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JEE Main 2019 · 9 April · Shift II

JEE Main 2019 Mathematics question on Definite Integrals and Areas

The area (in sq units) of the region A={(x,y):y22≤x≤y+4}A = \left\{(x, y): \frac{y^2}{2} \leq x \leq y + 4\right\} is
  1. A.
    30
  2. B.
    533\frac{53}{3}
  3. C.
    16
  4. D.
    18
Show answer and solution

Correct option: D

Solution

Find intersection of y²/2 = y + 4: y² = 2y + 8, so y² - 2y - 8 = 0, giving y = 4 and y = -2.

Area = ∫ from y = -2 to y = 4 of [(y + 4) - y²/2] dy

= [y²/2 + 4y - y³/6] from -2 to 4

At y = 4: 8 + 16 - 64/6 = 24 - 32/3 = 40/3
At y = -2: 2 - 8 + 8/6 = -6 + 4/3 = -14/3

Area = 40/3 - (-14/3) = 54/3 = 18.

[JEE Main 2019 · 9 April · Shift II]

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