# Math Help - [SOLVED] integral in terms of pi

1. ## [SOLVED] integral in terms of pi

Hello,

Having established that $\pi = 2\int^1_{-1} \! \sqrt{1 - x^2} \, dx$,
I need to express $\int^2_{-2} \! (x - 3)\sqrt{4 - x^2} \, dx$ in terms of pi by using the properties of intergrals (linearity with respect to the integrand, additivity with respect to the interval of integration, invariance under translation, and expansion/ contraction of the interval of integration).

I can get as far as $\int^1_{-1} \! 4x \sqrt{1 - x^2} \, dx - 6\pi$, and according to the book, the answer is -6pi, so if I'm right so far, the first integral evaluates to 0. I just don't know how to show this.

2. Never mind, I got it