Expected value of a normally distributed random variable

How do you compute the expected value of a random variable? The problem I found asks; $$W = rV^3$$ where $r$ is a constant and $V$ is a normally distributed random variable with mean 6 and standard deviation 1. How can I compute $E[W]$?

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Write $V = \mu + \sigma Z$, where $Z$ is the standard normal random variable. Then $$\mathbb{E}(W) = r \mathbb{E}\left( \left(\mu + \sigma Z\right)^3 \right) = r \left( \mu^3 +3 \mu^2 \sigma \mathbb{E}(Z) + 3 \mu \sigma^2 \mathbb{E}(Z^2) + \sigma^3 \mathbb{E}(Z^3) \right)$$ Since $Z$ is symmetric, i.e. $Z \stackrel{d}{=} -Z$, it follows that $\mathbb{E}(Z) = 0$ and $\mathbb{E}(Z^3)=0$. It only remains to evaluate $$\mathbb{E}(Z^2) = \int_{-\infty}^\infty z^2 \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-z^2/2} \mathrm{d} z = \sqrt{\frac{2}{\pi}} \int_0^\infty z^2 \mathrm{e}^{-z^2/2} \mathrm{d} z \stackrel{z^2 = x}{=} \frac{1}{\sqrt{2\pi}} \underbrace{\int_0^\infty \sqrt{x} \mathrm{e}^{-x/2} \mathrm{d} x}_{2^{3/2} \Gamma(3/2) = \sqrt{2\pi}} = 1$$ Now combine these results to readily obtain $\mathbb{E}(W)$.
+1 To avoid the integral, note that $\mathbb{E}(Z^2)$ is the sum of its variance and the square of its mean, i.e. (as $Z$ has a standard normal distribution) $1+0^2=1$. –  Henry Oct 19 '12 at 7:03
$E[W] = rE[V^{3}]$ since $r$ is a constant. You can obtain the information about higher moments of the normal distribution here http://en.wikipedia.org/wiki/Normal_distribution See "Raw moment"