Notice that it's divided by c and therefore it's going,

the voltage and the capacitor's going to be a continuous function of time.

Let's consider this example, that c is two Fahrens the initial

voltage is t equals zero is minus two vaults so,

that means that for us t nought is going to be zero.

And then, our current is given a shown.

Now, the voltage that goes as with this, is shown here.

We start off at minus two because that's the given initial condition at the origin.

And then we see, because the current is constant from zero to two,

our voltage has this ramp characteristic.

So the only challenge for

us is to determine what is the maximum value of the ramp.

We can figure that out because at t equals two the integral will have captured all

of this area which is two times four or eight.

But don't forget your going to be dividing by c which is two.

So the voltage over this interval of time has to increment by four,

eight divided by two.

And that's what it does.

It goes from minus two to plus two.

And then, since the current is zero after that point,

the integral doesn't capture anymore area, and the voltage stays constant.

So we notice that this is the voltage is continuous,

even though, there is a discontinuity in the current, but

because it's a capacitor, the voltage is continuous at that point.