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Solve this: pefinite Integration Objective Type Each question has four choices a, b, c, and d, out of which only one is correct. 1 lim Sn is equal to log2 c. log8 h d 1 2 log 4 None of these

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$$\begin{gathered} {\mathrm{S}}_{\mathrm{n}}=\left[\frac{1}{1+\sqrt{\mathrm{n}}}+\frac{1}{2+\sqrt{2\mathrm{n}}}+....+\frac{1}{\mathrm{n}+\sqrt{{\mathrm{n}}^2}}\right] \\ {\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{\mathrm{r}+\sqrt{\mathrm{r} \mathrm{n}}} \\ =\sum _{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{{\displaystyle \frac{\mathrm{r}+\sqrt{\mathrm{r} \mathrm{n}}}{\mathrm{n}}}}.\frac{1}{\mathrm{n}} \\ =\sum _{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{{\displaystyle \frac{\mathrm{r}}{\mathrm{n}}+\sqrt{\frac{\mathrm{r}}{\mathrm{n}}}}}.\frac{1}{\mathrm{n}} \\ \Rightarrow \underset{\mathrm{n}\to \infty }{\lim } {\mathrm{S}}_{\mathrm{n}}=\underset{\mathrm{n}\to \infty }{\lim } \sum _{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{{\displaystyle \frac{\mathrm{r}}{\mathrm{n}}+\sqrt{\frac{\mathrm{r}}{\mathrm{n}}}}}.\frac{1}{\mathrm{n}} \\ \frac{1}{\mathrm{n}}\to \mathrm{dx} \mathrm{and} \frac{\mathrm{r}}{\mathrm{n}} \mathrm{as} \mathrm{x}, \mathrm{we} \mathrm{get} \\ \underset{\mathrm{n}\to \infty }{\lim } \sum _{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{{\displaystyle \frac{\mathrm{r}}{\mathrm{n}}+\sqrt{\frac{\mathrm{r}}{\mathrm{n}}}}}.\frac{1}{\mathrm{n}}={\int }_0^1 \frac{1}{\mathrm{x}+\sqrt{\mathrm{x}}}.\mathrm{dx} \\ ={\int }_0^1 \frac{1}{\sqrt{\mathrm{x}}}\left(\frac{1}{\sqrt{\mathrm{x}}+1}\right).\mathrm{dx} \\ =2{\int }_0^1 \frac{1}{2\sqrt{\mathrm{x}}}\left(\frac{1}{\sqrt{\mathrm{x}}+1}\right).\mathrm{dx} \\ \mathrm{Set} \sqrt{\mathrm{x}}=\mathrm{t} \mathrm{which} \mathrm{gives} \frac{1}{2\sqrt{\mathrm{x}}}.\mathrm{dx}=\mathrm{dt} \Rightarrow \int \frac{1}{2\sqrt{\mathrm{x}}}\left(\frac{1}{\sqrt{\mathrm{x}}+1}\right).\mathrm{dx}=\int \frac{1}{\mathrm{t}+1}.\mathrm{dt}=\ln \left(\mathrm{t}+1\right)=\ln \left(\sqrt{\mathrm{x}}+1\right) \\ \Rightarrow 2{\int }_0^1 \frac{1}{2\sqrt{\mathrm{x}}}\left(\frac{1}{\sqrt{\mathrm{x}}+1}\right).\mathrm{dx}=2{\left[\ln \left(\sqrt{\mathrm{x}}+1\right)\right]}_0^1 \\ =2\left[\ln \left(2\right)-\ln \left(1\right)\right] \\ =2\ln \left(2\right) \\ =\ln \left(2^2\right) \\ =\ln \left(4\right) \end{gathered}$$

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