Definite Integration
PYP_JEE_ADV_2024_P2
Grade None
Question:
PARAGRAPH II
Let $f : [0, \pi/2] \to [0, 1]$ be the function defined by $f(x) = \sin^2 x$ and let $g : [0, \pi/2] \to [0, \infty)$ be the function defined by $g(x) = \sqrt{\dfrac{\pi x}{2} - x^2}$.
The value of $\dfrac{16}{\pi^3} \int_{0}^{\pi/2} f(x)g(x) dx$ is ___.
Step-by-Step Solution
Key Concept: Using the relation between integrals from symmetric cancellation and identifying standard geometric shapes (like circles/semicircles) in integrals.
From Q16, we have:
$$2 \int_{0}^{\pi/2} f(x)g(x) dx = \int_{0}^{\pi/2} g(x) dx \implies \int_{0}^{\pi/2} f(x)g(x) dx = \dfrac{1}{2} \int_{0}^{\pi/2} g(x) dx$$
Let's evaluate $\int_{0}^{\pi/2} g(x) dx$:
$$J = \int_{0}^{\pi/2} \sqrt{\dfrac{\pi x}{2} - x^2} dx = \int_{0}^{\pi/2} \sqrt{\dfrac{\pi^2}{16} - \left(x - \dfrac{\pi}{4}\right)^2} dx$$
Substitute $v = x - \dfrac{\pi}{4}$:
$$J = \int_{-\pi/4}^{\pi/4} \sqrt{\left(\dfrac{\pi}{4}\right)^2 - v^2} dv$$
This integral represents the area of a semicircle of radius $R = \dfrac{\pi}{4}$:
$$J = \dfrac{1}{2} \pi R^2 = \dfrac{1}{2} \pi \left(\dfrac{\pi}{4}\right)^2 = \dfrac{\pi^3}{32}$$
Thus:
$$\int_{0}^{\pi/2} f(x)g(x) dx = \dfrac{1}{2} J = \dfrac{\pi^3}{64}$$
Now calculate the required value:
$$\dfrac{16}{\pi^3} \int_{0}^{\pi/2} f(x)g(x) dx = \dfrac{16}{\pi^3} \left(\dfrac{\pi^3}{64}\right) = \dfrac{16}{64} = 0.25$$
Thus, the answer is 0.25.
Correct Answer: 0.25