Single Variable Calculus, Chapter 5, Review Exercises, Section Review Exercises, Problem 44

Suppose that a particle moves along a line with velocity function $v(t) = t^2 - t$, where $v$ is measured in meters per second.
a.) Find the displacement of the particle during the time period $0 \leq t \leq 5$.

$
\begin{equation}
\begin{aligned}
\int^{t_2}_{t_1} v(t) dt &= s(t_2) - s(t_1)\\
\\
\int^{t_2}_{t_1} v(t) dt &= \int^5_0 \left(t^2 - t \right) dt\\
\\
\int^5_0 \left(t^2 - t \right) dt &= \int^5_0 t^2 dt - \int^5_0 t dt\\
\\
\int^5_0 \left(t^2 - t \right) dt &= \left[ \frac{t^{2+1}}{2+1} - \frac{t^{1+1}}{1+1} \right]^5_0\\
\\
\int^5_0 \left(t^2 - t \right) dt &= \left[ \frac{t^3}{3} - \frac{t^2}{2} \right]^5_0\\
\\
\int^5_0 \left(t^2 - t \right) dt &= \frac{(5)^3}{3} - \frac{(5)^2}{2} - \frac{(0)^3}{3} + \frac{(0)^2}{2}\\
\\
\int^5_0 \left(t^2 - t \right) dt &= \frac{125}{3} - \frac{25}{2}\\
\\
\int^5_0 \left(t^2 - t \right) dt &= \frac{175}{6} m\\
\\
\int^5_0 \left(t^2 - t \right) dt &= 29.17m
\end{aligned}
\end{equation}
$

b.) Find the distance traveled during this time period
Note that $v(t) = t^2 - t = t(t - 1)$, and $t(t-1) = 0$ then $t = 0$ and $t = 1$

$
\begin{equation}
\begin{aligned}
\int^5_0 |v(t)| dt = \int^5_0 \left| t^2 - t \right| dt &= \int^1_0 - \left( t^2 - t \right) dt + \int^5_1 \left( t^2 - t \right) dt\\
\\
\int^5_0 |v(t)| dt = \int^5_0 \left| t^2 - t \right| dt &= \int^1_0 \left( -t^2 + t \right) dt + \int^5_1 \left( t^2 - t \right) dt\\
\\
\int^5_0 |v(t)| dt = \int^5_0 \left| t^2 - t \right| dt &= \left[ \frac{-t^{2+1}}{2+1} + \frac{t^{1+1}}{1+1} \right]^1_0 + \left[ \frac{t^{2+1}}{2+1} - \frac{t^{1+1}}{1+1} \right]^5_1\\
\\
\int^5_0 |v(t)| dt = \int^5_0 \left| t^2 - t \right| dt &= \left[ \frac{-t^3}{3} + \frac{t^2}{2} \right]^1_0 + \left[ \frac{t^3}{3} - \frac{t^2}{2} \right]^5_1\\
\\
\int^5_0 |v(t)| dt = \int^5_0 \left| t^2 - t \right| dt &= \left[ \frac{-(1)^3}{3} + \frac{(1)^2}{2} + \frac{(0)^3}{3} - \frac{(0)^2}{2} \right] + \left[ \frac{(5)^3}{3} - \frac{(5)^2}{2} - \frac{(1)^3}{3} + \frac{(1)^2}{2} \right]\\
\\
\int^5_0 |v(t)| dt = \int^5_0 \left| t^2 - t \right| dt &= -\frac{1}{3} + \frac{1}{2} + \frac{125}{3} - \frac{25}{2} - \frac{1}{3} + \frac{1}{2}\\
\\
\int^5_0 |v(t)| dt = \int^5_0 \left| t^2 - t \right| dt &= \frac{59}{2}m\\
\\
\int^5_0 |v(t)| dt = \int^5_0 \left| t^2 - t \right| dt &= 29.5m
\end{aligned}
\end{equation}
$