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

To estimate the definite integral using a trapezoidal sum, calculate the areas of trapezoids formed by the given data points. Add up all the trapezoid areas to find the estimated definite integral. In terms of water flow, the estimate represents the total amount of water that flowed during the given time interval.

Step-by-step explanation:

To estimate the definite integral from 0 to 8 of R(t), dt using a trapezoidal sum, we need to calculate the areas of trapezoids formed by the given data points. The formula for the trapezoid area is (b1 + b2) * h / 2, where b1 and b2 are the lengths of the parallel sides and h is the height or interval width.

First, let's calculate the lengths of the parallel sides (base values).

For the first interval (0 to 2), b1 = R(0) = 1.95 gallons per hour, and b2 = R(2) = 2.5 gallons per hour.

Next, we calculate the height/interval width (h) which is 2 - 0 = 2 hours.

Now, we can find the area of the first trapezoid: (1.95 + 2.5) * 2 / 2 = 2.225 gallons.

Proceeding similarly for the remaining intervals, we find:

Second interval area: (2.5 + 2.8) * 1 = 5.3 gallons.

Third interval area: (2.8 + 4) * 4 = 27.2 gallons.

Fourth interval area: (4 + 4.26) * 1 = 8.26 gallons.

Now, add up all the trapezoid areas: 2.225 + 5.3 + 27.2 + 8.26 = 43.985 gallons.

The estimated definite integral from 0 to 8 of R(t), dt using the trapezoidal sum is approximately 43.985 gallons.

In terms of water flow, this estimate represents the total amount of water that flowed during the given time interval. It provides an approximate measure of the total volume of water that passed through the system from time 0 to time 8.

Answer 2

`\displaystyle\int_0^8 R(t) \, dt`


`\begin{array}l 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
`(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.