It's been a while since I calculused. I have this exponential probability distribution function:
$ 5e^{-5x} $
I see that Wolfram does the definite/indefinite integral as
Question: why is it ($ 1 - e^{-5x} $) for the definite but ($ -e^{5x} $) for the indefinite integral? I know this is like Calculus 101 but I can't remember and not sure how to figure it out.
The indefinite integral $\int 5e^{-5x} dx$ evaluates to $-e^{-5x} + C$, where $C$ is an arbitrary constant of integration (WA writes this as a light grey "+ constant").
The definite integral $\int_a^b 5e^{-5x} dx$ is evaluated as $[-e^{-5x}]_a^b = -e^{-5b} - (-e^{-5b}) = e^{-5b} - e^{-5a}$. You don't normally leave the $+C$ in for definite integrals because it will cancel when you evaluate the antiderivative at the two endpoints. When $a = 0$ and $b = 2$ you get $\int_0^2 5e^{-5x} dx = e^0 - e^{-10} = 1 - e^{-10}$.