Is the Integral Manipulation Explanation Correct?

[longrob]

This wasn't obvious to me.

From my book. We have,
$$p(x)=\begin{cases}<br /> \frac{1}{2} & -2\leq x\leq-1\,\textrm{or}\;1\leq x\leq2,\\<br /> 0 & \text{otherwise~}.\end{cases}$$

So, the kth moment is given by
$$M_{k}=\frac{1}{2}\int_{-2}^{-1}x^{k}dx+\frac{1}{2}\int_{1}^{2}x^{k}dx$$

So, obviously,
$$M_{k}=(1+(-1)^{k})\frac{1}{2}\int_{1}^{2}x^{k}dx$$

Is this a correct explanation of this step:
Since p(x) is an even function, it follows that the left integral is exactly equal to the right integral when k is even, and exactly equal to -1 times the right integral when k is odd. Can it be explained better/ more rigorously ? Will this explanation always apply to any even real-valued function of x ?

 

[mathman]

Your analysis is correct. It will apply to any even function, implicitly assuming the integral exists. The function doesn't have be real-valued, just even. Also the integral being zero applies to any odd function, with the same implicit assumption.