The waveform you have drawn as "waveform 2" does not look correct because it is drawn to look like a triangle.
Since during the first interval the voltage is a ramp, that means the current for that interval will be a parabola involving t^2, so it wont be a ramp with a straight line but the line will be curved gently upwards until the end of the interval.
For the second interval the voltage is a negative pulse so for that section the current will be a negative slope ramp.
Is that enough help or do you want to see the exact solution?
Seriously, I don't get it. Did I do the integration the correct way? Was there something wrong with the formula used? I'm not well versed in calculus but I'm learning it.
The guy whose work I have with me has told me that it was this way the instructor solved it. My answer, 2 mA, for 0< t <0.1 is the same as of that someone else. The guy has the expression i(t)=4mA - 20mA(t) for the interval 0.1< t <0.2, which to me seems wrong.
I don't know how else I can integrate the v(t)=-2/1000 over the interval 0.1< t <0.2. Could you please help me with it?
For the first interval, we have a voltage ramp that goes from 0 to 4v in time 0 to 0.1 seconds.
The slope therefore is 4/0.1=40 as you noted.
This makes the equation for voltage:
V(t)=40*t
Integrating that with respect to t we get;
40*t^2/2+K=20*t^2+K
with K=0 we have:
20*t^2.
This is a parabola with center at the origin.
v=L*di/dt
v dt=L di
1/L*v dt=di
1/L*40*t dt=di
1/L*20*t^2=i
with L=100H and t=0.1 sec, we get:
i=2ma
So you got the right answer i think, but your writing is hard to read, sorry. If you type out your results i can see it better.
You want to do the second interval or you want to see that too?
Intuitively when you see a ramp voltage you expect a parabola current, when you see a pulse voltage you expect a ramp current (inductor).