The point of intersection of the
Parabolas y² = 4x and x² = 4y are (0, 0) and (4, 4)

Now, the area of the region OAQBO bounded by curves ${\mathit{y}}^2$ = 4x and ${\mathit{x}}^2$ = 4y
$\underset{0}{\overset{4}{\int }}\left(2\sqrt{\mathit{x}}.\frac{{\mathit{x}}^2}{4}\right)$ dx = ${\left[2\frac{{\mathit{x}}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{{\mathit{x}}^3}{12}\right]}_0^4$ = $\frac{32}{3}-\frac{16}{3}$ = $\frac{16}{3}$ sq units ………..(i)
Again, the area of the region OPQAO bounded by the curves ${\mathit{x}}^2$ = 4y, x = 0, x = 4 and the x-axis,
$\underset{0}{\overset{4}{\int }}\frac{{\mathit{x}}^2}{4}$ dx = ${\left[\frac{{\mathit{x}}^3}{12}\right]}_0^4$ = $\left(\frac{64}{12}\right)$ = $\frac{16}{3}$ sq units ……….(ii)
Similarly, the area of the region OBQRO bounded by the curve ${\mathit{y}}^2$ = 4x, the y-axis, y = 0 and y = 4
$\underset{0}{\overset{4}{\int }}\frac{{\mathit{y}}^2}{4}$ dy = ${\left[\frac{{\mathit{y}}^3}{12}\right]}_0^4$ = $\frac{16}{3}$ sq units ... (iii)
From (i), (ii), and (iii) it is concluded that the area of the region OAQBO = area of the region OPQAO = area of the region OBQRO, i.e., area bounded by parabolas
${\mathit{y}}^2$ = 4x and ${\mathit{x}}^2$ = 4y divides the area of the square into three equal parts.