Question

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Answer 1

✪ᴀɴsᴡᴇʀ✪

$\red{\boxed{\lim \limits_{n \to \infty} \frac{1}{{n}^{2}} \sum \limits_{k=0}^{n-1}[k\int_{k}^{k+1}\sqrt{(x-k)(k+1-x)}dx] = \frac{\pi} {16}}}$

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

$S = \lim \limits_{n \to \infty} \frac{1}{{n}^{2}} \sum \limits_{k=0}^{n-1}[k\int_{k}^{k+1}\sqrt{(x-k)(k+1-x)}dx]$

$\lim \limits_{n \to \infty} \frac{1}{n} \sum \limits_{k=0}^{n-1}[\frac{k}{n} \int_{k}^{k+1}\sqrt{(x-k)(k+1-x)}dx]$

$\lim \limits_{n \to \infty} \frac{1}{n} \sum \limits_{k=0}^{n-1}[\frac{k}{n}I]$

where

$I = \int_{k}^{k+1}\sqrt{(x-k)(k+1-x)}dx$

$I = \int_{k}^{k+1}\sqrt{[x-k][1-(x-k)]}dx$

Substitute $(x-k) = {sin} ^{2}(\theta)$

$\to dx = 2sin(\theta)cos(\theta)d\theta$

As $x \to k, \theta \to 0$

As $x \to k+1, \theta \to \frac{\pi }{2}$

$I = \int_{0}^{\frac{\pi }{2}}\sqrt{[{sin} ^{2}(\theta )][1-({sin} ^{2}(\theta ))]}2sin(\theta )cos(\theta )d\theta$

$I = \int_{0}^{\frac{\pi }{2}}\sqrt{{sin} ^{2}(\theta ){cos}^{2}(\theta )}2sin(\theta )cos(\theta )d\theta$

$I = \int_{0}^{\frac{\pi }{2}} 2sin(\theta ) cos(\theta )sin(\theta )cos(\theta )d\theta$

$I = \frac{1}{2} \int_{0}^{\frac{\pi }{2}} {[2sin(\theta ) cos(\theta )]}^{2 }d\theta$

$I = \frac{1}{2} \int_{0}^{\frac{\pi }{2}}{sin}^{2 }(2 \theta )d\theta$

$I = \frac{1}{4} \int_{0}^{\frac{\pi }{2}}2{sin}^{2 }(2 \theta )d\theta$

$I = \frac{1}{4} \int_{0}^{\frac{\pi }{2}}[1-cos(4 \theta )] d\theta$

$I = \frac{1}{4} [\theta - \frac{sin(4 \theta )}{4}]_{0}^{\frac{\pi }{2}}$

$I = \frac{1}{4} [(\frac{\pi }{2} - 0) - \frac{sin(2 \pi) - sin(0)}{4}]$

$I = \frac{1}{4} [\frac{\pi }{2} - \frac{0- 0}{4}]$

$I = \frac{\pi }{8}$

Using this in expression of S, we get

$S = \lim \limits_{n \to \infty} \frac{1}{n} \sum \limits_{k=0}^{n-1}[\frac{k}{n}\frac{\pi }{8} ]$

$S = \frac{\pi }{8} \lim \limits_{n \to \infty} \frac{1}{n} \sum \limits_{k=0}^{n-1}[\frac{k}{n}]$

$S = \frac{\pi }{8} \int_{0}^{1 }xdx$

$S = \frac{\pi }{8}[\frac{{x}^{2 }}{2}] _{0}^{1 }$

$S = \frac{\pi }{8}[\frac{{1}^{2}- {0}^{2}}{2}]$

$S = \frac{\pi }{16}$

Answer 2

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