Question

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

Answer 1

We know that
${\int }_a^{b}f\left(x\right)={lim}_{h\to 0}h\left[f\right(a)+f(a+h)+f(a+2h)+...+f\{a+(n-1\left)h\right\}]$
where, nh=b -a
Here and a=2, b=3 and nh=b -a=3-2=1
$f\left(x\right)=x^2,f\left(2\right)=2^2,f(2+h)=(2+h{)}^{2}f(2+2h)=(2+2h{)}^2,.....,f(2+(n-1\left)h\right)=(2+(n-1)h{)}^2\therefore {\int }_2^3{x}^{2}dx={lim}_{h\to 0}h[f\left(2\right)+f(2+h)+f(2+2h)+....+f(2+(n-1\left)h\right)]$
$={lim}_{h\to 0}h[2^2+(2+h{)}^2+(2+2h{)}^2+...+(2+(n-1)h{)}^2]={lim}_{h\to 0}h[2^2+(2^2+h^2+4h)+(2^2+2^2{h}^2+8h)+...+(2^2+(n-1{)}^2{h}^2+4(n-1)h\left)\right][\therefore (a+b{)}^2=a^2+b^2+2ab]={lim}_{h\to 0}h[(2^2+2^2+...+ntimes)+h^2(1^2+2^2+3^2)+....+(n-1{)}^2+4h(1+2+...+(n-1)\left)\right]={lim}_{h\to 0}\{4nh+h^3\frac{(n-1)n(2n-1)}{6}+4h^2\frac{(n-1)n}{2}\}\left[\begin{array}{c}\because \sum n^2=\frac{n(n+1)(2n+1)}{6}\\ \therefore \sum (n-1{)}^2=\frac{1}{6}(n-1\left)n\right(2n-1)and\sum (n-1)=\frac{n(n-1)}{2}\end{array}\right]$

$={lim}_{h\to 0}\{4nh+\frac{(hn-h)hn(2nh-h)}{6}+2(nh-h\left)\right(nh\left)\right\}={lim}_{h\to 0}\{4\times 1+\frac{(1-h)1(2\times 1-h)}{6}+2(1-h\left)1\right\}[\therefore nh=1]={lim}_{h\to 0}[4+\frac{(1-h)(2-h)}{6}+2(1-h\left)\right]=4+\frac{2}{6}+2=6+\frac{1}{3}=\frac{19}{3}$