Consider this example.

Suppose the voltage across a two Henry inductor is as shown above, and

it will be given that the current at t = 0 is 0.5 Amps.

The current function of time is shown below.

It starts at 0.5 Amps at t = 0, because that was the initial condition given,

and because the voltage has zero area up until two,

it means that your current waveform is going to be constant until two seconds.

Then you're going to be integrating that square, giving you a ramp in current.

And the pick of the ramp will have to rise up by 0.5.

Why is it 0.5?

Because the area of the box, of the voltage, is one.

But remember, you're dividing by L, which is two, so

your current is only going to increment by one-half.

And then there's no more area to be accumulated after three seconds, and so

the current will remain constant from three on.

A consequence of the integral I V characteristic

is that the current is continuous.

In spite of the voltage having step changes, as you see in the top graph,

the current is continuous at times two and three and every where else.

Now we will discuss the DC steady state for inductors.