Show that


$\int_{0}^{b} x d x$


It is known that,

$\int_{a}^{b} f(x) d x=(b-a) \lim _{n \rightarrow \infty} \frac{1}{n}[f(a)+f(a+h)+\ldots+f(a+(n-1) h)]$, where $h=\frac{b-a}{n}$

Here, $a=a, b=b$, and $f(x)=x$

$\therefore \int_{a}^{b} x d x=(b-a) \lim _{n \rightarrow \infty} \frac{1}{n}[a+(a+h) \ldots(a+2 h) \ldots a+(n-1) h]$

$=(b-a) \lim _{n \rightarrow \infty} \frac{1}{n}[(a+a+a+\ldots+a)+(h+2 h+3 h+\ldots+(n-1) h)]$

$=(b-a) \lim _{n \rightarrow \infty} \frac{1}{n}[n a+h(1+2+3+\ldots+(n-1))]$

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

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

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

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

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

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


$=(b-a)\left[\frac{2 a+b-a}{2}\right]$



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