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JEE Main 2023 · 24 Jan · Shift 2

JEE Main 2023 Mathematics question on Definite Integrals and Areas

∫324334489−4x2 dx\int_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}} \frac{48}{\sqrt{9-4x^2}}\, dx is equal to
  1. A.
    π6\dfrac{\pi}{6}
  2. B.
    π3\dfrac{\pi}{3}
  3. C.
    2π2\pi
  4. D.
    π2\dfrac{\pi}{2}
Show answer and solution

Correct option: C

Solution

Rewrite the integral by factoring: 9 - 4x² = 9(1 - (2x/3)²), so √(9-4x²) = 3√(1-(2x/3)²).

The integral becomes ∫ 48/[3√(1-(2x/3)²)] dx = 16 ∫ dx/√(1-(2x/3)²).

Substitute u = 2x/3, du = 2/3 dx, so dx = (3/2)du:

= 16 · (3/2) ∫ du/√(1-u²) = 24 [arcsin(u)] = 24 [arcsin(2x/3)]

Evaluate from x = 3√2/4 to x = 3√3/4:

At x = 3√3/4: 2x/3 = √3/2, arcsin(√3/2) = π/3
At x = 3√2/4: 2x/3 = √2/2, arcsin(√2/2) = π/4

Result = 24(π/3 - π/4) = 24 · π/12 = 2π

[JEE Main 2023 · 24 Jan · Shift 2]

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