A) \[\frac{3\pi }{8}\] sq. unit
B) \[\frac{5\pi }{8}\] sq. unit
C) \[\frac{\pi }{2}\] sq. unit
D) \[\frac{\pi }{8}\] sq. unit
Correct Answer: C
Solution :
[c] required area = area of the shaded region = 4 (area of the shaded region in first quadrant) \[=4\int_{0}^{1/\sqrt{2}}{({{y}_{1}}-{{y}_{2}})dx=4\int_{0}^{1/\sqrt{2}}{(\sqrt{1-{{x}^{2}}}-x)dx}}\] \[=4\left[ \frac{1}{2}\times \sqrt{1-{{x}^{2}}}+\frac{1}{2}{{\sin }^{-1}}x-\frac{{{x}^{2}}}{2} \right]_{0}^{1/\sqrt{2}}=\frac{\pi }{2}\] sq. unitYou need to login to perform this action.
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