Question

Can you check my work?
v(t)=t^3-2t^2+1 s(0)=1
Find the position function of the particle

Answer 1

\[\int\limits_{0}^{t}(t^3-2t^2+1)dt=s(t)-0\]

Answer 2

s(t)-1

Answer 3

\[\frac{ t^4 }{ 4 }-\frac{2t^3 }{ 3 }+t-1=s(t)-1\]

Answer 4

\[s(t)=\frac{ t^4 }{ 4 }-\frac{ 2t^3 }{ 3 }+t\]

Answer 5

that is right

Answer 6

but if i use indefinite integral i get s(t)=t^2/2-2t^3/3+t+1

Answer 7

v(t) is the velocity if you integrate that you would get the position ( you do not need limits from 0 to t)