Question

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

• A. 1
• B. 2
• C. 3
• D. 8

Answer 1

Answer: (D)

$L =\underset{\delta x \rightarrow 0}{\text{Lim}} \delta x \left( x _1+ x _2+ x _3+\ldots . .+ x _{ n }\right)$

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