JEE Main 2023 — Area Under Curves Question with Solution

Given,

$A$ be the area of the region

$\left\{(x,y):y\ge x^2,y\ge (1-x{)}^2,y\le 2x(1-x)\right\}$

Now solving $y=x^2 \& y=2x\left(1-x\right)$ we get, $x=0, \frac{2}{3}$

And solving $y={\left(1-x\right)}^2 \& y=2x\left(1-x\right)$ we get, 

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

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

$\Rightarrow x=1, \frac{1}{3}$

Now on plotting the diagram of the above region we get,

Now from the above diagram area of the shaded region will be,

$A={\int }_{\frac{1}{3}}^{\frac{2}{3}}\left(2x-2x^2\right)dx-\left\{{\int }_{\frac{1}{3}}^{\frac{1}{2}}{\left(1-x\right)}^{2}dx+{\int }_{\frac{1}{2}}^{\frac{2}{3}}{x}^{2}dx\right\}$

$\Rightarrow A={\left[x^2-\frac{2x^3}{3}\right]}_{\frac{1}{3}}^{\frac{2}{3}}-\left\{{\left[\frac{{\left(x-1\right)}^3}{3}\right]}_{\frac{1}{3}}^{\frac{1}{2}}+{\left[\frac{x^3}{3}\right]}_{\frac{1}{2}}^{\frac{2}{3}}\right\}$

$\Rightarrow A=\frac{5}{108}$

$\therefore A=\frac{5}{108}\Rightarrow 540 A=\frac{5}{108}\times 540=25$