Finding the displacement and the distance traveled by a particle from the velocity function

To find the displacement we simply integrate the velocity function
\[\text{Displacement}=\int_{-1}^{5}t^3-3t^2dt=\frac{t^4}{4}-t^3 |_{-1}^{5}\]
\[=(\frac{5^4}{4}-5^3)-(\frac{1}{4}+1)=\frac{5^4-4\cdot 5^3-5}{4}=\frac{120}{4}=30.\]

To find the distance traveled we should integrate absolute value of the velocity function: 
\[\text{Distance traveled}=\int_{-1}^{5}|t^3-3t^2|dt.\]
We have
\[t^3-3t^2=t^2(t-3).\]
Hence $|t^3-3t^2|=t^2 (t-3)=t^3-3t^2  \text{for}  t>3$, and $|t^3-3t^2|=-t^2 (t-3)=-t^3+3t^2  \text{for}  t<3.$ Thus 
\[\text{Distance traveled}=\int_{-1}^{5}|t^3-3t^2|dt=\int_{-1}^{3}-t^3+3t^2dt+\int_3^5t^3-3t^2 dt\]
\[=(-\frac{t^4}{4}+t^3)\big|_{-1}^{3}+(\frac{t^4}{4}-t^3)\big|_3^5\]
\[=(-\frac{3^4}{4}+3^3)-(-\frac{1}{4}-1)+(\frac{5^4}{4}-5^3)-(\frac{3^4}{4}-3^3)\]
\[=2(-\frac{3^4}{4}+3^3)+(\frac{1}{4}+1)+(\frac{5^4}{4}-5^3)\]
\[=2(-\frac{81}{4}+27)+(\frac{1}{4}+1)+(\frac{625}{4}-125)=46.\]