Question

The area of the region bounded by the curve x² = 4y and the straight line x = 4y – 2 is ______.

पर्याय

• `3/8` sq.units

• `5/8` sq.units

• `7/8` sq.units

• `9/8` sq.units

Answer 1

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