MHTCET 11 Apr 2026 Shift 1 Paper

Concept:The integral represents the area under the curve $y = \sqrt{9-x^2}$ from $x=0$ to $x=3$.Explanation:The equation $y = \sqrt{9-x^2}$ can be squared to get $x^2 + y^2 = 9$.This is a circle with center at the origin and radius $r = 3$.Since $y$ is positive, the curve gives only the upper half of the circle.The limits $x = 0$ to $x = 3$ cover the right quarter of that upper half.So the integral equals the area of one-fourth of the full circle.Area of the full circle is $\pi r^2 = \pi(3)^2 = 9\pi$.Therefore, the required area is $\frac{1}{4} \times 9\pi = \frac{9\pi}{4}$.Answer:$\int_{0}^{3} \sqrt{9-x^{2}} dx = \frac{9\pi}{4}$Correct option: B. $\frac{9\pi}{4}$