Question

Evaluate the definite integral as limit of sums:
${\int }_0^3{x}^{2}dx$

Answer 1

${\int }_2^3{x}^{2}dx$

${\int }_a^{b}f\left(x\right)dx$

$=(b-a)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h\left)\right]$

Putting $a=2$ and $b=3$

$h=\frac{b-a}{n}=\frac{3-2}{n}=\frac{1}{n}$

$f\left(x\right)=x^3$

${\int }_2^3{x}^{2}dx$

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

$=\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(2)+f(2+h)+f(2+2h)+...+f(2+(n-1)h\left)\right]$

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

$=\underset{n\to \infty }{lim}\frac{1}{n}(2^2+(2^2+h^2+4h)+(2^2+4h^2+8h)+...+(2^2+\left(\right(n-1)h{)}^2+4(n-1\left)h\right))$

$=\underset{n\to \infty }{lim}\frac{1}{n}\left[\right[\underset{}{2^2+2^2+...+2^2}]+[h^2+2^2{h}^2+...+(n-1{)}^2{h}^2]+[4h+8h+...+4(n-1\left)h\right]$

$=\underset{n\to \infty }{lim}\frac{1}{n}[4n+\frac{h^2(n-1)\left(n\right)(2n-1)}{6}+\frac{4h(n-1)\left(n\right)}{2}]$

$[\because 1^2+2^2+...n^2=\frac{n(n+1)(2n+1)}{6}\&1+2+3+...+n=\frac{n(n+1)}{2}]$

$=\underset{n\to \infty }{lim}n[4n+{\left(\frac{1}{n}\right)}^2\frac{(n-1)\left(n\right)(2n-1)}{6}+\left(\frac{1}{n}\right)\frac{4(n-1)\left(n\right)}{2}][\because h=\frac{1}{n}]$

$=\underset{n\to \infty }{lim}[\frac{4n}{n}+\frac{(n-1)\left(n\right)(2n-1)}{6n^3}+\frac{2(n-1)\left(n\right)}{n^2}]$

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

$=4+\frac{(1-0)(2-0)}{6}+\frac{2(1-0)}{1}$

$=4+\frac{1}{3}+2$

$=\frac{19}{3}$