Prove that the curves $y^2 = 4x$ and $x^2 = 4y$ divide the area of the square bounded by $x = 0$, $x = 4$, $y = 4$ and $y = 0$ into three equal areas.
The two parabolas intersect at the points $O \left(0,0\right)$ and $B \left(4,4\right)$.
The given parabolas and the lines intersect at the point $B \left(4,4\right)$
OABC is the square bounded by $x=0$, $x=4$, $y=4$ and $y=0$.
Let the area bounded by the curve $x^2 = 4y$, the X axis and the line $x = 4$ be $A_1$.
Then $A_1 = \left|I_1\right| = \displaystyle \int \limits_{0}^{4} y \; dx$ where $y = \dfrac{x^2}{4}$
$\begin{aligned}
\therefore \; I_1 & = \int \limits_{0}^{4} \dfrac{x^2}{4} \; dx \\\\
& = \dfrac{1}{4} \times \left[\dfrac{x^3}{3}\right]_{0}^{4} = \dfrac{64}{12} = \dfrac{16}{3}
\end{aligned}$
$\therefore$ $\;$ $A_1 = \dfrac{16}{3}$ sq units $\;\;\; \cdots \; (1)$
Let the area bounded by the curve $y^2 = 4x$, the X axis and the line $x = 4$ be $A_4$.
Then $A_4= \left|I_4\right| = \displaystyle \int \limits_{0}^{4} y \; dx$ where $y = 2 \sqrt{x}$
$\begin{aligned}
\therefore \; I_4 & = \int \limits_{0}^{4} 2 \sqrt{x} \; dx \\\\
& = 2 \times \dfrac{2}{3} \times \left[\left(x\right)^{3/2}\right]_{0}^{4} = \dfrac{4}{3} \times 8 = \dfrac{32}{3}
\end{aligned}$
$\therefore$ $\;$ $A_4 = \dfrac{32}{3}$ sq units $\;\;\; \cdots \; (2)$
$\therefore$ $\;$ Area $A_2 = A_4 - A_1 = \dfrac{32}{3} - \dfrac{16}{3} = \dfrac{16}{3}$ sq units $\;\;\; \cdots \; (3)$ $\hspace{2em}$ [From equations $(1)$ and $(2)$]
Let the area bounded by the curve $x^2 = 4y$, the Y axis and the line $y = 4$ be $A_5$.
Then $A_5 = \left|I_5\right| = \displaystyle \int \limits_{y=0}^{y=4} x \; dy$ where $x = 2 \sqrt{y}$
$\begin{aligned}
\therefore \; I_5 & = \int \limits_{0}^{4} 2 \sqrt{y} \; dy \\\\
& = 2 \times \dfrac{2}{3} \times \left[\left(y\right)^{3/2}\right]_{0}^{4} = \dfrac{4}{3} \times 8 = \dfrac{32}{3}
\end{aligned}$
$\therefore$ $\;$ $A_5 = \dfrac{32}{3}$ sq units $\;\;\; \cdots \; (4)$
$\therefore$ $\;$ Area $A_3 = A_5 - A_2 = \dfrac{32}{3} - \dfrac{16}{3} = \dfrac{16}{3}$ sq units $\;\;\; \cdots \; (5)$ $\hspace{2em}$ [From equations $(3)$ and $(4)$]
$\therefore$ $\;$ We have from equations $(1)$, $(3)$ and $(5)$, $A_1 = A_2 = A_3 = \dfrac{16}{3}$ sq units
$\implies$ The curves $y^2 = 4x$ and $x^2 = 4y$ divide the area of the square bounded by $x = 0$, $x = 4$, $y = 4$ and $y = 0$ into three equal areas.