Answer
(a) $\frac{175}{6}$
(b) $\frac{59}{2}$
Work Step by Step
(a) We can find the displacement:
$\int_{0}^{5}(t^2-t)~dt$
$= (\frac{t^3}{3}-\frac{t^2}{2})~\vert_{0}^{5}$
$= [\frac{(5)^3}{3}-\frac{(5)^2}{2}]-[\frac{(0)^3}{3}-\frac{(0)^2}{2}]$
$= (\frac{125}{3}-\frac{25}{2})-(0)$
$= \frac{250}{6}-\frac{75}{6}$
$= \frac{175}{6}$
(b) $\int_{0}^{5}\vert~v(t)~\vert~dt~~$ is the distance traveled by the particle.
On the interval $0 \lt t \lt 1$, the function $t^2-t$ is negative.
On the interval $1 \lt t \lt 5$, the function $t^2-t$ is positive.
We can find the distance traveled by the particle:
$\int_{0}^{5}\vert t^2-t\vert~dt$
$= \int_{0}^{1} (t-t^2)~dt+\int_{1}^{5} (t^2-t)~dt$
$= (\frac{t^2}{2}-\frac{t^3}{3})~\vert_{0}^{1}+(\frac{t^3}{3}-\frac{t^2}{2})~\vert_{1}^{5}$
$= [\frac{(1)^2}{2}-\frac{(1)^3}{3}]-[\frac{(0)^2}{2}-\frac{(0)^3}{3}]+[\frac{(5)^3}{3}-\frac{(5)^2}{2}]-[\frac{(1)^3}{3}-\frac{(1)^2}{2}]$
$= (\frac{1}{2}-\frac{1}{3})-(0)+ (\frac{125}{3}-\frac{25}{2})- (\frac{1}{3}-\frac{1}{2})$
$= \frac{1}{6}+\frac{250}{6}-\frac{75}{6}+\frac{1}{6}$
$= \frac{177}{6}$
$= \frac{59}{2}$