The velocity of a particle moving along the x-axis is v(t) = t^2 + 2t + 1, with t measured in minutes and v(t) measured in feet per minute. To the nearest foot find the total distance traveled by the particle from t = 0 to t = 2 minutes.
4 answers
I took the definite integral of t^2+2t+1 from [0,2] and got 9. Is this correct? I'm not sure if I'm doing it correctly
not so.
∫[0,2] t^2+2t+1 dt
= t^3/3 + t^2 + t [0,2]
= (8/3 + 4 + 2)-0
= 26/3
∫[0,2] t^2+2t+1 dt
= t^3/3 + t^2 + t [0,2]
= (8/3 + 4 + 2)-0
= 26/3
Yes I got 26/3 as the exact answer but in terms of it being to the nearest foot it would be 9 right?
I'd say so, yes.