# Prove that the curves y2 = 4x and x2 = 4y divide the area of the square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts. ?

Jitender Singh IIT Delhi
8 years ago
Ans:
Area of square ‘A’:
$A = 4 \times 4 =16$
Area b/w curves ‘a’:
$a = \int_{0}^{4}(2\sqrt{x} - \frac{x^{2}}{4})dx$
$a = (\frac{4x^{3/2}}{3}-\frac{x^{3}}{12})_{0}^{4} = \frac{32}{3}-\frac{16}{3} = \frac{16}{3}$
Area of rest two parts:
$A -a = \frac{32}{3}$
which is equally divided into 16/3 & 16/3.
Thanks & Regards
Jitender Singh
IIT Delhi