Question

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

Answer 1

We know that,
${\int }_a^{b}f\left(x\right)dx=(b-a)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(a)+f(a+h)+...+f(a+(n-)h\left)\right]$,
where $h=\frac{b-a}{n}$
Here, $a=a,b=b$, and $f\left(x\right)=x$
$\therefore {\int }_a^{b}xdx=(b-a)\underset{n\to \infty }{lim}\frac{1}{n}[a+(a+h)....(a+2h)....a+(n-1\left)h\right]$
$=(b-a)\underset{n\to \infty }{lim}\frac{1}{n}\left[\right(a\underset{ntimes}{+a+}a+...+a)+(h+2h+3h+....+(n-1)h\left)\right]$
$=(b-a)\underset{n\to \infty }{lim}\frac{1}{n}[na+h(1+2+3+....+(n-1)\left)\right]$
$=(b-a)\underset{n\to \infty }{lim}\frac{1}{n}[na+h\left\{\frac{(n-1)\left(n\right)}{2}\right\}]$
$=(b-a)\underset{n\to \infty }{lim}\frac{1}{n}(na+\frac{n(n-1)h}{2}]$
$=(b-a)\underset{n\to \infty }{lim}\frac{n}{n}[a+\frac{(n-1)h}{2}]$
$=(b-a)\underset{n\to \infty }{lim}[a+\frac{(n-1)h}{2}]$
$=(b-a)\underset{n\to \infty }{lim}[a+\frac{(n-1)(b-a)}{2n}]$
$=(b-a)\underset{n\to \infty }{lim}[a+\frac{(1-\frac{1}{n})(b-a)}{2}]$
$=(b-a)[a+\frac{(b-a)}{2}]$
$=(b-a)\left[\frac{2a+b-a}{2}\right]$
$=\frac{(b-a)(b+a)}{2}$
$=\frac{1}{2}(b^2-a^2)$