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I am afraid that both waveforms are wrong.
But waveform 2 is closer to the right one. Only its trace between 0.1s to 0.2s is right. Its first part (from 0 to 0.1s) should be as you did. That is i(t) ~ t^2.
Your work is right but first you need to integrate from t0=0 to t=t so that you can get i(t) as a function of t. Then you find the value at 0.1s at which you got 2A which will be used to calculate the initial value (also at t=0, hence not equal to 2A) in the function of the second part (see the following line).
For the second part (from 0.1s to 0.2s), i(t) ~ -t (with the initial value at t=0, hint: you will find it equal to 4A) and as you already got, i(0.2s) = 0A.
Try to repeat your work to get the full function of i(t) for the two parts... Good Luck
Consider that inductor current equals time integral of voltage. A linearly rising voltage thus must necessarily correspond to a quadratic increasing current.
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