The area enclosed by the ellipse x^2/16 + y^2/9 = 1 is given by : Option 1: 3 _0^4 sqrt(9 - x^2) dx Option 2: 3/4 _0^4 sqrt(9 - x^2) dx Option 3: 3 _0^4 sqrt(16 - x^2) dx Option 4: 3/4 _0^4 sqrt(16 - x^2) dx

The answer is '3 _0^4 sqrt(16 - x^2) dx'. View full solution:

CUET Mathematics - The area enclosed by the ellipse x^2/16 + y^2/9 = 1 is given by : | PYQs + Solutions

CUET Mathematics 2023 22 May Shift 3

Calculus

Easy

$3 \int_0^4 \sqrt{9 - x^2} \, dx$

$\frac{3}{4} \int_0^4 \sqrt{9 - x^2} \, dx$

$3 \int_0^4 \sqrt{16 - x^2} \, dx$

$\frac{3}{4} \int_0^4 \sqrt{16 - x^2} \, dx$

✅ Correct Option: 3