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## Homework Statement

Find q(t) for the following function.

http://imgur.com/gTtYRd0

We are also given that q(0) = -6 nC.

## Homework Equations

[tex]

q(t) = \int \mathrm{i(t)}\, \mathrm{d}t [/tex]

## The Attempt at a Solution

This is more of a question about why the steps to the solution are correct, as opposed to how to get to the solution, so I apologize if I posted this in the wrong section.

I would have thought that to find q(t) from x = 0 to x = 5 you would just find i(t) on that interval and integrate it. My teacher, however, showed us the following as the solution.

[tex]

q(t) = \int_{-\infty}^0 \mathrm{i(t)}\, \mathrm{d}t + \int_0^t \mathrm{i(t)}\, \mathrm{d}t

\\

\int_{-\infty}^0 \mathrm{i(t)}\, \mathrm{d}t = q(0)

\\

q(t) = q(0) + \int_0^t \mathrm{i(t)}\, \mathrm{d}t

[/tex]

I do not understand why he is including the integral with the lower limit of negative infinity. He does the same for the interval x = 5 to x = 14.

[tex]

q(t) = \int_{-\infty}^{5e-6} \mathrm{i(t)}\, \mathrm{d}t + \int_{5e-6}^t \mathrm{i(t)}\, \mathrm{d}t

\\

\int_{-\infty}^{5e-6} \mathrm{i(t)}\, \mathrm{d}t = q(5e-6)

\\

q(t) = q(5e-6) + \int_{5e-6}^t \mathrm{i(t)}\, \mathrm{d}t

[/tex]

My guess would be that he is treating i(t) = 0, when t < 0 (although he never explicitly said this), and then doing this.

[tex]

\int_{-\infty}^0 \mathrm{i(t)}\, \mathrm{d}t = q(0) - q({-\infty})

[/tex]

And since i(t) is zero at negative infinity, the integral would just equal q(0). I just can't seem to understand why this is necessary, and why I could not just ignore q(0) and q(5e-6) while integrating. Any help would be much appreciated.

EDIT: I think I figured out what I was doing wrong! Since q(t) gives you the total charge at a given time, you need to add the total charge from to the previous interval to the equation of the next, otherwise it is assuming the charge starts from zero? Maybe I'm just grasping at straws...

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