Chapter 8.1, Problem 27E

To determine

The number of cartoons of milk were filled by the machine over the 10 hrs period using trapezoidal rule .

Answer to Problem 27E

The value is about $1,156.5$ cartons of milk.

Explanation of Solution

Given information:

The given table shows a machine fills milk cartons with milk at an approximately constant rate with $n=10$ :

  

Formula:

The Trapezoidal rule is

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}\approx \frac{1}{2}\Delta x\left[f\left(x_0\right)+2f\left(x_2\right)+2f\left(x_3\right)+\dots +2f\left(x_{n-1}\right)+f\left(x_n\right)\right]$

Where $\Delta x=\frac{b-a}{n},a=x_0,b=x_n$

Let $t$ be the number of hours since 8am. The time in the table then correspond to $t=0$ to $t=10$ .

From given we know that $n=10$ , $a=0,b=10$

  $\begin{array}{l}\Delta x=\frac{b-a}{n}\\ \Rightarrow \Delta x=\frac{10-0}{10}=1\end{array}$

Substituting $\Delta x$ and the values of $f\left(x_i\right)$ from the given table in trapezoidal rule,

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}\approx \frac{1}{2}\Delta x\left[f\left(x_0\right)+2f\left(x_2\right)+2f\left(x_3\right)+\dots +2f\left(x_{n-1}\right)+f\left(x_n\right)\right]$

  $\begin{array}{l}=\frac{1}{2}\left(1\right)\left[120+2\left(110\right)+2\left(115\right)+2\left(119\right)+2\left(120\right)+2\left(120\right)+2\left(115\right)+2\left(112\right)+2\left(110\right)+121\right]\\ =\frac{1}{2}\left[120+220+230+238+240+240+230+224+220+121\right]\\ =\frac{1}{2}\left(2313\right)\\ =1156.5\end{array}$

Therefore, approximately $1156.5$ cartons of milk were filled over this 10-hours period.