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AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

Section: Mathematics

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Question : 76 of 160

Marks: +1, -0

\lim\limits_{n\rightarrow\infty} \frac{1}{n} \left\{ \sin^5\left(\frac{\pi}{6n}\right) + \sin^5\left(\frac{2\pi}{6n}\right) \right.

\left\{+\sin^5\left(\frac{3\pi}{6n}\right) + \dots + \sin^5\left(\frac{\pi}{2}\right) \right\} =

$\frac{8}{15\pi}$
$\frac{8}{5\pi}$
$\frac{32}{5\pi}$
$\frac{16}{5\pi}$

Solution:

Consider the expression,

\lim\limits_{n\rightarrow\infty} \frac{1}{n} \left\{ \sin^5\left(\frac{\pi}{6n}\right) + \sin^5\left(\frac{2\pi}{6n}\right) \right.

\left\{+\sin^5\left(\frac{3\pi}{6n}\right) + \dots + \sin^5\left(\frac{\pi}{2}\right) \right\}

Simplify the above,

\lim\limits_{n\rightarrow\infty} \frac{1}{n} \left\{ \sin^5\left(\frac{\pi}{6n}\right) + \sin^5\left(\frac{2\pi}{6n}\right) \right.

\left\{ +\sin^5\left(\frac{3\pi}{6n}\right) + \dots + \sin^5\left(\frac{3n\pi}{6n}\right) \right\}

= \lim\limits_{n\rightarrow\infty} \frac{1}{n} \sum\limits_{r=1}^{3n} \sin^5\left(\frac{r\pi}{6n}\right)

= \int\limits_{0}^{3} \sin^5\left(\frac{\pi}{6}x\right) dx

Let,

\frac{\pi}{6}x = t

Then for upper limit,

x=3,\; t=\frac{\pi}{2}

And for lower limit

x=0,\; t=0

and

dx = \frac{6}{\pi} dt

So,

\int\limits_{0}^{3} \sin^5\left(\frac{\pi}{6}x\right) dt = \frac{6}{\pi} \int\limits_{0}^{\pi/2} \sin^5(t) dt

= \frac{6}{\pi} \left( \frac{4(2)}{5(3)(1)} \right)

= \frac{16}{5\pi}

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