Chapter 5 - Integration - Review Exercises - Page 395: 45

$$\eqalign{ & {\text{The displacement is 0}}{\text{.}} \cr & {\text{The distance traveled is }}\frac{{20}}{\pi }. \cr} $$

$$\eqalign{ & {\text{Velocity given }}v\left( t \right) = 5\sin \pi t,{\text{ initial point }}s\left( 0 \right) = 0 \cr & {\text{The displacemente between }}t = 0{\text{ and }}t = 2{\text{ is}} \cr & s\left( t \right) = \int_0^2 {v\left( t \right)} dt \cr & s\left( t \right) = \int_0^2 {5\sin \pi t} dt \cr & s\left( t \right) = 5\int_0^2 {\sin \pi t} dt \cr & s\left( t \right) = \frac{5}{\pi }\int_0^2 {\sin \pi t\left( \pi \right)} dt \cr & {\text{Integrating, recall that }}\int {\sin u} du = - \cos u + C \cr & s\left( t \right) = \frac{5}{\pi }\left[ { - \cos \pi t} \right]_0^2 \cr & s\left( t \right) = - \frac{5}{\pi }\left( {\cos 2\pi - \cos 0} \right) \cr & s\left( t \right) = - \frac{5}{\pi }\left( {1 - 1} \right) \cr & s\left( t \right) = 0 \cr & {\text{Then, the displacement is 0}}{\text{.}} \cr & \cr & {\text{The distance traveled during this interval is given by}} \cr & \int_0^2 {\left| {v\left( t \right)} \right|dt} \cr & = \int_0^2 {\left| {5\sin \pi t} \right|dt} \cr & = 5\int_0^2 {\left| {\sin \pi t} \right|dt} \cr & {\text{By the properties of absolute value }}\int_0^2 {\left| {\sin \pi t} \right|dt} = 2\int_0^1 {\sin \pi tdt} \cr & = 5\left( {2\int_0^1 {\sin \pi tdt} } \right) \cr & = 10\int_0^1 {\sin \pi tdt} \cr & {\text{Integrating}} \cr & = 10\left[ { - \frac{1}{\pi }\cos \pi t} \right]_0^1 \cr & = - \frac{{10}}{\pi }\left[ {\cos \pi - \cos 0} \right] \cr & = - \frac{{10}}{\pi }\left( { - 1 - 1} \right) \cr & = \frac{{20}}{\pi } \cr & {\text{Then, the distance traveled is }}\frac{{20}}{\pi }. \cr} $$