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Question
Find the area of the region bounded by the curve x2 = 4y and the line x = 4y – 2.
Solution
Curve x2 = 4y ...(1)
Line x = 4y – 2 ...(2)
Solving (1) and (2)
x = x2 – 2
`\implies` x2 – x – 2 = 0
(x – 2)(x + 1) = 0
∴ x = 2, – 1
∴ y = `x^2/4 = 1, 1/4`
Points of intersection are `A(2, 1), B(-1, 1/4)`
The shaded region is required.
Let PQ to be the elementary strip.
So area of shaded region
= `int_-1^2 (y_"line" - y_"curve")dx`
= `int_-1^2 (x + 2)/4 - x^2/4dx`
= `1/4[x^2/2 + 2x - x^3/3]_-1^2`
= `1/4[(4/2 + 4 - 8/3)] - [(1/2 - 2 + 1/3)]`
= `1/4[6 - 8/3 - 1/2 + 2 - 1/3]`
= `1/4[6 - 9/3 - 1/2 + 2]`
= `1/4[6 - 3 - 1/2 + 2]`
= `1/4 xx 9/2`
= `9/8` sq.units
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