How are these two integrals related?

Question

How to express the integral $$\int_{-2}^{2} (x-3) \sqrt{4-x^2} \ dx $$ in terms of the integral $$ \int_{-1}^{1} \sqrt{1-x^2} \ dx?$$ I know that the latter integral is equal to $\pi / 2$.

We can't use substitution. We can only use the following two results:

1. $$ \int_{a}^{b} f(x) \ dx = \int_{a+c}^{b+c} f(x-c) \ dx.$$

And

1. $$ \int_{a}^{b} f(x) \ dx = \frac{1}{k} \int_{ka}^{kb} f(\frac{x}{k}) \ dx$$ for any $k \neq 0$.

Answer 1

Using $k=\frac{1}{2}$ in the second rule, we get

$$\begin{align} \int\limits_{-2}^{2} (x-3) \sqrt{4-x^2} \ dx &=2\int\limits_{-1}^{1} (2x-3) \sqrt{4-4x^2} \ dx \\&=2\int\limits_{-1}^{1} 2(2x-3) \sqrt{1-x^2} \ dx \\&=4\int\limits_{-1}^{1} (2x-3) \sqrt{1-x^2} \ dx \\&=8\int\limits_{-1}^{1}x\sqrt{1-x^2} \ dx-12\int\limits_{-1}^{1}\sqrt{1-x^2}\ dx \\&=-12\int\limits_{-1}^{1}\sqrt{1-x^2}\ dx \end{align} $$ where the last equality occurs because $x\sqrt{1-x^2}$ is an odd function and we are integrating from $-1$ to $1$, so $\int_{-1}^{1}x\sqrt{1-x^2} \ dx=0$.

Answer 2

$\textbf{Hint:}$

Use the second formula for $k=\frac{1}{2}$.You get linear combination of integral $\int_{-1}^{1} \sqrt{1-x^2} dx$ and integral $\int_{-1}^{1} -2x\sqrt{1-x^2} dx$ Integral $\int_{-1}^{1} x\sqrt{1-x^2} dx$ is easy to calculate: note that $((1-x^2)^{\frac{3}{2}})'=-3x\sqrt{1-x^2}$.