The answer is: There is a total of 5.797 gallons pumped during the given period.
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!
