[tex]\displaystyle\int_0^8 R(t) \, dt[/tex]
[tex]\begin{array}{l | c c c c c}
t \text{ (hours)} & 0 & 2 & 3 & 7 & 8 \\
R(t) \text{ (gallons per hour)} & 1.95 & 2.5 & 2.8 & 4 & 4.26
\end{array}[/tex]
The area of a trapezoid is given by [tex]\frac{1}{2}(a+b)(h)[/tex]. If we take the height to be the change in x, then [tex]1/2\big[R(t_i) + R(t_{i-1})\big](\Delta t)[/tex] is the area of one trapezoid for one subinterval with borders [tex]t_i [/tex] and [tex]t_{i-1}[/tex]. Therefore, we approximate as follows, using the data to dictate our trapezoid bases and heights:
[tex]\displaystyle\int_0^8 R(t) \, dt \\ \\
\approx \tfrac{1}{2} (R(2) + R(0))(2-0) + \tfrac{1}{2} (R(3) + R(2))(3-2) \\ \\
{}\qquad + \tfrac{1}{2} (R(7) + R(3))(7-3) + \tfrac{1}{2} (R(8) + R(7))(8-7) \\ \\
&= \tfrac{1}{2}(2.5 + 1.95)(2) + \tfrac{1}{2} (2.8+2.5)(1) \\ \\
{}\qquad + \tfrac{1}{2} (4 + 2.8 )( 5) + \tfrac{1}{2} (4.26+4 )(1) \\ \\
&= 24.83 \text{ gallons}[/tex]
From t = 0 hours to t = 8 hours, the total amount of water that flows out is approximately 24.83 gallons.
