Let (f: [ 0, (π)/(2)] arrow[ 0,1]) be the function defined by (f(x)= ^2 x) and let (g:[ 0, (π)/(2)] arrow[ 0, ∞) ) be the function defined by (g(x)= (π x)/(2)-x^2 ). The value of (16)/(π^3) ∫_0^ (π)/(2) f(x) g(x) d x is

Correct answer: 0 Now I_1=∫_0^ (π)/(2) f(x) · g(x) d x=(1)/(2) ∫_0^ (π)/(2) g(x) d x (it is explained in previous question solution) i.e. (1)/(2) ∫_0^ (π)/(2) ((π)/(4))^2-(x-(π)/(4))^2 dx Using ∫ a^2-x^2 =(1)/(2)(x a^2-x^2 +a^2 ^ -1 ((x)/(a)))+C (1)/(2)[((x-(π)/(4)))/(2) (π x)/(2)-x^2 +((π^2)/(2))/(2) ^ -1 ((x-(π)/(4))/((π)/(4)))]_0^ π / 2 aligned & (1)/(2)[(0+(π^3)/(64))-(0+((-π^3)/(64)))] & (1)/(2) × (π^3)/(32) aligned Now (16)/(π^3) × (π^3)/(64)=(1)/(4)=0.25

JEE Advanced 2024Paper 2

Exam: JEE Advanced
Year: 2024
Paper: Paper 2
Subject: Mathematics
Chapter: Definite Integration

JEE Advanced 2024 · Paper 2 · Mathematics · Definite Integration

Let (f: [ 0, (π)/(2)] arrow[ 0,1]) be the function defined by...

Let (f:

\left[

\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(x)=

\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

Correct answer

Step-by-step solution

Now

I_1=\int_0^{\frac{\pi}{2}} f(x) \cdot g(x) d x=\frac{1}{2} \int_0^{\frac{\pi}{2}} g(x) d x

(it is explained in previous question solution) i.e.

\frac{1}{2} \int_0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2} \mathrm{dx}

Using

\int \sqrt{a^2-x^2}=\frac{1}{2}\left(x \sqrt{a^2-x^2}+a^2 \sin ^{-1}\left(\frac{x}{a}\right)\right)+C \Rightarrow \frac{1}{2}\left[\frac{\left(x-\frac{\pi}{4}\right)}{2} \sqrt{\frac{\pi x}{2}-x^2}+\frac{\frac{\pi^2}{2}}{2} \sin ^{-1}\left(\frac{x-\frac{\pi}{4}}{\frac{\pi}{4}}\right)\right]_0^{\pi / 2}

\begin{aligned} & \Rightarrow \frac{1}{2}\left[\left(0+\frac{\pi^3}{64}\right)-\left(0+\left(\frac{-\pi^3}{64}\right)\right)\right] \\ & \Rightarrow \frac{1}{2} \times \frac{\pi^3}{32}\end{aligned}

Now

\frac{16}{\pi^3} \times \frac{\pi^3}{64}=\frac{1}{4}=0.25$

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