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JEE Main 2019 · 11 Jan · Shift I

JEE Main 2019 Mathematics question on Definite Integrals and Areas

The area (in sq units) of the region bounded by the curve x2=4yx^2 = 4y and the straight line x=4y−2x = 4y - 2 is
  1. A.
    78\frac{7}{8}
  2. B.
    98\frac{9}{8}
  3. C.
    54\frac{5}{4}
  4. D.
    34\frac{3}{4}
Show answer and solution

Correct option: B

Solution

Find intersections: x² = 4y and x = 4y - 2, so 4y = x + 2, giving x² = x + 2, i.e. x² - x - 2 = 0, so x = -1 and x = 2.

Area = ∫from -1 to 2 [(x+2)/4 - x²/4] dx = (1/4)∫[(x+2-x²)]dx

= (1/4)[x²/2 + 2x - x³/3] from -1 to 2

= (1/4)[(2 + 4 - 8/3) - (1/2 - 2 + 1/3)]

= (1/4)[(10/3) - (-7/6)]

= (1/4)[20/6 + 7/6] = (1/4)(27/6) = 27/24 = 9/8.

[JEE Main 2019 · 11 Jan · Shift I]

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