Biased/Unbiased estimator I'm trying to solve a statistic exam and i got lost with this exercise.
1) Consider a sample from a continuos probability distribution with density:
$$
f(x) = \begin{cases} (1+\theta x)/8 & \text{if }-4<x<4, \\ 0 & \text{otherwise.} \end{cases}
$$               
a) Is $\bar\theta = \frac{3}{16}X$ (this $X$ is the sample median) an unbiased estimator of $\theta$?
Where should i start? Should i integrate the function, and compare the expected value of it with the expected value of the estimator?
How do i integrate if i don't know which value should i assign to $\theta$? (should i integrate leaving it as a constant?)
I hope you understand, i translated the exercise, maybe some term are different in english. 

Often one uses a lower-case $x$ for the argument to the density function or to the c.d.f., and a capital $X$ to refer to the random variable whose density it is.  But you've said your capital $X$ is the sample median, and that suggests a sample size more than $1$. So I am uncertain of the meaning of the question.
If we assume you mean simple a sample of one observation, then one can find an expected value by integrating:
$$
\mathbb E X = \int_{-4}^4 xf(x)\,dx = \int_{-4}^4 x\frac{1+\theta x}{8}\,dx
$$
$$
= \frac18 \left(\int_{-4}^4 \frac x8\,dx + \theta \int_{-4}^4 x^2\,dx \right)
$$
The first integral is readliy seen to be $0$ (you're integrating an odd function over an interval symmetric about $0$).  The second one comes to $16\theta/3$.
So $\frac{3}{16}\mathbb E X = \theta$.  The expression $\frac3{16}X$ is "observable" i.e. does not depend on $\theta$, and its expected value remains equal to $\theta$ if $\theta$ changes.
