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 2 int 0 π/2 f(x) g(x) d x-int 0 π/2 g(x) d x is

Mathematics · Definite Integration

JEE Advanced 2024 — Paper 2 — Question 38

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 $2 \int_{0}^{\frac{\pi}{2}} f(x) g(x) d x-\int_{0}^{\frac{\pi}{2}} g(x) d x$ is $\qquad$

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Step-by-step solution

$\quad I_{1}=2 \int_{0}^{\frac{\pi}{2}} f(x) \cdot g(x) d x=2 \int_{0}^{\frac{\pi}{2}} \sin ^{2} x \cdot g(x) d x$ $I_{1}=2 \int_{0}^{\frac{\pi}{2}} \cos ^{2} x \cdot g(x) d x$ $2 I_{1}=2 \int_{0}^{\frac{\pi}{2}} g(x) d x$ $I_{1}-\int_{0}^{\frac{\pi}{2}} g(x) d x=0$

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

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