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Question
Find the area of the region included between y2 = 9x and y = x
Solution
Given that, y2 = 9x .....(i)
And y = x .....(ii)
Solving equation. (i) and (ii)
We have x2 = 9x
⇒ x2 – 9x = 0
x(x – 9) = 0
∴ x = 0, 9
Required area = `int_0^9 sqrt(9x) "d"x - int_0^9 x "d"x`
= `3int_0^9 sqrt(x) "d"x - int_0^9 x "d"x`
= `3 * 2/3 [x^(3/2)]_0^9 - 1/2 [x^2]_0^9`
= `2[(9)^(3/2) - 0] - 1/2 [(9)^2 - 0]`
= `2(27) - 1/2 (81)`
= `54 - 81/2`
= `(108 - 81)/2`
= `27/2` sq.units
Hence, the required area = `27/2` sq.units
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