$\int_0^\infty dx\;\beta e^{-\beta x}=1$? $$\displaystyle\lim_{p\to \infty} \int_0^pdx\;e^{-x}=- e^{-x}|_0^p=0+e^0=1$$
My textbook states that $\int_0^\infty dx\;\beta e^{-\beta x}=1$, but I get that 
$$\displaystyle\lim_{p\to \infty} \int_0^pdx\;\beta e^{-\beta x}=-\beta e^{-\beta x}|_0^p=0+\beta e^0=\beta$$
What have I done wrong (or what have I not considered)?
 A: We have
$$
\left(\color{blue}{-e^{-\beta x}}\right)'=\beta e^{-\beta x}
$$ giving
$$
\lim_{p\to \infty} \int_0^pdx\;\beta e^{-\beta x}=\lim_{p\to \infty}\left(\color{blue}{-e^{-\beta x}}\frac{}{}|_0^p\right)=0+ e^{-0}=1
$$
A: $$\int_0^pβe^{-βx}dx=\int_0^p[-e^{-βx}]'dx =\big[ e^{-βx}\big]_0^p=-e^{-βp}+1$$
Thus : 
$$\lim_{p\to \infty} \int_0^pdx\;\beta e^{-\beta x}= \lim_{p \to \infty} (-e^{-βp}+1)=1$$
A: Take another look at
$\displaystyle\lim_{p\to \infty} \int_0^pdx\;\beta e^{-\beta x}; \tag 1$
Note that
$\displaystyle \lim_{p\to \infty} \int_0^pdx\;\beta e^{-\beta x} = \lim_{p\to \infty} \beta \int_0^pdx\; e^{-\beta x} = \beta \lim_{p\to \infty} \int_0^pdx\; e^{-\beta x}, \tag 2$
where I have pulled $\beta$ out of the integral, and then out of the limit, since it is a constant. 
Now
$\displaystyle \int_0^pdx\; e^{-\beta x} = -\dfrac{1}{\beta}\left (e^{-\beta x} \right \vert_0^p, \tag 3$
since the antiderivative of $e^{-\beta x}$ is $(-1 / \beta)e^{-\beta x}$:
$\left (-\dfrac{1}{\beta} e^{-\beta x} \right)' = \left (-\dfrac{1}{\beta} \right ) (-\beta) e^{-\beta x} = e^{-\beta x}; \tag 4$
so from (3):
$\displaystyle \int_0^pdx\; e^{-\beta x} = -\dfrac{1}{\beta}\left (e^{-\beta x} \right \vert_0^p = -\dfrac{1}{\beta}(e^{-\beta p} - e^{-\beta (0)}) = -\dfrac{1}{\beta}(e^{-\beta p} - 1), \tag 5$
thus if $\beta > 0$, 
$\displaystyle \lim_{p\to \infty} \int_0^pdx\; e^{-\beta x} = \lim_{p\to \infty}-\dfrac{1}{\beta}(e^{-\beta p} - 1) = \dfrac{1}{\beta}; \tag 6$
using this result in (2) yields
$\displaystyle \lim_{p\to \infty} \int_0^pdx\;\beta e^{-\beta x} = 1. \tag 7$
The error in the text of the question was neglecting the factor of $-1 / \beta$ in the primitive of $e^{-\beta x}$.  Also, remember that we need $\beta > 0$ for the limits to make sense, so that (1) is defined.
A: $$\int_{0}^{p}\beta e^{-\beta x}dx=-e^{-\beta x}|_{0}^{p}=e^{-\beta x}|_{0}^{p}=1-e^{-\beta p}$$
