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4 4 votes
Solution. We have

$$
E(|X|)=\int_{-\infty}^{\infty}|x| \frac{1}{4} e^{-|x| / 2} \mathrm{~d} x=2 \int_{0}^{\infty} x \frac{1}{4} e^{-x / 2} \mathrm{~d} x=\int_{0}^{\infty} x \frac{1}{2} e^{-x / 2} \mathrm{~d} x=2
$$

where the last equality follows by observing that the integral is nothing but the expected value of an exponentially distributed random variable with mean 2 .
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