Question related to Integration and Probability Density Functions My question is regarding integration questions related to the probabilities of continuous random variables.
If X = 0  to 5 is represented by f1(x) and X=5 to 10 is represented by f2(x) and we want P(0<=X<=10). Would the answer be integral over 0 to 5 for f1(x)  + integral of 5 to 10 for f2(x)? That is I am confused when we are to calculate probabilities that are represented by 2 sets of functions. Also are the endpoints exclusive or inclusive to probability calculations? If I calculate the integral of 0-10 will 0 and 10 be included as probabilities? I am guessing though that for continuous RV's P(0<=X<=10) is the same as any combination of <=,>=,<,> as after all they cannot take discrete values.
Any help would be appreciated.
 A: Yes to your first question:  You have it right about integrating each function in the range in which it applies.  For example:
$$
f(x) =\left\{ \begin{array}{cc} \frac{x^2}{144} & 0\leq x<6 \\
\frac{x}{64} & 6 \leq x \leq 10 \end{array} \right. 
$$
Then 
$$P(2<x<7) = \int_2^6  \frac{x^2}{144} dx + \int_6^7 \frac{x}{64} dx
$$
For the question about discrete values:  If the functions $f(x)$ involve no $\delta$-functions, that is, if the distribution has no discrete properties, then inclusion or exclusion of the endpoints does not matter.
A: If $X$ admits a density $f$ which a piecewise continuous function over $[0,10]$ such that 
$$
f(x) =
\begin{cases}
f_1(x),  & \text{if $0\leq x \leq 5$,} \\[2ex]
f_2(x), & \text{if $5\leq x \leq 10,$}
\end{cases}
$$ then
$$
P(0\leq X\leq10)=\int_0^5f_1(x) \:dx+\int_5^{10}f_2(x) \:dx.
$$ Since $f_1$ and $f_2$ are continuous, you get 
$$
P(0< X\leq10)=P(0< X<10)=P(0\leq X\leq10)=\cdots,
$$ resulting from
$$
\int_a^af_1(x) \:dx=0=\int_b^bf_2(x) \:dx, \quad a,b \in [0,10].
$$
A: We have $p \{x\} = \int_{\{x\}} f(t) dt = 0$, so you have
$p [a,b] = p[a,b) = \dots$, that is, the addition of removal of single (in fact countable, if you want) point 
will not change the value of the probability.
In your case, you can define $f(x) = \begin{cases} f_1(x), & x \in [0,5) \\
f_2(x), & x \in [5,10] \end{cases}$ and so the probability will be
$p [0,10] = \int_{[0,10]} f(t) dt$.
