#### Question

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

#### Solution

A_{1}, A_{2}, A_{3 }are the areas denoted in the figure. We need to prove A_{1} = A_{2 }= A_{3}.

`A_1 = ∫_0^4y1dx`

`= ∫_0^4x^2/4dx `

`=1/4[x^3/3]_0^4`

`=16/3 sq. units`

`A_2=∫_0^4(y_2−y_1)dx`

`=∫_0^4(sqrt(4x)−x^2/4)dx`

`= [4/3x^3/2−x^3/12]_0^4`

`=16/3sq. units`

`A_3=area bounded by y^2=4x, y=0 and y=4`

`=∫_0^4x_1dy=∫_0^4y^2/4dy=1/4[y^3/3]_0^4=16/3`

`therefore A_1=A_2 =A_3 =16/3 sq. units`

Thus, *y*^{2} = 4*x* and *x*^{2} = 4*y* divide the area of square bounded by* x* = 0, *x* = 4, *y* = 4 and* y* = 0 into three equal parts.

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Solution Prove that the curves y^2 = 4x and x^2 = 4y divide the area of square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts. Concept: Area of the Region Bounded by a Curve and a Line.