Verified answer

The total distance traveled in the first 8 seconds requires you to note that the particle changes direction when v(t) = 0, which is at 6 seconds. So you have to separately compute the positive displacement from t = 0 to 6 and add it to the positive displacement from t = 6 to 8.

∫₀⁶ (12-2t) dt = 12t-t² ]₀⁶ = 36

+ [12t-t²]₆⁸ = (96-64)-(72-36) = 32-36 = -4, but as I said it needs to be positive, so 4.

Therefore, total distance traveled from t = 0 to 8 seconds = 36+4=40 feet.

First off, since the problem says "include appropriate units" and fails to do so, itself, when writing "time interval [0,8]" -- you should beat up the teacher. It should say [0,8s], for example. I don't know if this is 8 seconds, 8 minutes, 8 hours, etc., otherwise. Just a note. Of course I'll assume 8 seconds.

        ∫₀⁸ (12-2t) dt

        ∫₀⁸ 12 dt - ∫₀⁸ 2t dt

        12 ∫₀⁸ dt - 2 ∫₀⁸ t dt

        12 t|₀⁸ - t²|₀⁸

        12 (8-0) - (8²-0²)

        96 - 64 = 32

        32 feet

So I agree with you. But I also have no idea what "total distance traveled" means, just like you. You aren't told how much distance was traveled outside of the [0s,8s] period. All you can say is what the displacement was during that interval.

The question doesn't make sense to me, either. I think you've done all you can.