Question:
If $f: \mathbf{R} \rightarrow \mathbf{R}$ is given by $f(\mathrm{x})=\mathrm{x}+1$, then the value of $\lim _{\mathrm{n} \rightarrow \infty} \frac{1}{\mathrm{n}}\left[f(0)+f\left(\frac{5}{\mathrm{n}}\right)+f\left(\frac{10}{\mathrm{n}}\right)+\ldots .+f\left(\frac{5(\mathrm{n}-1)}{\mathrm{n}}\right)\right]$ is :
Correct Option: , 4
Solution:
$I=\sum_{r=0}^{n-1} f\left(\frac{5 r}{n}\right) \frac{1}{n}$
$I=\int_{0}^{1} f(5 x) d x$
$I=\int_{0}^{1}(5 x+1) d x$
$I=\left[\frac{5 x^{2}}{2}+x\right]_{0}^{1}$
$I=\frac{5}{2}+1=\frac{7}{2}$