Question

FRIEND'S SOLVE IT......

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Answer 1

Answer:

see the attachment given.....

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Answer 2

We're asked to evaluate,

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\dfrac{1}{n^2}\sum_{k=0}^{n-1}\left[k\int\limits_k^{k+1}\sqrt{(x-k)(k+1-x)}\ dx\right]\quad\quad\dots(1)$

Let us consider the integral.

$\displaystyle\longrightarrow I=\int\limits_k^{k+1}\sqrt{(x-k)(k+1-x)}\ dx$

$\displaystyle\longrightarrow I=\int\limits_k^{k+1}\sqrt{-(x-k)(x-(k+1))}\ dx$

$\displaystyle\longrightarrow I=\int\limits_k^{k+1}\sqrt{-\left(x-\left(\dfrac{k+(k+1)}{2}+\dfrac{k-(k+1)}{2}\right)\right)\left(x-\left(\dfrac{k+(k+1)}{2}-\dfrac{k-(k+1)}{2}\right)\right)}\ dx$

$\displaystyle\longrightarrow I=\int\limits_k^{k+1}\sqrt{-\left(x-\left(\dfrac{2k+1}{2}-\dfrac{1}{2}\right)\right)\left(x-\left(\dfrac{2k+1}{2}+\dfrac{1}{2}\right)\right)}\ dx$

$\displaystyle\longrightarrow I=\int\limits_k^{k+1}\sqrt{-\left(x-\dfrac{2k+1}{2}+\dfrac{1}{2}\right)\left(x-\dfrac{2k+1}{2}-\dfrac{1}{2}\right)}\ dx$

$\displaystyle\longrightarrow I=\int\limits_k^{k+1}\sqrt{\left(\dfrac{1}{2}+x-\dfrac{2k+1}{2}\right)\left(\dfrac{1}{2}-x+\dfrac{2k+1}{2}\right)}\ dx$

$\displaystyle\longrightarrow I=\int\limits_k^{k+1}\sqrt{\left(\dfrac{1}{2}+\left(x-\dfrac{2k+1}{2}\right)\right)\left(\dfrac{1}{2}-\left(x-\dfrac{2k+1}{2}\right)\right)}\ dx$

$\displaystyle\longrightarrow I=\int\limits_k^{k+1}\sqrt{\left(\dfrac{1}{2}\right)^2-\left(x-\dfrac{2k+1}{2}\right)^2}\ dx$

$\displaystyle\longrightarrow I=\int\limits_k^{k+1}\sqrt{\dfrac{1}{4}-\dfrac{(2x-2k-1)^2}{4}}\ dx$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\limits_k^{k+1}\sqrt{1-(2x-2k-1)^2}\ dx\quad\quad\dots(2)$

Let,

$\longrightarrow u=2x-2k-1$

$\longrightarrow dx=\dfrac{1}{2}\ du$

For $x=k,$

$\longrightarrow u=2k-2k-1$

$\longrightarrow u=-1$

For $x=k+1,$

$\longrightarrow u=2(k+1)-2k-1$

$\longrightarrow u=1$

Then (2) becomes,

$\displaystyle\longrightarrow I=\dfrac{1}{4}\int\limits_{-1}^1\sqrt{1-u^2}\ du\quad\quad\dots(3)$

Let,

$\displaystyle\longrightarrow u=\sin v$

$\displaystyle\longrightarrow du=\cos v\ dv$

For $u=-1,$ $v=-\dfrac{\pi}{2}.$ For $u=1,$ $v=\dfrac{\pi}{2}.$

Then (3) becomes,

$\displaystyle\longrightarrow I=\dfrac{1}{4}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos v\,\sqrt{1-\sin^2v}\ dv$

$\displaystyle\longrightarrow I=\dfrac{1}{4}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2v\ dv\quad\left[\,\because\,1-\sin^2\theta=\cos^2\theta\,\right]$

$\displaystyle\longrightarrow I=\dfrac{1}{8}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(1+\cos(2v)\right)dv\quad\left[\,\because\ 1+\cos(2\theta)=2\cos^2\theta\,\right]$

$\displaystyle\longrightarrow I=\dfrac{1}{8}\left[v+\dfrac{\sin(2v)}{2}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$

$\displaystyle\longrightarrow I=\dfrac{1}{8}\left[\left(\dfrac{\pi}{2}+\dfrac{\sin\left(2\cdot\frac{\pi}{2}\right)}{2}\right)-\left(-\dfrac{\pi}{2}+\dfrac{\sin\left(2\cdot-\frac{\pi}{2}\right)}{2}\right)\right]$

$\displaystyle\longrightarrow I=\dfrac{1}{8}\left[\dfrac{\pi}{2}+\dfrac{\pi}{2}\right]$

$\displaystyle\longrightarrow I=\dfrac{\pi}{8}$

Hence (1) becomes,

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\dfrac{1}{n^2}\sum_{k=0}^{n-1}\dfrac{k\pi}{8}$

$\displaystyle\longrightarrow L=\dfrac{\pi}{8}\lim_{n\to\infty}\dfrac{1}{n^2}\sum_{k=0}^{n-1}k$

$\displaystyle\longrightarrow L=\dfrac{\pi}{8}\lim_{n\to\infty}\dfrac{1}{n^2}\cdot\dfrac{n(n-1)}{2}$

$\displaystyle\longrightarrow L=\dfrac{\pi}{16}\lim_{n\to\infty}\dfrac{n-1}{n}$

Applying L'hospital's Rule,

$\displaystyle\longrightarrow L=\dfrac{\pi}{16}\lim_{n\to\infty}\dfrac{1}{1}$

$\displaystyle\longrightarrow\underline{\underline{L=\dfrac{\pi}{16}}}$

Hence $\bf{\dfrac{\pi}{16}}$ is the answer.