A) \[e\]
B) \[e-1\]
C) \[1-e\]
D) \[e+1\]
Correct Answer: B
Solution :
\[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{\frac{1}{n}}{{e}^{r/n}}=\int_{0}^{1}{{{e}^{x}}}dx\] Now, \[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{\frac{1}{n}}{{e}^{r/n}}=\int_{0}^{1}{{{e}^{x}}}dx=[{{e}^{x}}]_{0}^{1}=e-1\]
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