A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{6}\]
Correct Answer: D
Solution :
\[\int_{1/4}^{1/2}{\frac{dx}{\sqrt{x-{{x}^{2}}}}=\int_{1/4}^{1/2}{\frac{dx}{\sqrt{{{\left( \frac{1}{2} \right)}^{2}}-{{\left( x-\frac{1}{2} \right)}^{2}}}}}}=\left[ {{\sin }^{-1}}\left( \frac{\frac{2x-1}{2}}{1/2} \right) \right]_{1/4}^{1/2}\] \[=[{{\sin }^{-1}}(2x-1)]_{1/6}^{1/2}=\pi /6\].
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