Chapter 4 - Integration - 4.4 Exercises - Page 290: 97

$$\left( {\text{a}} \right)d = 0{\text{ft}},{\text{ }}\left( {\text{b}} \right)T = \frac{{63}}{2}{\text{ft}}$$

$$\eqalign{ & {\text{Let the velocity }}v\left( t \right) = {t^3} - 10{t^2} + 27t - 18,{\text{ in feet per second, }} \cr & {\text{for 1}} \leqslant t \leqslant {\text{7}} \cr & \left( {\text{a}} \right){\text{The displacement in feet is given by:}} \cr & d = \int_a^b {f\left( t \right)} dt \cr & d = \int_1^7 {\left( {{t^3} - 10{t^2} + 27t - 18} \right)} dt \cr & d = \left[ {\frac{{{t^4}}}{4} - \frac{{10{t^3}}}{3} + \frac{{27{t^2}}}{2} - 18t} \right]_1^7 \cr & d = \left[ {\frac{{{{\left( 7 \right)}^4}}}{4} - \frac{{10{{\left( 7 \right)}^3}}}{3} + \frac{{27{{\left( 7 \right)}^2}}}{2} - 18\left( 7 \right)} \right] \cr & - \left[ {\frac{{{{\left( 1 \right)}^4}}}{4} - \frac{{10{{\left( 1 \right)}^3}}}{3} + \frac{{27{{\left( 1 \right)}^2}}}{2} - 18\left( 1 \right)} \right] \cr & {\text{simplifying}} \cr & d = - \frac{{91}}{{12}} + \frac{{91}}{{12}} \cr & d = 0{\text{ft}} \cr & \cr & \left( {\text{b}} \right){\text{ The total distance traveled is given by:}} \cr & T = \int_a^b {\left| {f\left( t \right)} \right|} dt \cr & T = \int_1^7 {\left| {{t^3} - 10{t^2} + 27t - 18} \right|} dt \cr & {t^3} - 10{t^2} + 27t - 18 = 0 \cr & \left( {t - 1} \right)\left( {t - 3} \right)\left( {t - 5} \right) = 0 \cr & {\text{Where:}} \cr & {t^3} - 10{t^2} + 27t - 18 < 0{\text{ for }}\left( { - \infty ,1} \right) \cup \left( {3,6} \right) \cr & {t^2} - t - 12 \geqslant 0{\text{ for }}\left[ {1,3} \right] \cup \left[ {6,\infty } \right) \cr & {\text{then}} \cr & T = \int_1^3 {\left( {{t^3} - 10{t^2} + 27t - 18} \right)} dt - \int_3^6 {\left( {{t^3} - 10{t^2} + 27t - 18} \right)} dt \cr & + \int_6^7 {\left( {{t^3} - 10{t^2} + 27t - 18} \right)} dt \cr & {\text{Integrating each one by a calculator :}} \cr & T = \frac{{16}}{3} + \frac{{63}}{4} + \frac{{125}}{{12}} \cr & T = \frac{{63}}{2}{\text{ft}} \cr} $$