Answer
$${V_{avg}} \approx 0.53183{\text{ liter}}$$
Work Step by Step
$$\eqalign{
& {\text{The volume is given by }} \cr
& V = 0.1729t + 0.1522{t^2} - 0.0374{t^3},{\text{ }}t{\text{ is in seconds}} \cr
& {\text{Let }}V = 0{\text{ to find the interval of the positive cycle}} \cr
& 0.1729t + 0.1522{t^2} - 0.0374{t^3} = 0 \cr
& t\left( {0.0374{t^2} - 0.1522t - 0.1729} \right) = 0 \cr
& {\text{By using the quadratic formula}} \cr
& t = 0,{\text{ }}t \approx - 0.9255,{\text{ }}t \approx 4.9950 \cr
& {\text{Using }}t = 0{\text{ and }}t = 5{\text{ }}\left( {t{\text{ is positive}}} \right) \cr
& \cr
& {\text{The average volume of air is given by:}} \cr
& {V_{avg}} = \frac{1}{{5 - 0}}\int_0^5 {\left( {0.1729t + 0.1522{t^2} - 0.0374{t^3}} \right)} dt \cr
& {\text{Integrating}} \cr
& {V_{avg}} = \frac{1}{5}\left[ {\frac{{0.1729{t^2}}}{2} + \frac{{0.1522{t^3}}}{3} - \frac{{0.0374{t^4}}}{4}} \right]_0^5 \cr
& {V_{avg}} = \frac{1}{5}\left[ {\frac{{0.1729{{\left( 5 \right)}^2}}}{2} + \frac{{0.1522{{\left( 5 \right)}^3}}}{3} - \frac{{0.0374{{\left( 5 \right)}^4}}}{4}} \right] - 0 \cr
& {V_{avg}} \approx 0.53183{\text{ liter}} \cr} $$
