JEE Advanced · Mathematics · 27. Definite Integration
Paragraph: 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 \(g(x)=\sqrt{\frac{\pi x}{2}-x^2}\).
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
- A 0
- B 15
- C 12
- D 10
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
\(I=2 \int_0^{\frac{\pi}{2}} \underbrace{\sin ^2 x \cdot \sqrt{\frac{\pi x}{2}-x^2}}_{I_1}-\int_0^{\frac{\pi}{2}} g(x) d x\)
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 \(I=2 I_1-\int_0^{\frac{\pi}{2}} g(x)=0\)
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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