Q. $\int_{0}^{a}\sqrt{a^{2} - x^{2}}\ dx$ is equal to

A) $\pi a^{2}$
B) $\frac{1}{2}\pi a^{2}$
C) $\frac{1}{3}\pi a^{2}$
D) $\frac{1}{4}\pi a^{2}$
Explanation

(d)
$\int_{0}^{a}\sqrt{a^{2} - x^{2}}\ dx = \left\lbrack \frac{x}{2}\sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2}\sin^{- 1}\left( \frac{x}{a} \right) \right\rbrack_{0}^{a}$
$= \left\lbrack 0 + \frac{a^{2}}{2}\sin^{- 1}(1) - 0 - \frac{a^{2}}{2}\sin^{- 1}(0) \right\rbrack$
$= \frac{a^{2}}{2}.\frac{\pi}{2} - 0 = \frac{a^{2}\pi}{4}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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