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Q.
The area bounded by the curve $y = \cos\, x$, the line joining $(- \pi/4, \cos (- \pi/4))$ and $(0, 2)$ and the line joining $( \pi/4, \cos ( \pi/4))$ and $(0, 2)$ is
KVPYKVPY 2013
Solution:
Given, $y = \cos\, x$
Equation of line joining
$\left(\frac{- \pi}{4}, \cos \left(\frac{- \pi}{4}\right)\right)$ and $\left( 0, 2\right)$ is
$ y - 2 = \frac{ 2- 1\sqrt{2}}{0+ \pi 4} \left( x - 0\right) $
$\Rightarrow y -2 = \left(8 - 2\sqrt{2}\right)x$
$ \Rightarrow y =\frac{\left(8 -2\sqrt{2}\right)}{\pi}x +2$
Equation of line joining $\left( \frac{\pi}{4}, \cos \frac{\pi}{4}\right)$ and $ \left(0, 2\right)$ is
$ y -2 = \frac{2-\frac{1}{\sqrt{2}}}{0-\frac{\pi}{4}} \left(x -0\right) $
$\Rightarrow y = \left(\frac{-8+2\sqrt{2}}{\pi}\right)x + 2$
Graph of given curve, $y = \cos\,x$, and line are
Area of shaded region $= 2$ area of curve $APCA$
$=2 \int\limits_{0}^{\pi/4} \left[\frac{\left(-8 +2\sqrt{2}\right)}{\pi} x + 2 - \cos\, x\right]dx $
$ = 2\left[\frac{-8x^{2}}{2\pi} +\frac{2\sqrt{2}x^{2}}{2\pi} +2x - \sin \,x\right]_{0}^{\pi/4} $
$=2\left[\frac{-4}{\pi} \left(\frac{\pi}{4}\right)^{2} + \frac{\sqrt{2}}{\pi}\left(\frac{\pi}{4}\right)^{2} + \frac{2\pi}{4} -\frac{1}{\sqrt{2}}\right] $
$=2\left[\frac{\pi}{4}\left(\frac{-4}{4} +\frac{\sqrt{2}}{4} +2\right) - \frac{1}{\sqrt{2}}\right] $
$= \left(\frac{4+\sqrt{2}}{8}\right)\pi - \sqrt{2}$
