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Question
Find the area of the region lying in the first quadrant bounded by the region y = 4x2, x = 0, y = 0 and y = 4
Solution
The equation of a parabola given is y = 4x2
⇒ x2 = `y/4`
⇒ x = `sqrt(y)/2`
∴ x = `sqrt(y/4)`
Also y varies from 0 to 4.
Area of the region lying in the first quadrant
A = `int_"a"^"b" x "d"y`
= `int_0^4 sqrt(y)/2 "d"y`
= `1/2 int_0^4 (y)^(1/2) "d"y`
= `1/2 [y^(3/2)/(3/2)]_0^4`
= `1/2 xx 2/3 (y^(3/2))_0^4`
= `1/3 {(4)^(3/2) - (0)}`
= `1/3 xx 2/3 [y^(3/2)]_0^4`
= `1/3 {(4)^(3/2) - (0)}`
= `1/3 [4sqrt(4)]`
= `1/3 [4(2)]`
A = `8/3` sq.units
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