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JEE Main 2019 · 8 April · Shift I

JEE Main 2019 Mathematics question on Definite Integrals and Areas

The area (in sq units) of the region A={(x,y)∈R×R∣0≤x≤3, 0≤y≤4, y≤x2+3x}A = \{(x, y) \in R \times R \mid 0 \leq x \leq 3,\ 0 \leq y \leq 4,\ y \leq x^2 + 3x\} is
  1. A.
    536\dfrac{53}{6}
  2. B.
    88
  3. C.
    596\dfrac{59}{6}
  4. D.
    263\dfrac{26}{3}
Show answer and solution

Correct option: C

Solution

The region requires 0 ≤ x ≤ 3, 0 ≤ y ≤ 4, and y ≤ x² + 3x.

Since x² + 3x ≥ 4 when x² + 3x - 4 ≥ 0, i.e., (x+4)(x-1) ≥ 0, so for x ≥ 1 the parabola is at or above y = 4. Thus:

- For 0 ≤ x ≤ 1: the effective upper bound on y is min(x² + 3x, 4) = x² + 3x (since x² + 3x ≤ 4 here).
- For 1 ≤ x ≤ 3: the effective upper bound is min(x² + 3x, 4) = 4.

Area = ∫₀¹ (x² + 3x) dx + ∫₁³ 4 dx

∫₀¹ (x² + 3x) dx = [x³/3 + 3x²/2]₀¹ = 1/3 + 3/2 = 2/6 + 9/6 = 11/6

∫₁³ 4 dx = 4 × 2 = 8 = 48/6

Total = 11/6 + 48/6 = 59/6

[JEE Main 2019 · 8 April · Shift I]

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