Question

Use a trapezoidal sum with the four sub-intervals indicated by the data in the table to estimate the definite integral from 0 to 8 of R of t, dt. Using correct units, explain the meaning of your answer in terms of water flow. Give 3 decimal places in your answer.
t(hours) 0 | 2 | 3 | 7 | 8
R(t) (gallons per hour) 1.95 | 2.5 | 2.8 | 4 | 4.26

Answer 1

   $\displaystyle\int_0^8 R(t) \, dt$

   $\begin{array}{l | c c c c c} t \text{ (hours)} & 0 & 2 & 3 & 7 & 8 \\ R(t) \text{ (gallons per hour)} & 1.95 & 2.5 & 2.8 & 4 & 4.26 \end{array}$

The area of a trapezoid is given by $\frac{1}{2}(a+b)(h)$. If we take the height to be the change in x, then $1/2\big[R(t_i) + R(t_{i-1})\big](\Delta t)$ is the area of one trapezoid for one subinterval with borders $t_i$ and $t_{i-1}$. Therefore, we approximate as follows, using the data to dictate our trapezoid bases and heights:

   $\displaystyle\int_0^8 R(t) \, dt \\ \\ \approx \tfrac{1}{2} (R(2) + R(0))(2-0) + \tfrac{1}{2} (R(3) + R(2))(3-2) \\ \\ {}\qquad + \tfrac{1}{2} (R(7) + R(3))(7-3) + \tfrac{1}{2} (R(8) + R(7))(8-7) \\ \\ &= \tfrac{1}{2}(2.5 + 1.95)(2) + \tfrac{1}{2} (2.8+2.5)(1) \\ \\ {}\qquad + \tfrac{1}{2} (4 + 2.8 )( 5) + \tfrac{1}{2} (4.26+4 )(1) \\ \\ &= 24.83 \text{ gallons}$

From t = 0 hours to t = 8 hours, the total amount of water that flows out is approximately 24.83 gallons.