Chapter 4 - Integration - 4.4 Exercises - Page 290: 96

$$\left( {\text{a}} \right)d = - \frac{{56}}{3}{\text{ft}},{\text{ }}\left( {\text{b}} \right)T = \frac{{79}}{3}{\text{ft}}$$

$$\eqalign{ & {\text{Let the velocity }}v\left( t \right) = {t^2} - t - 12,{\text{ in feet per second, }} \cr & {\text{for 1}} \leqslant t \leqslant {\text{5}} \cr & \left( {\text{a}} \right){\text{The displacement in feet is given by:}} \cr & d = \int_a^b {f\left( t \right)} dt \cr & d = \int_1^5 {\left( {{t^2} - t - 12} \right)} dt \cr & d = \left[ {\frac{{{t^3}}}{3} - \frac{{{t^2}}}{2} - 12t} \right]_1^5 \cr & d = \left[ {\frac{{{{\left( 5 \right)}^3}}}{3} - \frac{{{{\left( 5 \right)}^2}}}{2} - 12\left( 5 \right)} \right] - \left[ {\frac{{{{\left( 1 \right)}^3}}}{3} - \frac{{{{\left( 1 \right)}^2}}}{2} - 12\left( 1 \right)} \right] \cr & {\text{simplifying}} \cr & d = - \frac{{185}}{6} + \frac{{73}}{6} \cr & d = - \frac{{56}}{3}{\text{ft}} \cr & \cr & \left( {\text{b}} \right){\text{ The total distance traveled is given by:}} \cr & T = \int_a^b {\left| {f\left( t \right)} \right|} dt \cr & T = \int_1^5 {\left| {{t^2} - t - 12} \right|} dt \cr & {t^2} - t - 12 = 0 \cr & \left( {t - 4} \right)\left( {t + 3} \right) = 0 \cr & {\text{Where:}} \cr & {t^2} - t - 12 < 0{\text{ for }}\left( { - 3,4} \right) \cr & {t^2} - t - 12 \geqslant 0{\text{ for }}\left( { - \infty , - 3} \right] \cup \left[ {4,\infty } \right) \cr & {\text{then}} \cr & T = \int_1^5 {\left| {{t^2} - t - 12} \right|} dt \cr & T = \int_1^4 {\left[ { - \left( {{t^2} - t - 12} \right)} \right]} dt + \int_4^5 {\left( {{t^2} - t - 12} \right)} dt \cr & T = \int_1^4 {\left( {12 + t - {t^2}} \right)} dt + \int_4^5 {\left( {{t^2} - t - 12} \right)} dt \cr & T = \left[ {12t + \frac{1}{2}{t^2} - \frac{1}{3}{t^3}} \right]_1^4 + \left[ {\frac{1}{3}{t^3} - \frac{1}{2}{t^2} - 12t} \right]_4^5 \cr & {\text{Evaluating}} \cr & T = \frac{{45}}{2} + \frac{{23}}{6} \cr & T = \frac{{79}}{3}{\text{ft}} \cr} $$