Find the $P(X>1)$ for the given pdf? A part of this question asks me to find the $\Pr(X>1)$ given that $$f_X(x) = \begin{cases}\frac{1}{\sqrt{4x}} & 1 <x<4 \cr 0 & \text{otherwise} \end{cases}$$
I solved this by taking the integral from $1$ to $\infty$ of the $\frac{1}{\sqrt{4x}}$ function. However, I was getting $\infty$ as my answer. But then I realized that the pdf has a restriction $1< x<4$. Does that mean even if they want the $\Pr(x>1)$ I can only go 4 and not $\infty$ because of the restriction. Then, since this is a pdf, the integral over the whole pdf would be one? 
I just want to make sure I'm thinking this through correctly. Thanks. 
 A: If we believe that the function $f(x)=\frac{1}{2\sqrt{x}}$ on the interval $(1,4)$, and $f(x)=0$ elsewhere really is the probability density function for a random variable $X$, then yes, automatically  $\Pr(X\gt 1)=1$.
If we wish to check that the function $f(x)$ really is a density function, note that it is non-negative, and calculate
$$\int_1^4 \frac{1}{2x^{1/2}}\,dx.$$
The result is indeed $1$. For an antiderivative of $\frac{1}{2\sqrt{x}}$ is $x^{1/2}$, and $4^{1/2}-1^{1/2}=1$. 
A: If the PDF has a restriction of $1<x<4$ then for $x\geq4\, \ and \ x\leq1,\ f(x)=0$ so you would be correct in taking the integral from 1 to 4, and you are correct in saying that will be equal to 1 (given the PDF is defined correctly).
A: That is correct:  another way to think about it is that the density is expressed as a piecewise function.  If we integrate over all real values, we find $$\int_{x=-\infty}^\infty f_X(x) \, dx = \int_{x=-\infty}^1 0 \, dx + \int_{x=1}^4 \frac{1}{\sqrt{4x}} \, dx + \int_{x=4}^\infty 0 \, dx = 1.$$  We also immediately (and trivially) see that $\Pr[X > 1] = 1$, since the probability that $X \le 1$ is zero:  there is no density below $1$.
