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Question
Using integration, find the area bounded by the curve x2 = 4y and the line x = 4y − 2.
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Solution
The area bounded by the curve, x2 = 4y, and line, x = 4y − 2, is represented by the shaded area OBAO.
Let A and B be the points of intersection of the line and parabola.
Coordinates of point A are (-1, 1/4).
Coordinates of point B are (2, 1).
We draw AL and BM perpendicular to x-axis.
It can be observed that,
Area OBAO = Area OBCO + Area OACO … (1)
Then, Area OBCO = Area OMBC − Area OMBO
`=int_0^2(x+2)/4dx-int_0^2x^2/4dx`
`=1/4[x^2/2+2x]_0^2-1/4[x^3/3]_0^2`
`=-1/4[(-1)^2/2+2(-1)]-[1/4((-1)^3/3)]`
`=7/24`
Therefore, required area = `(5/6+7/24)=9/8 units`
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