## Thomas' Calculus 13th Edition

we use integration by parts with u=$\sqrt{4-x^2})$ and dv =1/2 to find $\int \sqrt{4- x^2}dx$. Geometrically, the region R is a quarter of a circle of radius 2 with a triangle of area 2 removed, giving area $\pi-$ 2.
$\int^{}_{} \int^{}_{R}dA=\int^{2}_{0} \int^{\sqrt{4-x^2}}_{2-x}1dydx$ =$\int^{2}_{0}(\sqrt{4-x^2}-(2-x))dx$ =$(\frac{x}{2}\sqrt{4-x^2}+2sin^{-1}\frac{x}{2}+\frac{x^2}{2}-2x)]^2_0$ =$\pi-2$ The region R is shaed in given graph below