Chapter 4 - Integration - 4.4 Exercises - Page 291: 102

$$27{\text{ units}}$$

$$\eqalign{ & x\left( t \right) = \left( {t - 1} \right){\left( {t - 3} \right)^2},{\text{ 0}} \leqslant t \leqslant {\text{5}} \cr & {\text{Differentiating}} \cr & x'\left( t \right) = \frac{d}{{dt}}\left[ {\left( {t - 1} \right){{\left( {t - 3} \right)}^2}} \right] \cr & x'\left( t \right) = 2\left( {t - 1} \right)\left( {t - 3} \right) + {\left( {t - 3} \right)^2} \cr & {\text{Let }}x\left( t \right) = 0 \cr & 2\left( {t - 1} \right)\left( {t - 3} \right) + {\left( {t - 3} \right)^2} = 0 \cr & \left( {t - 3} \right)\left[ {2\left( {t - 1} \right) + \left( {t - 3} \right)} \right] = 0 \cr & \left( {t - 3} \right)\left( {2t - 2 + t - 3} \right) = 0 \cr & \left( {t - 3} \right)\left( {3t - 5} \right) = 0 \cr & \left( {t - 3} \right)\left( {3t - 5} \right) < 0{\text{ for }}\left( {\frac{5}{3},3} \right) \cr & \left( {t - 3} \right)\left( {3t - 5} \right) > 0{\text{ for }}\left( { - \infty ,\frac{5}{3}} \right] \cup \left[ {3,\infty } \right) \cr & {\text{The total distance traveled is given by:}} \cr & {T_d} = \int_0^5 {\left| {x'\left( t \right)} \right|} dt \cr & {\text{Therefore,}} \cr & {T_d} = \int_0^5 {\left| {\left( {t - 3} \right)\left( {3t - 5} \right)} \right|} dt \cr & {T_d} = \int_0^5 {\left| {3{t^2} - 5t - 9t + 15} \right|} dt \cr & {T_d} = \int_0^5 {\left| {3{t^2} - 14t + 15} \right|} dt \cr & {T_d} = \int_0^{5/3} {\left( {3{t^2} - 14t + 15} \right)} dt - \int_{5/3}^3 {\left( {3{t^2} - 14t + 15} \right)} dt \cr & + \int_3^5 {\left( {3{t^2} - 14t + 15} \right)} dt \cr & {\text{Integrating}} \cr & {T_d} = \left[ {{t^3} - 7{t^2} + 15t} \right]_0^{5/3} - \left[ {{t^3} - 7{t^2} + 15t} \right]_{5/3}^3 + \left[ {{t^3} - 7{t^2} + 15t} \right]_3^5 \cr & {T_d} = \frac{{275}}{{27}} - \left( { - \frac{{32}}{{27}}} \right) + 16 \cr & {T_d} = \frac{{739}}{{27}} \approx 27{\text{ units}} \cr} $$