Question

Prove that the curves $y^2=4x$ and $x^2=4y$ divide the area of the square bounded by $x=0,x=4,y=0$ and $y=4$ into three equal parts.

Answer 1

To prove:

$A_1=A_2=A_3=\frac{A}{3}$

[$A$=Area of square]

$A=4\times 4=16sq.units$

$A_1={\int }_0^4\frac{x^2}{4}dx$

$\therefore A_1=\frac{16}{3}sq.units$

$A_2={\int }_0^4\frac{y^2}{4}dy$

$\therefore A_2=\frac{16}{3}sq.units$

$A_3=A-(A_1+A_2)$

$A_3=16-2\times \frac{16}{3}$

$\therefore A_3=\frac{16}{3}sq.units$

$A_1=A_2=A_3=\frac{16}{3}=\frac{A}{3}$

Hence proved.