Question

Use a definite integral to find the area of the region between the given curve and the x-axis on the interval $[0,b]$:
$y=\sqrt{b^2-x^2}$

Answer 1

$A={\int }_0^{b}ydx={\int }_0^b\sqrt{b^2-x^2}dx$

$={\left[\frac{x}{2}\sqrt{b^2-x^2}+\frac{b^2}{2}{\sin }^{-1}\frac{x}{b}\right]}_0^b$

$=\frac{b^2}{2}\frac{\pi }{2}$