The area bounded by the curves y2=4x and x2=4y is:
Solutions · 1
Official solution
The curves intersect at (0,0) and (4,4).
\[\text{Area} = \int_0^4\left(2\sqrt x - \frac{x^2}{4}\right)dx = \left[\frac{4}{3}x^{3/2}-\frac{x^3}{12}\right]_0^4 = \frac{32}{3}-\frac{16}{3} = \frac{16}{3}\]