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Question
Prove that the curves y2 = 4x and x2 = 4y divide the area of square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts.
Solution
A(OABC) = 4 × 4 = 16 sq. units
From, y2 = 4x and x2 = 4y
`(x^2/4)^2 = 4x`
or `x^4/16 = 4x`
or x4 - 64x = 0
or x(x3 - 64) = 0
or x = 0 or x = 4
when x = 0, y = 0
x = 4, y = 4
Point of intersection of the two parabolas is (0, 0) and (4, 4).
Area of part III = `int_0^4 y dx` (parabola x2 = 4y)
= `int_0^4 (x^2)/4 dx = [1/4 x^3/3]_0^4`
= `1/12(64 - 0)`
= `64/12`
= `16/3` sq.units
Area of I = Area of square - Area of II and III
= `16 - int_0^4 sqrt(4x) dx`
= `16 - (2 × 2)/3 [x^(3/2)]_0^4`
= `16 - 32/3` sq. units
= `16/3` sq. units
Area of II = Area of square - Area of I - Area of III
= 16 - `16/3 - 16/3` sq. units
= `16/3` sq. units
The two curves divide the square into three equal parts.
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