How to find the total distance travelled in a kinematics integration question? 
I completed question a which is $\int _0^3\:t^2-6t+5 = -3$ or more specifically $3m$ to the left
Now isnt distance the absolute value of the displacement? $\left|\int _{t1}^{t2}\:v\left(t\right)\right|=\left|x1\:-x2\right|$
So for question b i did this:
$\left|\int _0^3\:t^2-6t+5\right| = 3$
and apparently this is incorrect. I am a bit confused what I did wrong, everything seems logical to me. So if you can try question b) that will be great and please point out what I am missing. Thnkas

Some definitions:

*

*$$\text{average speed} = \frac{\text{distance travelled}}{\text{time
   elapsed}}$$

*$$\text{speed at time }t=\left| v(t)\right|$$

*$$\text{average value of } f(t) \text{ over }
   [t_1, t_2]=\frac{\int_{t_1}^{t_2}f(t)\,\mathrm{d}t}{t_2-t_1}$$
Therefore
\begin{align}\text{distance travelled over }[t_1, t_2] &= \text{average speed}\times\text{time elapsed}\\
&=\frac{\int_{t_1}^{t_2}\left| v(t)\right|\,\mathrm{d}t}{t_2-t_1} \times \left(t_2-t_1\right)\\
&=\boxed{\int_{t_1}^{t_2}\left| v(t)\right|\,\mathrm{d}t}\\
&\neq \left| {\int_{t_1}^{t_2} v(t)\,\mathrm{d}t } \right|\\
&=\text{magnitude of displacement over }[t_1, t_2].\end{align}
The magnitude of displacement is the shortest distance between Positions $1$ and $2$, so can be smaller than the distance travelled from Position $1$ to $2$ (what an odometer measures).
