Expectation

$\displaystyle E(X)$	$\displaystyle =$	$\displaystyle \int_{-\infty}^{\infty} x f(x) dx$
	$\displaystyle =$	$\displaystyle \int_{-\infty}^{-1} 0dx + \int_{-1}^{0}x(1+x)dx + \int_{0}^{1}x(1-x)dx + \int_{1}^{\infty} 0dx$
	$\displaystyle =$	$\displaystyle \left[\frac{x^2}{2} + \frac{x^3}{3}\right]_{-1}^{0} + \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_{0}^{1} = 0$

$\displaystyle E(X^2)$	$\displaystyle =$	$\displaystyle \int_{-\infty}^{\infty} x^2 f(x) dx$
	$\displaystyle =$	$\displaystyle \int_{-\infty}^{-1} 0dx + \int_{-1}^{0}x^2(1+x)dx + \int_{0}^{1}x^2(1-x)dx + \int_{1}^{\infty} 0dx$
	$\displaystyle =$	$\displaystyle \left[\frac{x^3}{3} + \frac{x^4}{4}\right]_{-1}^{0} + \left[\frac{x^3}{3} - \frac{x^4}{4}\right]_{0}^{1}$
	$\displaystyle =$	$\displaystyle -(-\frac{1}{3} + \frac{1}{4}) + \frac{1}{3} - \frac{1}{4} = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}$