Question

Evaluate the given definite integrals as limit of sums:
${\int }_2^3{x}^{2}dx$

Answer 1

It is known that,
${\int }_a^{b}f\left(x\right)dx=(b-a)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(a)+f(a+h)+f(a+2h)......f\{a+(n-1\left)h\right\}]$,
where $h=\frac{b-a}{n}$
Here, $a=2,b=3$, and $f\left(x\right)=x^2$
$\therefore {\int }_2^3{x}^{2}dx=(3-2\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(2)+f(2+\frac{1}{n})+f(2+\frac{2}{n})....f\{2+(n-1\left)\frac{1}{n}\right\}]$
$=\underset{n\to \infty }{lim}\frac{1}{n}\left[\right(2{)}^2+{(2+\frac{1}{n})}^2+{(2+\frac{2}{n})}^2+....{(2+\frac{(n-1)}{n})}^2]$
$=\underset{n\to \infty }{lim}\frac{1}{n}[2^2+\{2^2+{\left(\frac{1}{n}\right)}^2+2.2\cdot \frac{1}{n}\}+...+\left\{\right(2{)}^2+\frac{(n-1{)}^2}{n^2}+2.2\cdot \frac{(n-1)}{n}\}]$
$=\underset{n\to \infty }{lim}\frac{1}{n}\left[\right(2^2+\underset{ntimes}{......}+2^2)+\{{\left(\frac{1}{n}\right)}^2+{\left(\frac{2}{n}\right)}^2+.....+{\left(\frac{n-1}{n}\right)}^2\}+2.2\cdot \{\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+....+\frac{(n-1)}{n}\}]$
$=\underset{n\to \infty }{lim}\frac{1}{n}[4n+\frac{1}{n^2}\{1^2+2^2+3^2....+(n-1{)}^2\}+\frac{4}{n}\{1+2+...+(n-1\left)\right\}]$
$=\underset{n\to \infty }{lim}\frac{1}{n}[4n+\frac{1}{n^2}\left\{\frac{n(n-1)(2n-1)}{6}\right\}+\frac{4}{n}\left\{\frac{n(n-1)}{2}\right\}]$
$=\underset{n\to \infty }{lim}\frac{1}{n}[4n+\frac{n(1-\frac{1}{n})(2-\frac{1}{n})}{6}+\frac{4n-4}{2}]$
$=\underset{n\to \infty }{lim}[4+\frac{1}{6}(1-\frac{1}{n})(2-\frac{1}{n})+2-\frac{2}{n}]$
$=4+\frac{2}{6}+2=\frac{19}{3}$