<h2>
Hello!</h2>
The answer is: There is a total of 5.797 gallons pumped during the given period.
<h2>
Why?</h2>
To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)
The given function is:

So, the integral will be:

So, integrating we have:

Performing a change of variable, we have:

Then, substituting, we have:



![\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]](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B9%7D%20%5Cint%5Climits%5E4_0%20%7B%281%20%29%5C%20du-%20%5Cfrac%7B5%7D%7B9%7D%20%5Cint%5Climits%5E4_0%20%7B%28%5Cfrac%7B1%7D%7Bu%20%7D%29%5C%20du%3D%5Cfrac%7B5%7D%7B9%7D%20%28u-lnu%29%2F%5B0%2C4%5D)
Reverting the change of variable, we have:
![\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B9%7D%20%28u-lnu%29%2F%5B0%2C4%5D%3D%5Cfrac%7B5%7D%7B9%7D%28%281%2B3t%29-ln%281%2B3t%29%29%2F%5B0%2C4%5D)
Then, evaluating we have:
![\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](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B9%7D%28%281%2B3t%29-ln%281%2B3t%29%29%5B0%2C4%5D%3D%28%5Cfrac%7B5%7D%7B9%7D%28%281%2B3%284%29-ln%281%2B3%284%29%29%29-%28%5Cfrac%7B5%7D%7B9%7D%28%281%2B3%280%29-ln%281%2B3%280%29%29%29%3D%5Cfrac%7B5%7D%7B9%7D%2810.435%29-%5Cfrac%7B5%7D%7B9%7D%281%29%3D5.797)
So, there is a total of 5.797 gallons pumped during the given period.
Have a nice day!