n}\left[n\left(\frac{(1-\frac{1}{n})(2-\frac{1}{n})}{6}\right)\right]$
$={lim}_{n\to \propto }\left[\left(\frac{(1-\frac{1}{n})(2-\frac{1}{n})}{6}\right)\right]$
$=\frac{(1-0)(2-0)}{6}$
$=\frac{1}{3}$