Application of Integrals — Class 12 Maths Solution

For the point of intersection, solve y $^2 = 4x$ and ${x^2} = 4y$.
$\Rightarrow$ ${\left( {\frac{{{x^2}}}{4}} \right)^2} = 4x$

$\Rightarrow$ ${x^4} = {4^3}x = x = 0,4$

Area bounded between curves
$= \int_0^4 {\left( {\sqrt {4x} - \frac{{{x^2}}}{4}} \right)} dx = \left[ {2 \cdot \frac{{{x^{3/2}}}}{{\frac{3}{2}}} - \frac{{{x^3}}}{{12}}} \right]_0^4$

$= \frac{4}{3}{(4)^{3/2}} - \frac{{{{\left( 4 \right)}^3}}}{{12}} = \frac{{32}}{3} - \frac{{16}}{3} = \frac{{16}}{3}$