Remember that velocity is the derivative of displacement with respect to time. So to "undo" that derivative and get back displacement, we have to integrate over the time period.
There's one caveat though: the question asks for distance, not displacement and this particle actually turns around at t=1t=1 second (see if you can figure out how I know that). But here's the trick: integrate the absolute value of the function. That will cause the negative parts of the integral to become positive and we end up calculating the whole distance traveled as opposed to effective distance (i.e. displacement).
t=5 t=1 t=5
∫ |8t-8| dt= ∫ - (8t-8) dt + ∫ (8t-8) dt
t=0 t=0 t=1
t=1 t=5
=[8t-4t^2] + [4t^2-8t]
t=0 t=1
=4+64 = 65 meters