Answer
$\int_{a}^{b}I(t)~dt~~$ represents the total amount of charge that flows through the wire from time $a$ until time $b$.
Work Step by Step
We can state the Net Change Theorem as follows:
The integral of a rate of change is the net change:
$\int_{a}^{b}F'(x)~dx = F(b)- F(a)$
$~~I(t)~~$ is the current in the wire, which is the rate at which charge flows through the wire each second.
Therefore, $~~\int_{a}^{b}I(t)~dt~~$ represents the total amount of charge that flows through the wire from time $a$ until time $b$.
