QJEEM23Mad7cd: JEE Main 2023 Mathematics - Class 12 JEEM23Mad7cd Solution: Let be the area of the region . Then is equal to :

$y = {x^2}$ and $y = 2x(1 - x)$

$\Rightarrow {x^2} = 2x - 2{x^2}$

$\Rightarrow x = 0,x = {2 \over 3}$

Now,

$y = {(1 - x)^2}$ and $y = 2x(1 - x)$

$\Rightarrow 1 + {x^2} - 2x = 2x - 2{x^2}$

$\Rightarrow 3{x^2} - 4x + 1 = 0$

$x = 1,x = {1 \over 3}$

$\therefore$ $A = \int\limits_{{1 \over 3}}^{{2 \over 3}} {(2x - 2{x^2})dx - \left\{ {\int\limits_{{1 \over 3}}^{{1 \over 2}} {{{(1 - x)}^2}dx + \int\limits_{{1 \over 2}}^{{2 \over 3}} {{x^2}dx} } } \right\}}$

= $\left( {{x^2} - {{2{x^3}} \over 3}} \right)_{{1 \over 3}}^{{2 \over 3}} - \left\{ {\left( {{{{{\left( {x - 1} \right)}^3}} \over 3}} \right)_{{1 \over 3}}^{{1 \over 2}} + \left( {{{{x^3}} \over 3}} \right)_{{1 \over 2}}^{{2 \over 3}}} \right\}$

$= {5 \over {108}}$

$\therefore$ $540A = 25$