
well since they want displacement (vector) not distance (scalar) watch out for sign reversals. Like when is v = 0 ???
t^2  3t +2 = 0
(t 1)(t2) = 0
so do three intervals
(a) t = 0 to t = 1
(b) t = 1 to t = 2
(c) t = 2 to t = 5

integral from p to q of (t^2+3t2 ) dt =t^3/3 + 3t^2/2  2 t at q  at p
= (q^3+p^3) + (3/2)(q^2p^2) 2(qp)
(a) from p=0 to q=1
= 1 + 3/2  2 = 3 + 3/2 = 3/2
(b) etc etc etc

I am still getting it wrong

since they want displacement, ignore sign changes.
∫[0,5] t^2+3t2 dt = 86/5

86/5 isn't right

in that case there is a mistake in the problem or in the answer key. All it wants is just a straight integral. You can see that the particle moves
left for 1 second
right for 1 second
left for 3 more seconds. It will wind up to the left of where it started.

displacement has to include sign changes
if you start at home, walk north a mile, then south a mile
your distance is TWO miles
your displacement is ZERO miles

its asking for total distance traveled so the answer wont be negative

Yes, there must be sign changes

It asked for the NET distance, not the TOTAL distance
Yes total is to the left

Huh? It said
Find the displacement (net distance covered)
not distance traveled.
But, in that case, you want
∫[0,1] (t^2+3t2) dt + ∫[1,2] (t^2+3t2) dt  ∫[2,5] (t^2+3t2) dt
= 5/6 + 1/6 + 27/2
= 29/2

answer key said incorrect to 29/2

or that could be just ∫[0,5] t^2+3t2 dt = 29/2
but to do the actual evaluation, you need to split it up as above.

well, something is wrong. We've done it both ways.
Maybe the answer key wants 14.5 instead of 29/2
You're on your own now, buddy.

could it be you made a mistake?

I think the question is poorly worded and it is hard to say. Obviously the displacement is negative (you end up to the left), but the net distance traveled might mean the absolute value of the displacement. Whoever wrote the question does not do physics.

I think it is asking for the absolute value

I am not so sure about that, displacement, unlike distance, includes direction.