are integration limit to find expected value from a pdf inclusive (or exclusive)?

Question

This is my probability density function (pdf) $$pdf = e^{-\frac{r}{\lambda}} \frac{1}{\lambda}$$

I want to find the expected value (EV) from $0<r \leq r_0$, $r$ is the random variable.

My first attempt is $EV_1 - EV_2$ (see below calculation), but this does not work.

The solution from the paper is $EV_3 + EV_4 = \lambda - \lambda e^{-\frac{r_0}{\lambda}}$

my questions:

1. Are integration limits used to find expected value from a pdf inclusive (or exclusive)? it seems like the lower limit is inclusive but the upper limit is exclusive. Any reference that I can look about this?
2. What is the interpretation of $EV_4$?

Here are some values that I calculated. $$EV_1 = \int_0^\infty r \left(e^{-\frac{r}{\lambda}} \frac{1}{\lambda} \right) dr = \lambda $$

$$EV_2 = \int_{r_0}^\infty r \left(e^{-\frac{r}{\lambda}} \frac{1}{\lambda} \right) dr = r_0 e^{-\frac{r_0}{\lambda}} + \lambda e^{-\frac{r_0}{\lambda}} $$

$$EV_3 = \int_{0}^{r_0} r \left(e^{-\frac{r}{\lambda}} \frac{1}{\lambda} \right) dr = \lambda - r_0 e^{-\frac{r_0}{\lambda}} - \lambda e^{-\frac{r_0}{\lambda}} $$

$$EV_4 = \int_{r_0}^\infty r_0 \left(e^{-\frac{r}{\lambda}} \frac{1}{\lambda} \right) dr = r_0 e^{-\frac{r_0}{\lambda}} $$

This question is actually a physical formulation that I am reading. SO, if you need more explanation about the problem, let me know.

EDIT: These are the 3 references I used. They are discussing the same thing in the screenshots I attached.

1. https://doi.org/10.1080/00018735200101151

2. https://doi.org/10.1098/rspa.1950.0077

3. https://doi.org/10.1098/rspa.1950.0107

Answer 1

From the references you give, it is now clear that you want the mean of a censored distribution rather than a truncated distribution. Specifically, if you have a random variable $r$ with pdf $\frac{\exp \left(-\frac{r}{\lambda }\right)}{\lambda }$, then you want the mean of the censored variable $s$ which is $s=r$ if $0<s\leq r_0$ and $s=r_0$ if $r>r_0$.

$$E(s)=\int_0^{r_0} \frac{r \exp \left(-\frac{r}{\lambda }\right)}{\lambda } \, dr + \int_{r_0}^{\infty } \frac{r_0 \exp \left(-\frac{r}{\lambda }\right)}{\lambda } \, dr$$

$$=\left(\lambda -e^{-\frac{r_0}{\lambda }} (\lambda +r_0)\right)+\left(r_0 e^{-\frac{r_0}{\lambda }}\right)=\lambda -\lambda e^{-\frac{r_0}{\lambda }}$$