Answer:
118.5 m
Step-by-step explanation:
I'm not sure what you mean by distance, displacement distance or total distance, so I'm gonna do both.
Displacement =
[tex]\int\limits^4_1 {s(t)} \, dt \\\int\limits^4_1 {5t^2+t+2} \, dt \\\frac{5}{3}(4)^3+\frac{1}{2}(4)^2 + 2(4) - (\frac{5}{3}(1)^3+\frac{1}{2}(1)^2 + 2(1) )\\ = 118.5\\[/tex]
total distance is just absolute value when using integral
[tex]\int\limits^4_1 {| s(t) |} \, dt[/tex]
= 118.5
luckily it didn't matter for this problem, but there is a significant different between displacement and total distance, so remember to use absolute value signs when needed
