<s>[INST] Given that $\pi=2\int_{-1}^1\sqrt{1-x^2}dx$ use the properties of the integral to compute $\int_{-2}^{2}(x-3)\sqrt{4-x^2}dx$ in terms of $\pi$. This is part c of the exercise set 2.4 in Apostol's Calculus 2nd Ed. We are only supposed to use the properties of the integral such as: linearity, additivity with respect to the interval of integration, invariance under translation, expansion or contraction of the interval of integration, or any consequences of them. A hint would be helpful.
 [/INST] By linearity
$$\int_{-2}^{2} (x-3) \sqrt{4-x^2}dx = \int_{-2}^{2} x \sqrt{4-x^2}dx - 3 \int_{-2}^{2} \sqrt{4-x^2}dx$$
But $\int_{-b}^b f(x) dx = 0$ for odd $f(x)$ (Optional Exercise 25), and $f(x) = x \sqrt{4-x^2}$ is odd.
\begin{align}
\int_{-2}^{2} (x-3) \sqrt{4-x^2}dx &= - 3 \int_{-2}^{2} \sqrt{4-x^2}dx\\
&= -6 \int_{-1}^{1} \sqrt{4-(2x)^2}dx \qquad \mbox{Contraction, k = 1/2,  Thm 1.19}\\
&= -12 \int_{-1}^{1} \sqrt{1-x^2}dx \qquad \mbox{Homogeneous property Thm 1.3}\\
&= -6 \pi
\end{align}
</s>