(1) Using a trapezoidal sum, we approximate the integral by
[tex]\displaystyle\int_0^8R(t)\,\mathrm dt\approx\frac{(R(0)+R(2))(2-0)+(R(3)+R(2))(3-2)+(R(7)+R(3))(7-3)+(R(8)+R(7))(8-7)}2[/tex]
[tex]\displaystyle\int_0^8R(t)\,\mathrm dt\approx24.830[/tex]
measured in gallons of water.
(2) Notice that the table shows the specific values of [tex]R(t)[/tex] increasing, meaning [tex]R'(t)>0[/tex]. There's not enough evidence to suggest that we have [tex]R'(t)=0[/tex] at any time.
