QJEEM18Mae071: JEE Main 2018 Mathematics - Class 12 JEEM18Mae071 Solution: The area (in sq. units) of the region {x R : x 0, y 0, y x 2…

y = $\sqrt x$

y = x $-$ 2

$\therefore\,\,\,$ $\sqrt x$ = x $-$ 2

$\Rightarrow$ $\,\,\,$ x = x² $-$ 4x + 4

x² $-$ 5x + 4 = 0

x² $-$ 4x $-$ x + 4 = 0

$\Rightarrow$ $\,\,\,$ x(x $-$ 4) $-$ (x $-$ 4) = 0

$\Rightarrow$ $\,\,\,$ (x $-$ 4) (x $-$ 1) = 0

$\therefore\,\,\,$ x = 4, 1

and y = 2, $-$ 1

$\therefore\,\,\,$ Their point of intersection (4, 2) and (1, $-$ 1)



Required area is shown in the shaded figure.

$\therefore\,\,\,$ Required area

= $\int\limits_0^2 {\sqrt x \,dx + \int\limits_2^4 {\left( {\sqrt x - x + 2} \right)\,dx} }$

= $\int\limits_0^4 {\sqrt x \,dx + \int\limits_2^4 {\left( {2 - x} \right)\,dx} }$

= $\left[ {{2 \over 3}{x^{{3 \over 2}}}} \right]_0^4 + \left[ {2x - {{{x^2}} \over 2}} \right]_2^4$

= ${2 \over 3}\left( 8 \right)$ + 2 $\left( {4 - 2} \right)$ $-$ ${1 \over 2}$ (16 $-$ 4)

= ${{16} \over 3}$ + 4 $-$ 6

= ${{10} \over 3}$