Integral Calculus> options 1/35 1/14 1/10 1/5...

options

1/35

1/14

1/10

1/5

Aditya Kartikeya , 10 Years ago

Grade 10

1 Answers

Jitender Singh

Last Activity: 10 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.