Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 56

A particle moves along a line so that its velocity at time $t$ is $v(t) = t^2 - 2t - 8$ (measured in meters per second).
a.) Find the displacement of the particle during the time period $1 \leq t \leq 6$ from the formula.

$
\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^6_1 \left( t^2 - 2t - 8 \right) dt\\
\\
\int \left( t^2 - 2t - 8 \right) dt &= \int t^2 dt - 2 \int t d t - \int 8 dt\\
\\
\int \left( t^2 - 2t - 8 \right) dt &= \frac{t^{2+1}}{2+1} - 2 \left( \frac{t^{1+1}}{1+1} \right) - 8 \left( \frac{t^{0+1}}{0+1} \right)\\
\\
\int \left( t^2 - 2t - 8 \right) dt &= \frac{t^3}{3} - \frac{\cancel{2}t^2}{\cancel{2}} - 8t\\
\\
\int \left( t^2 - 2t - 8 \right) dt &= \frac{t^3}{3} - t^2 - 8t\\
\\
\int^6_1 \left( t^2 - 2t - 8 \right) dt &= \frac{(6)^3}{3} - (6)^2 - 8 (6) - \left[ \frac{(1)^3}{3} - (1)^2 - 8 (1) \right]\\
\\
\int^6_1 \left( t^2 - 2t - 8 \right) dt &= \frac{216}{3}-36-48-\frac{1}{3}+1+8\\
\\
\int^6_1 \left( t^2 - 2t - 8 \right) dt &= 72-36-48-\frac{1}{3}+9\\
\\
\int^6_1 \left( t^2 - 2t - 8 \right) dt &= \frac{-10}{3} \text{ meters or } -3.33 \text{ meters}
\end{aligned}
\end{equation}
$


This means the particle moved $3.33$m toward the left.
b.) Find the distance traveled during this time period
Note that $v(t) = t^2 - 2t - 8 = (t- 4)(t+2)$ and then $(t-4)(t+2) = 0$
$t = 4$ and $ t = -2$
Only $t = 4$ is in the interval $[1,6]$, thus, the distance traveled is...

$
\begin{equation}
\begin{aligned}
\int^6_1 |v(t) | dt &= \int^4_1 - v(t) dt + \int^6_4 v(t) dt\\
\\
\int^6_1 |v(t) | dt &= \int^4_1 \left( -t^2 + 2t + 8 \right) dt + \int^6_4 \left( t^2-2t - 8 \right) dt\\
\\
\int^6_1 |v(t) | dt &= \left[ -\frac{t^3}{3} + t^2 + 8t \right]^4_1 + \left[ \frac{t^3}{3} - t^2 - 8t \right]^6_4\\
\\
\int^6_1 |v(t) | dt &= \frac{-(4)^3}{3} + (4)^2 + 8 (4) - \left[ \frac{-(1)^3}{3} + (1)^2 + 8(1) \right] + \left[ \frac{(6)^3}{3} - (6) - 8(6) - \left[ \frac{(4)^3}{3} - (4)^2 - 8(4) \right] \right]\\
\\
\int^6_1 |v(t) | dt &= \frac{-64}{3} + 16 +32 + \frac{1}{3} - 1 - 8 + 72 - 36 - 48 - \frac{64}{3} + 16 + 32\\
\\
\int^6_1 |v(t) | dt &= \frac{98}{3} \text{ meters } \qquad \text{ or } \qquad \int^6_1 |v(t) | dt = 32.67 \text{ meters}
\end{aligned}
\end{equation}
$