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Question
The area of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 is ______.
Options
`3/8` sq.units
`5/8` sq.units
`7/8` sq.units
`9/8` sq.units
Solution
The area of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 is `9/8` sq.units.
Explanation:
Given that: The equation of parabola is x2 = 4y .....(i)
And equation of straight line is x = 4y – 2 .....(ii)
Solving equation (i) and (ii)
We get y = `x^2/4`
x = `4(x^2/4) - 2`
⇒ x = x2 – 2
⇒ x2 – x – 2 = 0
⇒ x2 – 2x + x – 2 = 0
⇒ x(x – 2) + 1(x – 2) = 0
⇒ (x – 2)(x + 1) = 0
∴ x = –1, x = 2
Required area = `int_(-1)^2 (x + 2)/4 "d"x - int_(-1)^2 x^2/4 "d"x`
= `1/4 [x^2/2 + 2x]_-1^2 - 1/4 * 1/3 [x^3]_1^2`
= `1/4 [(4/2 + 4) - (1/2 - 2)] - 1/12[8 + 1]`
= `1/4 [6 + 3/2] - 1/12 [9]`
= `1/4 xx 15/2 - 3/4`
= `15/8 - 3/4`
= `9/8` sq.units
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