Question

Find the value of definite integrals, as the limit of a sum (by first principle).
${\int }_a^b{x}^{2}dx$

Answer 1

${\int }_a^b{x}^{2}dx$
${\int }_a^{b}f\left(x\right)dx=\underset{h\to 0}{lim}\left[f\right(a+h)+f(a+2h)+f(a+3h)+...+f(a+nh\left)\right]$
Here $a=a,b=b,nh=b-a$
${\int }_a^b{x}^{2}dx=\underset{h\to 0}{lim}h\left[f\right(a+h)+f(a+2h)+..+f(a+nh\left)\right]$
$=\underset{h\to 0}{lim}h\left[\right(a+h{)}^2+(a+2h{)}^2+...+(a+nh{)}^2]$
$=\underset{h\to 0}{lim}h\left[\right(a^2+h^2+2ah)+(a^2+2^2{h}^2+2.a^{2}h)+...+(a^2+n^2{h}^2+2anh\left)\right]$
$=\underset{h\to 0}{lim}h\left[\right(a^2+a^2+...+a^2)+h^2(1+2^2+...+n^2)+...+2ah(1+2+...+n\left)\right]$
$=\underset{h\to 0}{lim}h[a^2\times n+h^2\times \frac{n(2n+1)(n+1)}{6}+2ah\times \frac{n(n+1)}{2}]$
$=\underset{h\to 0}{lim}[a^{2}nh+h^3\frac{\left(n\right)(n+1)(2n+1)}{6}+a{h}^{2}n(n+1\left)\right]$
$=\underset{h\to 0}{lim}\left[a^2\right(b-a)+\frac{\left(nh\right)(nh+h)(2nh+h)}{6}+anh(nh+h\left)\right]$
$=\underset{h\to 0}{lim}\left[a^2\right(b-a)+\frac{(b-a)(b-a+h)\left\{2\right(b-a)+h\}}{6}+a(b-a\left)\right(\overline{b-a}+h\left)\right]$
$=a^2(b-a)+a(b-a)(b-a)+\frac{1}{3}(b-a)(b-a)(b-a)$
$=(b-a)a^2+a(b-a{)}^2+\frac{1}{3}(b-a{)}^3$
$=\frac{b-a}{3}[3a^2+3a(b-a)+(b-a{)}^2]$
$=\frac{b-a}{3}[3a^2+3ab-3a^2+b^2-2ab+a^2]$
$=\frac{b-a}{3}[b^2+ab+a^2]$
$=\frac{b^3-a^3}{3}$