JEE Advanced 2024 Paper 2 · Q16 · Definite Integrals

Question

PARAGRAPH "II": Let $f : \left[0, \dfrac{\pi}{2}\right] \to [0, 1]$ be the function defined by $f(x) = \sin^{2} x$ and let $g : \left[0, \dfrac{\pi}{2}\right] \to [0, \infty)$ be the function defined by $g(x) = \sqrt{\dfrac{\pi x}{2} - x^{2}}$.

The value of $\displaystyle 2\int_{0}^{\pi/2} f(x)\,g(x)\,dx - \int_{0}^{\pi/2} g(x)\,dx$ is ______.

SolveKar AI · Accepted Answer

Note $g(x) = \sqrt{\pi x/2 - x^{2}} = \sqrt{(\pi/4)^{2} - (x - \pi/4)^{2}}$ — a semicircle of radius $\pi/4$ centred at $x = \pi/4$, hence $g$ is symmetric: $g(\pi/2 - x) = g(x)$. Let $I = \int_{0}^{\pi/2} f(x) g(x)\,dx = \int_{0}^{\pi/2} \sin^{2} x\,g(x)\,dx$. Substitute $x 	o \pi/2 - x$: $I = \int_{0}^{\pi/2} \cos^{2} x\,g(x)\,dx$. Adding, $2I = \int_{0}^{\pi/2} g(x)\,dx$. Therefore $2I - \int_{0}^{\pi/2} g(x)\,dx = 0$.

JEE Advanced 2024 Paper 2 · Q16 · Definite Integrals

Solution