Let f: [0, π/2 ] rightarrow[0,1] be the function defined by f(x)=sin 2 x and let g: [0, π/2 ] rightarrow[0, ∞) be the function defined by g=sqrt π x/2-x 2 . The value of frac 16 π 3 int 0 π/2 f(x) g(x) d x is

Mathematics · Definite Integration

JEE Advanced 2024 — Paper 2 — Question 39

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

The value of $\frac{16}{\pi^{3}} \int_{0}^{\frac{\pi}{2}} f(x) g(x) d x$ is $\qquad$

Answer: 0.25

Numerical answer — enter this value.

Step-by-step solution

$\quad I=\frac{16}{\pi^{3}} \int_{0}^{\frac{\pi}{2}} f(x) g(x) d x=\frac{8}{\pi^{3}} \int_{0}^{\frac{\pi}{2}} g(x) d x$ $I=\frac{8}{\pi^{3}} \int_{0}^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^{2}-\left(\frac{\pi}{4}-x\right)^{2}} d x$ $\mathrm{I}=\frac{1}{4}$

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Exam: JEE Advanced 2024
Paper: Paper 2
Subject: Mathematics
Chapter: Definite Integration
Topic: Evaluation of Definite Integrals

Answer: 0.25

Step-by-step solution

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