The interval  $[0,4]$  is divided into n  equal  sub-intervals by the points  $x_0,x_1,x_2,\mathrm{....},x_{n-1},x_n$, where  $0=x_0<x_1<x_2<x_3\mathrm{...}<x_n=4$. If $\delta x=x_i-x_{i-1}$ for $i=1,2,3,\mathrm{....},n,$ then the value of  $\underset{\delta x\to 0}{\lim }\sum _{i=1}^n{x}_i\delta x$ is

If  $f\left(x\right)$  be a continuous function defined on the closed interval  $[a,b]$ and if the interval  $[a,b]$ be divided into  n equal parts each of width  h, we have
 $b-a=nh$, then
 $\underset{n\to \infty }{\lim }\sum _{r=0}^{n-1}\frac{1}{n}f\left(\frac{r}{n}\right)={\int }_0^{1}f\left(x\right)dx$
Here,  $a=0,b=4,h=\delta x=\frac{b-a}{n}=\frac{4}{n}$  and  $x_i=i\cdot h$

$\underset{\delta x\to 0}{\lim }\sum _{i=1}^n{x}_i\cdot \delta x=\underset{n\to \infty }{\lim }\sum _{i=1}^{n}i\left(\frac{4}{n}\right)\cdot \frac{4}{n}=16{\int }_0^{1}xdx=16{\left[\frac{x^2}{2}\right]}_0^1=16(\frac{1}{2}-0)=8$