Definite Integration Question 426
Question: The area enclosed by the curve $ y=\sqrt{4-x^{2}}, $
$ y\ge \sqrt{2}\sin ( \frac{x\pi }{2\sqrt{2}} ) $ , and the y-axis divides the x-axis in the ratio
Options:
A) $ \frac{{{\pi }^{2}}-8}{{{\pi }^{2}}+8} $
B) $ \frac{{{\pi }^{2}}-4}{{{\pi }^{2}}+4} $
C) $ \frac{\pi -4}{\pi -4} $
D) $ \frac{2{{\pi }^{2}}}{2\pi +{{\pi }^{2}}-8} $
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Answer:
Correct Answer: D
Solution:
[d] $ y=\sqrt{4-x^{2}},y=\sqrt{2}\sin ( \frac{x\pi }{2\sqrt{2}} ) $ intersect at $ x=\sqrt{2} $ Area to the left of y-axis is $ \pi $ Area to the right of y-axis $ =\int\limits_0^{\sqrt{2}}{( \sqrt{4-x^{2}}-\sqrt{2}\sin \frac{x\pi }{2\sqrt{2}} )dx} $
$ =( \frac{x\sqrt{4-x^{2}}}{2}+\frac{4}{2}{{\sin }^{-1}}\frac{x}{2} )_0^{\sqrt{2}}+( \frac{4}{\pi }\cos \frac{x\pi }{2\sqrt{2}} )_0^{\sqrt{2}} $
$ =( 1+2\times \frac{\pi }{4} )+\frac{4}{\pi }(0-1)=1+\frac{\pi }{2}-\frac{4}{\pi } $
$ =\frac{2\pi +{{\pi }^{2}}}{2\pi } $ Sq. units. $ \therefore $ Ratio $ =\frac{2{{\pi }^{2}}}{2\pi +{{\pi }^{2}}-8} $
BETA
