Chapter 6 - Applications of Integration - 6.1 Velocity and Net Change - 6.1 Exercises - Page 407: 13

$$\eqalign{ & a.{\text{ Positive direction for }}\left( {0,2} \right){\text{ and }}\left( {3,5} \right) \cr & {\text{ Negative direction for }}\left( {2,3} \right) \cr & b.{\text{ }}\frac{{275}}{{12}}{\text{m}} \cr & c.{\text{ }}\frac{{95}}{4}{\text{m}} \cr} $$

$$\eqalign{ & {\text{Let }}v\left( t \right) = {t^3} - 5{t^2} + 6t{\text{ on the interval }}0 \leqslant t \leqslant 5 \cr & \cr & a.{\text{ The graph of the velocity of the given interval is shown}} \cr & {\text{below }}\left( {{\text{Graph using geogebra}}} \right). \cr & *{\text{The motion of the object is in the positive direction when }} \cr & v\left( t \right) > 0.{\text{ then from the graph we see that}} \cr & v\left( t \right) > 0{\text{ on the intervals }}\left( {0,2} \right){\text{ and }}\left( {3,5} \right) \cr & *{\text{The motion of the object is in the negative direction when }} \cr & v\left( t \right) < 0.{\text{ then from the graph we see that}} \cr & v\left( t \right) < 0{\text{ on the interval}}\left( {2,3} \right) \cr & \cr & b.{\text{ The displacement of the object between }}t = a{\text{ and }}t = b \cr & {\text{where }}a > b{\text{ is :}} \cr & s\left( b \right) - s\left( a \right) = \int_a^b {v\left( t \right)} dt,{\text{ }}\left( {{\text{see page 399 on the book}}} \right) \cr & {\text{Therefore}}{\text{,}} \cr & {\text{For the given interval }}0 \leqslant t \leqslant 5 {\text{ }} \cr & s\left( 5 \right) - s\left( 0 \right) = \int_0^5 {\left( {{t^3} - 5{t^2} + 6t} \right)} dt \cr & {\text{Integrating}} \cr & s\left( 5 \right) - s\left( 0 \right) = \left[ {\frac{1}{4}{t^4} - \frac{5}{3}{t^3} + 3{t^2}} \right]_0^5 \cr & s\left( 5 \right) - s\left( 0 \right) = \left[ {\frac{1}{4}{{\left( 5 \right)}^4} - \frac{5}{3}{{\left( 5 \right)}^3} + 3{{\left( 5 \right)}^2}} \right] - \left[ 0 \right] \cr & s\left( 5 \right) - s\left( 0 \right) = \frac{{275}}{{12}} \cr & {\text{The displacement is }}\frac{{275}}{{12}}{\text{m}} \cr & \cr & c.{\text{ The distance traveled of the given interval is given by}} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = \int_0^5 {\left| {{t^3} - 5{t^2} + 6t} \right|} dt \cr & {\text{By the definition of absolute value and using the graph}} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = \int_0^2 {\left( {{t^3} - 5{t^2} + 6t} \right)} dt - \int_2^3 {\left( {{t^3} - 5{t^2} + 6t} \right)} dt \cr & + \int_3^5 {\left( {{t^3} - 5{t^2} + 6t} \right)} dt \cr & {\text{Integrating }} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = \left[ {\frac{1}{4}{t^4} - \frac{5}{3}{t^3} + 3{t^2}} \right]_0^2 - \left[ {\frac{1}{4}{t^4} - \frac{5}{3}{t^3} + 3{t^2}} \right]_2^3 \cr & + \left[ {\frac{1}{4}{t^4} - \frac{5}{3}{t^3} + 3{t^2}} \right]_3^5 \cr & {\text{Evaluating and simplifying}} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = \frac{8}{3} + \frac{5}{{12}} + \frac{{62}}{3} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = \frac{{95}}{4} \cr} $$