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options 1/35 1/14 1/10 1/5

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  1. 1/35
  2. 1/14
  3. 1/10
  4. 1/5

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Grade:10

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 years ago
Ans:
Hello Student,
Please find answer to your question below

L = \lim_{n\rightarrow \infty }\sum_{r=1}^{4n}\frac{\sqrt{n}}{\sqrt{r}(3\sqrt{r}+4\sqrt{n})^{2}}
L = \lim_{n\rightarrow \infty }\sum_{r=1}^{4n}\frac{\sqrt{n}}{n\sqrt{r}(3\frac{\sqrt{r}}{\sqrt{n}}+4)^{2}}
L = \lim_{n\rightarrow \infty }\sum_{r=1}^{4n}\frac{1}{n\frac{\sqrt{r}}{\sqrt{n}}(3\frac{\sqrt{r}}{\sqrt{n}}+4)^{2}}
L = \int_{0}^{4}\frac{1}{\sqrt{x}(3\sqrt{x}+4)^{2}}dx
3\sqrt{x} + 4 = t
\frac{3}{2\sqrt{x}}dx = dt
x=0\rightarrow t=4
x=4\rightarrow t=10
L = \frac{2}{3}\int_{4}^{10}\frac{1}{t^2}dt
L = \frac{2}{3}[\frac{-1}{t}]_{4}^{10}
L = \frac{2}{3}[\frac{1}{4}-\frac{1}{10}]
L = \frac{2}{3}[\frac{10}{40}-\frac{4}{40}]
L = \frac{2}{3}[\frac{6}{40}]
L = \frac{1}{10}
Option (3) is correct.

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