Question

The area enclosed by the curve y= V4-x² and the line y = |x| is is what​

Answer 1

$\large\underline{\sf{Solution-}}$

The given curves are

$\rm :\longmapsto\:y = \sqrt{4 - {x}^{2} }$

and

$\rm :\longmapsto\:y = |x|$

Now,

$\rm :\longmapsto\:y = \sqrt{4 - {x}^{2} }$

represents a semicircle whose center is (0, 0) and radius 2 units respectively.

and

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:y = |x| = \begin{cases} &\sf{ - x \: \: when \: x < 0} \\ \\ &\sf{ \: x \: \: when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

So, represents two lines passes through the origin.

$\rm :\longmapsto\: y_{1} = \sqrt{4 - {x}^{2} }$

$\rm :\longmapsto\: y_{2} = x$

$\rm :\longmapsto\: y_{3} = - x$

Now, point of intersection of

$\rm :\longmapsto\:y = \sqrt{4 - {x}^{2} } \: \: and \: \: y = x$

$\rm :\longmapsto\:x = \sqrt{4 - {x}^{2} }$

$\rm :\longmapsto\: {x}^{2} = 4 - {x}^{2}$

$\rm :\longmapsto\: 2{x}^{2} = 4$

$\rm :\longmapsto\: {x}^{2} = 2$

$\rm\implies \:x = \sqrt{2}$

$\rm\implies \:y = \sqrt{2}$

Similarly,

The point of intersection of

$\rm :\longmapsto\:y = \sqrt{4 - {x}^{2} } \: \: and \: \: y = - x \: is \: ( - \sqrt{2}, \sqrt{2})$

So, the required area bounded by the curves is

$\rm \:  =  \: 2\displaystyle\int_{0}^{ \sqrt{2} }( y_{1} - y_{2})$

$\rm \:  =  \: 2\displaystyle\int_{0}^{ \sqrt{2} }( \sqrt{4 - {x}^{2} } - x)$

$\rm \:  =  \: 2\bigg[\dfrac{x}{2} \sqrt{4 - {x}^{2} } + \dfrac{4}{2} {sin}^{ - 1}\dfrac{x}{2} - \dfrac{ {x}^{2} }{2} \bigg]_{0}^{ \sqrt{2} }$

$\rm \:  =  \: 2\bigg[\dfrac{ \sqrt{2} }{2} \sqrt{4 - 2 } + \dfrac{4}{2} {sin}^{ - 1}\dfrac{ \sqrt{2} }{2} - \dfrac{2}{2} \bigg] - 0$

$\rm \:  =  \: 2\bigg[\dfrac{ \sqrt{2} }{2} \sqrt{2} + 2 {sin}^{ - 1}\dfrac{1}{ \sqrt{2} } -1\bigg]$

$\rm \:  =  \: 2\bigg[1+ 2 \times \dfrac{\pi}{4} -1\bigg]$

$\rm \:  =  \: 2\bigg[ \dfrac{\pi}{2} \bigg]$

$\rm \:  =  \: \pi \: square \: units$