# What does it mean that the probability density function is proportional to a function?

I'm studying for SOA/CAS Exam P and I have a problem that says that $X$ is a continuous and positive random variable whose probability density function is proportional to: $$\frac{1}{(1+x)^5}$$ Where $0\lt x \lt \infty$ and I need to find $E(X)$. But how do I use the information that $X$ is proportional to that function? What does it mean exactly?

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It means that the density function is a constant times the given expression. So $$f_X(x)=\frac{c}{(1+x)^5}$$ for some constant $c$ (when $0\lt x\lt \infty$). The "positive random variable" part means that $f_X(x)=0$ elsewhere.
The constant $c$ must be chosen so that $\displaystyle \int_0^\infty f_X(x)\,dx=1$. After doing the calculation, you should end up with $c=4$.
After we know the constant $c$, we can find $E(X)$ by the usual $$E(X)=\int_0^\infty x\frac{c}{(1+x)^5}\,dx.$$ The standard approach is to make the substitution $u=1+x$. There is an easier way, by noting that $\dfrac{x}{(1+x)^5}=\dfrac{1}{(1+x)^4}-\dfrac{1}{(1+x)^5}$.
After doing the calculation, you should end up with $E(X)=\dfrac{1}{3}$.