The area enclosed by the curves y= 4-x^ 2 , y 2 ( x 2 2 ) and x- axis is divided by y…
Mathematics · JEE Advanced · NTA Exams — Integral Calculus
The area enclosed by the curves
\(y=\sqrt{4-x^{2}}, y \geq \sqrt{2} \sin \left(\frac{x \pi}{2 \sqrt{2}}\right)\) and \(x-\) axis is divided by \(y-\) axis in the ratio
\(y=\sqrt{4-x^{2}}, y \geq \sqrt{2} \sin \left(\frac{x \pi}{2 \sqrt{2}}\right)\) and \(x-\) axis is divided by \(y-\) axis in the ratio
- \(\frac{\pi^{2}-8}{\pi^{2}+8}\)
- \(\frac{\pi^{2}-4}{\pi^{2}+4}\)
- \(\frac{\pi-4}{\pi+4}\)
- \(\frac{2 \pi^{2}}{2 \pi+\pi^{2}-8}\)
Answer
(D) 2 ^ 2 2 + ^ 2 -8
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