A) \[1+\sqrt{5}\]
B) \[-1+\sqrt{5}\]
C) \[-1+\sqrt{2}\]
D) \[1+\sqrt{2}\]
Correct Answer: B
Solution :
Let\[I=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{r=1}^{2n}{\frac{r}{\sqrt{{{n}^{2}}+{{r}^{2}}}}}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{r=1}^{2n}{\frac{r/n}{\sqrt{1+{{(r/n)}^{2}}}}}\] Put \[\frac{r}{n}=x,\frac{1}{n}=dx,\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{2n}{=}\int_{0}^{2}{{}}\] \[\therefore \] \[I=\int_{0}^{2}{\frac{x}{\sqrt{1+{{x}^{2}}}}}dx=[\sqrt{1+{{x}^{2}}}]_{0}^{2}\] \[=\sqrt{5}-1\]
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