Inverse problem
Let us consider the download example from the point of view of the router
which connects you to the Internet.
The router knows what download rate $f^\prime(t)$[MB/s] you had at all
times during the download because it was sending you the data.
The router does not have access to your computer and therefore does
not know the actual file size $f(t)$ on your computer.
Nevertheless, the router can infer the information about the size
of the file $f(t)$ from the download rate $f^\prime(t)$ that it has.
The sum (integral) of the download rate between $t=0$ and $t=\tau$
corresponds to the total new downloaded data that appeared on your computer.
During this period, the change in the file size $\Delta f$
is given by the integral:
\[
\Delta f = f(\tau)-f(0) = \int_0^\tau f'(t)\; dt.
\]
The $\int$ sign is a mnemonic for sum.
Assuming that the file size starts from zero $f(0)=0$[MB] at $t=0$,
the router can use the integration procedure to find $f(\tau)$,
the file size on your computer at $t=\tau$:
\[
f(\tau) = \int_0^\tau f^\prime(t)\; dt
= \int_0^\tau 0.004 t \;dt
= \left[ 0.002t^2 \right]_{t=0}^{t=\tau}
= 0.002\,\tau^2 \quad \textrm{ [MB]}.
\]
As you can see, the router was able to compute
the file size at time $t=\tau$ from the information about the download rate $f'(t)$.
Fundamental theorem of calculus
The above example illustrates how the integral operation acts as the inverse operation of the derivative.
Given $f'(t)$, you can compute $f(\tau)$ as follows:
\[
f(\tau)
= \int_0^\tau f'(t) \; dt \ +\ f(0).
\]
This is one of the fundamental ideas of calculus.
