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Question
Find the area of the following region using integration ((x, y) : y2 ≤ 2x and y ≥ x – 4).
Solution
y2 ≤ 2x
y ≥ x – 4
y2 = 2x ...(i)
y – x + 4 = 0 or x – y = 4 ...(ii)
Put the value of x from (ii) in (i), we have
y2 = 2(y + 4)
y2 – 2y – 8 = 0
`\implies` y = 4, – 2
When y = 4, x = 4 + 4 = 8
When y = – 2, x = 4 – 2 = 2
Required area = `int_-2^4 (y + 4) dy - int_-2^4 y^2/2 dy`
= `[(y + 4)^2/2]_-2^4 - 1/2[y^3/3]_-2^4`
= `1/2 [64 - 4] - 1/6[64 + 8]`
= 30 – 12
= 18 sq. units.
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