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The area bounded by the curves $y=cosx$ and $y=sinx$ between the ordinates $x=0\; and \; x=\frac{3\pi }{2}\; is$

Option 1)
$4\sqrt{2}-2$

Option 2)
$4\sqrt{2}+2$

Option 3)
$4\sqrt{2}-1$

Option 4)
$4\sqrt{2}+1$

Answers (1)

As we learnt in

Area along x axis -

Let $y_{1}= f_{1}(x)\, and \, y_{2}= f_{2}(x)$ be two curve then area bounded between the curves and the lines

x = a and x = b is

$\left | \int_{a}^{b} \Delta y\, dx\right |= \left | \int_{a}^{b}\left ( y_{2}-y_{1} \right ) dx\right |$

- wherein

Where $\Delta y= f_{2}\left ( x \right )-f_{1}(x)$

$Area=\int_{0}^{\frac{\pi }{4}}(\cos x-\sin x)dx$ $+\int_{\frac{\pi }{4}}^{\frac{5\pi }{4}}(\sin x-\cos x)dx\\$ $+\int_{\frac{5\pi }{4}}^{\frac{3\pi }{2}}(\cos x-\sin x)dx$

$=\left [ \sin x+\cos x \right ]_{0}^{\frac{\pi }{4}}+\left [ -\cos -\sin x \right ]_{\frac{\pi }{4}}^{\frac{5\pi }{4}}+\left [ \sin x+\cos x \right ]_{\frac{5\pi }{4}}^{\frac{3\pi }{2}}$

$=\sqrt{2}^{-1}+\left [ \sqrt{2} +\sqrt{2}\right ]+1(-1)+\sqrt{2}$

$=4\sqrt{2}-2$

Option 1)

$4\sqrt{2}-2$

This option is correct

Option 2)

$4\sqrt{2}+2$

This option is incorrect

Option 3)

$4\sqrt{2}-1$

This option is incorrect

Option 4)

$4\sqrt{2}+1$

This option is incorrect

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