Why is $P(X>x)=\int_x^{\infty}f_X(y) \ dy?$ Can someone intuitively explain why it is that $$P(X>x)=\int_x^{\infty}f_X(y) \ dy?$$
I don't see how i can enterpret the integral as an area. Area of what? what is $f_X(y)$ in this case? why integrating w.r.t $y$ if we have a function of $x$ in the left side? If we multiply both sides by $-1$ we get
$$-(1-P(X<x))=-1+F_X(x)=\int_\infty^xf_X(y) \ dy,$$
but this does not make any sense :S.
 A: If you think about a descrete random variable $X$ with a probability distribution $f(X)$ I think it's very intuitive and clear that the probability of that random variable to be more (or less) than some value is to be defined as $$P(X\geq x_k)=\sum_{x_i\geq x_k}^nf(x_i)$$ if we want $P(X\leq x_k)$ we just invert the inequality sign. This mens, for example, that if we want to know the following 

What is the probability of getting more than 4 from the toss of a dice?

we can easily calculate. We know that the probability distribution of a dice toss is a uniform probability (to put it simply: every number on the dice has the same probability of coming out from a random unbiased toss). So our probability distribution is $$f(X) = \frac{1}{6}$$from this we can evaluate the answer to our problem $$P(X>4) = P(X=1)+P(X=2)+P(X=3)+P(X=4) = \sum_{k=0}^4 f(k) = {4\over 6} = {2\over 3}$$ 
Simple enough. There're many discrete distributions: Bernoulli, geometric, binomial, Pascal's et al. But what about if our random variable isn't discrete? (this case is very more interesting, e.g. Gaussian distribution). If the random variable is distributed with a continuous probability distribution what we have is that $$\underset{\uparrow \\ \text{discrete}}{f(x_i)}\longrightarrow \underset{\uparrow\text{continuous}}{f_X(x)dx} \\\underset{\uparrow \\ \text{discrete}}{\sum_i}\longrightarrow \underset{\uparrow\text{continuous}}{\int_X} $$ we can't in any way sum the single probabilities if the variable is continuous! You can clearly imagine that in an interval, for example $[0,2]$, there're infinity many points, so $P(X\in [0,2])$ has infinitely many values to sum up! This is why we calculate the area under the probability distribution: it's just the clear "continuation" from discrete sum to continuous sum. 
The reason behind the $y$ in the integral is just because the cumulative distribution, e.g. $P(X>x)$ is a function of $x$, so to not be confused between the variable on which we integrate and the variable on the limit of the integral, we just change it's name. It would be pretty confusing reading something like $$P(X>x) = \int_\color{blue}{x}^\infty f_X(\color{red}{x})dx$$ because the  red $x$ is not the same as the $blue$ x!If we change the name to the red $x$ it's clearly not a problem because it's a variable on which we are integrating on. After the integration that variable will disappear. Just change it's name, call it whatever, I personally like the $x'$ "notation" $$P(X>x) = \int_x^\infty f_X(x')dx'$$ but if you call it "Goofy", won't change much $$P(X>x) = \int_x^\infty f_X(\text{Goofy})d\text{Goofy}$$ Why the integral is the way of calculating an area is another story, too long to write down! But there're many good sources on the internet
A: $P\{X \leq x\}=\int_{-\infty} ^{x} f(t) \, dt$. Sinve $\{X>x\}$ is the complement of $\{X\leq x\}$ we get $P\{X>x\}=1-P\{X\leq x\}=1-\int_{-\infty} ^{x} f(t) \, dt=\int_x ^{\infty}  f(t) \, dt$ because $\int_{-\infty} ^{\infty}  f(t) \, dt=1$
A: The area is taken over the density function. If some points are more important for us they are denser in our analysis (the density function is higher in those points) so if you integrate the density function over some interval, you are exactly considering those points included by the region and throwing out the others. In this case $$P(x\in I)=\int_{x\in I}f_X(x)dx$$where $I$ is any arbitrary set such that the density function is measurable over it. For example here we have $$I=(x,\infty)$$for some $x\in\Bbb R$
