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

Kawal , 10 Years ago
Grade 11
anser 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

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