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Question
Find the area bounded by the curve y2 = 36x, the line x = 2 in first quadrant
Solution
Given equation of the curve is y2 = 36x
∴ y = `+- sqrt(36x)`
∴ y = `6sqrt(x)` .....[∵ In first quadrant, y > 0]
Required area = `int_0^2 y "d"x`
= `int_0^2 6sqrt(x) "d"x`
= `6[(x^(3/2))/(3/2)]_0^2`
= `4[(2)^(3/2) - 0]`
= `4(2sqrt(2))`
= `8sqrt(2)` sq.units
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