Question

Evaluate the integral
${\int }_0^a\sqrt{a^2-x^2}dx$

• A. $\frac{a^2}{4}$
• B. $\pi a^2$
• C. $\frac{\pi a^2}{2}$
• D. $\frac{\pi a^2}{4}$

Answer 1

Answer: (D)

$\frac{\pi a^2}{4}$

${\int }_0^a\sqrt{a^2-x^2}dx$

$\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)+c$

${\int }_0^a\sqrt{a^2-x^2}\cdot dx={[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)+c]}_0^a$

$=[\frac{a}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{a}{a}\right)+c]-[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{0}{a}\right)+c]$

$=\left[\frac{a^2}{2}{\sin }^{-1}\right(1)+c]-[\sqrt{a^2}+c]$

$=\frac{a^2}{2}\left(\frac{\pi }{2}\right)$

$=\frac{\pi a^2}{4}$