prove that the curves y2 = 4x andd x2 = 4y divide the area of the square bounded by x=0 - Maths - Application of Integrals

Question

prove that the curves y2 = 4x andd x2 = 4y divide the area of
the square bounded by x=0 x=4 and y = 4 and y = 0 in three equal
parts?

Gursheen kaur. · Accepted Answer

Area enclosed by the parabola $x^{2} = 4 y$

and the x-axis is obtained by integrating $\int \frac{x^{2}}{4} d x$ from limits 0 to 4 is

$x^{2} = 4 y \int_{0}^{4} y = \int_{0}^{4} \frac{x^{2}}{4} d x = {[\frac{x^{3}}{12}]}_{0}^{4} = [\frac{4^{3}}{12} - 0] = \frac{16}{3}$

Similarly,

Area enclosed by the parabola $y^{2} = 4 x$

and the y-axis is obtained by integrating $\int \frac{y^{2}}{4} d y$ from limits 0 to 4 is

$\int_{0}^{4} \frac{y^{2}}{4} d y = {[\frac{y^{3}}{12}]}_{0}^{4} = [\frac{4^{3}}{12} - 0] = \frac{16}{3}$

Now,

The third part = ${(4)}^{2} - \frac{16}{3} - \frac{16}{3} = \frac{16}{1} - (2 \times \frac{16}{3}) = \frac{16}{1} - \frac{32}{3} = \frac{48 - 32}{3} = \frac{16}{3}$

Thus,

The curves $y^{2} = 4 x$ and $x^{2} = 4 y$ divide the area of the square bounded by x=0 ,x=4 and y = 4 and y = 0 in three equal parts and each part $= \frac{16}{3}$

prove that the curves y2 = 4x andd x2 = 4y divide the area of the square bounded by x=0 x=4 and y = 4 and y = 0 in three equal parts?