Please reply soonlim(n->inf) 1/n ∑(r =1 to 2n) r/(n2+r2)^1/2

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Integral Calculus> Please reply soon lim(n->inf) 1/n ∑(r =1 ...

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lim(n->inf) 1/n ∑(r =1 to 2n) r/(n²+r²)^1/2

Kawal , 10 Years ago

Grade 11

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:Hello student, please find answer to your question
$L = \lim_{n\rightarrow \infty }\frac{1}{n}\sum_{r =1}^{2n}\frac{r}{\sqrt{n^{2}+r^{2}}}$
$L = \lim_{n\rightarrow \infty }\frac{1}{n}\sum_{r =1}^{2n}\frac{\frac{r}{n}}{\sqrt{1+(\frac{r}{n})^{2}}}$
$\frac{r}{n} = x$
$\lim_{n\rightarrow \infty }\frac{1}{n} = dx$
$L = \lim_{n\rightarrow \infty }\frac{1}{n}\sum_{r =1}^{2n}\frac{\frac{r}{n}}{\sqrt{1+(\frac{r}{n})^{2}}} = \int_{0}^{2}\frac{x}{\sqrt{1+x^{2}}}dx$
$L = \frac{1}{2} \int_{0}^{2}\frac{2x}{\sqrt{1+x^{2}}}dx$
$1+x^{2} = t$
$2xdx = dt$
$x = 0\rightarrow t =1$
$x = 2\rightarrow t =5$
$L = \frac{1}{2} \int_{1}^{5}\frac{1}{\sqrt{t}}dt$
$L = (\sqrt{t})_{1}^{5}$
$L = \sqrt{5}-1$