Answer
$\frac{2\pi}{3}+\sqrt 3-2$
Work Step by Step
$A=\int_{0}^{1}\frac{4-x}{\sqrt (4-x^{2})}dx$
$A=A1+A2$
$A1=\int_{0}^{1}\frac{4}{\sqrt (4-x^{2})}dx$
The denominator looks like $arcsin$.
$\int\frac{du}{\sqrt (a^{2}-u^{2})}=arcsin(\frac{u}{a})+C$
$a=2$
$u=x$
$du=dx$
$A1= 4\int_{0}^{1}\frac{1}{\sqrt (4-x^{2})}dx$
$A1=4[arcsin(\frac{x}{2})]_{0}^{1}$
$A1=4[arcsin\frac{1}{2}-arcsin0]$
$A1=4[\frac{\pi}{6}-0]$
$A1=\frac{2\pi}{3}$
$A2=\int_{0}^{1}\frac{-x}{\sqrt (4-x^{2})}dx$
Since there is a $x$ in the numerator, u-substitution seems like it could work.
let $u = 4-x^{2}$
$du=-2x$
$u(0)=4$
$u(1)=3$
$\frac{1}{2}du=-xdx$
$A2=\frac{1}{2}\int_{4}^{3}\frac{1}{\sqrt u}du$
$A2=\frac{1}{2}[2u^{\frac{1}{2}}]_{4}^{3}$
$A2=\frac{1}{2}[2\sqrt 3-2\sqrt 4]$
$A2=\sqrt 3-2$
$\frac{2\pi}{3}+\sqrt 3-2$
