Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the quantity 1 plus 3 times t with t measure in hours and and D(t) measured in gallons per hour. How much oil will the pumping stations deliver during the 4-hour period from t = 0 to t = 4? Give 3 decimal places.

Hello!The answer is:  There is a total of 5.797 gallons pumped during the given period.Why?To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)The given function is:[tex]D(t)=\frac{5t}{1+3t}[/tex]So, the integral will be:[tex]\int\limits^4_0 {\frac{5t}{1+3t}} \ dx[/tex]So, integrating we have:[tex]\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx[/tex]Performing a change of variable, we have:[tex]1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}[/tex]Then, substituting, we have:[tex]\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du[/tex][tex]\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )[/tex][tex]\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du[/tex][tex]\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4][/tex]Reverting the change of variable, we have:[tex]\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4][/tex]Then, evaluating we have:[tex]\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797[/tex]So, there is a total of 5.797 gallons pumped during the given period.Have a nice day!