Question

If $f\left(x\right)=x+1$, then find $\underset{n\to \infty }{\lim }\frac{1}{n}\left[f\left(0\right)+f\left(\frac{5}{n}\right)+f\left(\frac{10}{n}\right)+...+f\left(\frac{5\left(n-1\right)}{n}\right)\right]$

• A. $\frac{7}{2}$
• B. $\frac{3}{2}$
• C. $\frac{5}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A

$\frac{7}{2}$

Explanation for the correct option.

Find the limit of the given expression.

The limit $\underset{n\to \infty }{\lim }\frac{1}{n}\left[f\left(0\right)+f\left(\frac{5}{n}\right)+f\left(\frac{10}{n}\right)+...+f\left(\frac{5\left(n-1\right)}{n}\right)\right]$ can be written as:

$\underset{n\to \infty }{\lim }\frac{1}{n}\left[f\left(0\right)+f\left(\frac{5}{n}\right)+f\left(\frac{10}{n}\right)+...+f\left(\frac{5\left(n-1\right)}{n}\right)\right]=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=0}^{n-1}f\left(\frac{5r}{n}\right)$

And thus its limit is given as:

$\begin{array}{rcl}\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=0}^{n-1}f\left(\frac{5r}{n}\right)& =& {\int }_0^{1}f\left(5x\right)dx\\ & =& {\int }_0^1\left[5x+1\right]dx\mathbf{}\mathbf{}\left[f\left(x\right)=x+1\right]\\ & =& {\left[5\frac{x^2}{2}+x\right]}_0^1\\ & =& 5\times \frac{1^2}{2}+1-0-0\\ & =& \frac{5}{2}+1\\ & =& \frac{7}{2}\end{array}$

Hence, the correct option is A.