Paragraph: 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 ). Question: The value of 2 ∫_0^ (π)/(2) f(x)

Correct answer: 0 I=2 ∫_0^ (π)/(2) ^2 x · (π x)/(2)-x^2 _ I_1 -∫_0^ (π)/(2) g(x) d x Let I_1=∫_0^ (π)/(2) ^2 x ((π)/(4))^2-(x-(π)/(4))^2 (making perfect square) apply kings I_1=∫_0^ (π)/(2) ^2 x ((π)/(4))^2-((π)/(2)-x)^2 add both 2 I_1=∫_0^ (π)/(2) ((π)/(4))^2-(x-(π)/(4))^2 i.e. 2 I_1=∫_0^ (π)/(2) g(x) Now I=2 I_1-∫_0^ (π)/(2) g(x)=0

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

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

Paragraph: 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}

). Question: 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

Correct answer

Step-by-step solution

I=2 \int_0^{\frac{\pi}{2}} \underbrace{\sin ^2 x \cdot \sqrt{\frac{\pi x}{2}-x^2}}_{I_1}-\int_0^{\frac{\pi}{2}} g(x) d x

Let

I_1=\int_0^{\frac{\pi}{2}} \sin ^2 x \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2}

(making perfect square) apply kings

I_1=\int_0^{\frac{\pi}{2}} \cos ^2 x \sqrt{

\left(\frac{\pi}{4}\right)^2-\left(\frac{\pi}{2}-x\right)^2}

add both

2 I_1=\int_0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2}

i.e.

2 I_1=\int_0^{\frac{\pi}{2}} g(x)

Now

I=2 I_1-\int_0^{\frac{\pi}{2}} g(x)=0

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