The integral $\displaystyle \int_0^2\sqrt{4-x^2}dx$ is the area of $\displaystyle \frac{1}{4}$ of the circle. So the total area should be:
$\displaystyle 4\int_0^2\sqrt{4-x^2}dx$
The indefinite integral is $\displaystyle I(x)=\frac{1}{2}\sqrt{4-x^2}x+2sin^{-1}\left(\frac{x}{2}\right)$
Wolfram Mathematica Online Integrator
So did you calculate $\displaystyle 4[I(2)-I(0)=4\pi$
Your book must be wrong. The area of a circle with radius $\displaystyle 2$ is $\displaystyle 4\pi$. You can show this using integration as I did above.