The Easiest IIT JEE Problem - 2

Let us first find the points of intersection between these three parabolic curves. Substitution of either of the first two into the third yields:

$(x^2)^2 = 4x - 3 \Rightarrow x^4 - 4x + 3 = 0 \Rightarrow (x-1)^2 (x^2 + 2x + 3) = 0 \Rightarrow x = 1, -1 \pm \sqrt{2}i.$

With $x = 1$ being the only real root we now have $y = \pm 1$ . To find the bounded area, I will integrate with respect to y as this area is symmetric with respect to the x-axis:

$A = 2 \int_{0}^{1} \frac{y^2 + 3}{4} - \sqrt{y} dy \Rightarrow 2[\frac{y^3}{12} + \frac{3y}{4} - \frac{2}{3}y^{\frac{3}{2}}]|_{0}^{1} \Rightarrow \boxed{\frac{1}{3}}.$