Question

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

• A. 1
• B. 2
• C. 3
• D. 8

Answer 1

Answer: (D)

$L=\underset{\delta x\to 0}{\mathrm{Lim}}\delta x\left(x_1+x_2+x_3+\dots ..+x_n\right)$

$=\underset{n\to \infty }{\mathrm{Lim}}\frac{4}{n}\left[\frac{4}{n}+\frac{8}{n}+\frac{12}{n}+\dots \dots .+4\cdot \frac{n}{n}\right]\left(\delta x=\frac{4}{n}\right)$
$=\underset{n\to \infty }{\mathrm{Lim}}\frac{16}{n^2}(1+2+3+\dots \dots +n)=\underset{n\to \infty }{\mathrm{Lim}}\frac{16}{n^2}\cdot \frac{n(n+1)}{2}=8$