Question=autosave\#questio... ns that the particle moved approximately 30.00 meters to the left. that $v(t)=t^{2}-t-20=(t-5)(t+4)$ and so $v(t) \leq v \quad 0$ on the interval $[1,5]$ and $v(t)$ nus, from this equation, the distance traveled is $$\begin{aligned} \int_{1}^{7}|v(t)| d t & =\int_{1}^{5}[-v(t)] d t+\int_{5}^{7} v(t) d t \\ & =\int_{1}^{5}\left(-t^{2}+t+20\right) d t+\int_{5}^{7}\left(t^{2}-t-20\right) d t \end{aligned}$$ $$=\left[-\frac{t^{3}}{3}+\frac{t^{2}}{2}+20 t\right]_{1}^{5}+\left[\frac{t^{3}}{3}-\frac{t^{2}}{2}-20 t\right]$$ $$=\frac{725}{3}$$ Your answer is correct.