Let me preface by saying that I am completely ignorant of the math covered in Calculus 2, which for me includes integration. All I know is that the integral is the area under a curve, which I understand as the sum total of all data points up to a given value of $x$, and that the derivative of an integral is just the original function.
I'm working on a problem where I'm given $f '(t)$, and I know that $f(t)$ is the number of errors made by a shift worker from $t = 0$ to $t$. To me, $f(t)$ sounds like an integral, since it describes the total errors made by the worker till $t$. My question then is, would $f'(t)$ be some function (let's say $g$), whose integral is $f(t)$?
The difference in my analysis is that $g(t)$ would be the number of errors made at the instant $t$, while a typical derivative is be the rate that errors are occurring at $t$.
First of all I'll assume that this is a continous probability case and not discrete one, because if it's a discrete probability the integral interpretation doesn't apply. I'll also define cumulative function as $\mathbf F(t)$ and it's derivative as $\mathbf F'(t) = f(t)$.
Please note that nothing below applies if you are given an event probability table or a cumulative function that looks like this: $$\mathbf F(t) = \begin{cases} 0.2, \ 0\leq t \leq 2\\ 0.6, \ 2\leq t \leq 4\\ 1, \ 4\leq t \end{cases} $$ This is a discrete case and no integral can be applied to it.
By definition, a cumulative function $\mathbf F(t)$ returns a probability that something is less than or equal to $t$. In your case I suppose it would return a probability that there was no more than $t$ errors made by a worker. Taking a derivative of that function, in some sense, would give you a rate of change, but more importantly the derivative of cumulative function $\mathbf F(t)$ is probability density function $f(t)$. That is, by taking a derivative you still get a function.
You can find online images of cumulative and probability density functions plotted together. They tell slightly different story. While $\mathbf F(t)$ tells the probability of something happening no more than $t$, the probability density allows you to take a look at the probability that some value of $t$ lies in some region $[a,b]$. That probability is in fact defined as the integral $\int_a^b f(t) \ dt$.
You can ask "What is the probability that a worker makes at least 3 and no more than 5 mistakes?". You can calculate the integral $\int_3^5 f(t) \ dt$ to find out.
Alternatively, you can make use of that definition and only use the cumulative $\mathbf F(t)$. Since $\mathbf F(t)$ tells the probability that your worker performas no more than $t$ mistakes, asking $\mathbf F(5) - \mathbf F(3)$ would produce the same result as integrating. To make it more obvious $\mathbf F(5) - \mathbf F(3) = \int_0^5 f(t) \ dt \ - \int_0^3 f(t) \ dt = \int_3^5 f(t) \ dt$, thanks to the properties of an integral.
Using the language of math, here you're considering the probability that your worker performs anywhere from 0 to 5 mistakes (which of course includes making anywhere from 0 to 3 mistakes) and you subtract the probability that he makes 0 to 3 mistakes, leaving only the probability to make 3, 4 or 5 mistakes.