Question

Evaluate the definite integral as limit of sums:
${\int }_a^{b}xdx$

Answer 1

${\int }_a^{b}xdx$

We know that

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

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

Putting $a=a$ and $b=b$

$h=\frac{b-a}{n}$

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

Hence, we can write

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

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

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

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

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

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

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

$=(b-a)\underset{n\to \infty }{lim}(a+\frac{(n-1)h}{2})$

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

$[\because h=\frac{b-a}{n}]$

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

$=(b-a)(a+\frac{b-a}{2})$

$=\frac{(b-a)(b+a)}{2}$

$=(b-a)(a+\frac{b-a}{2})$

$=\frac{(b-a)(b+a)}{2}$

$=\frac{b^2-a^2}{2}$