Consider this example.

Suppose the voltage across a 2 Henry inductor is as shown above.

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

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 had zero area up until 2,

it means that your current wave form 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 peak 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 1, but

remember you're dividing by L, which is 2,

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

And then there's no more area to be accumulated after 3 seconds, and so

the current will remain constant from 3 on.

A consequence of the integral IV 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 2 and 3 and everywhere else.

Now, we will discuss the DC Steady State for inductors.