Chapter 6, Problem 33RE

(a)

To determine

To calculate:The upper estimate and lower estimate for the amount of oil being consumed by the generator during the week.

Answer to Problem 33RE

The lower estimate is 4.392 L and the upper estimate is 4.008 L

Explanation of Solution

Given information:

Oil Consumption on Pathfinder Island A diesel generator runs continuously, consuming oil at agradually increasing rate until it must be temporarily shut down to have the filters replaced.

  

Formula used:

The lower estimate and upper estimate method is used.

Calculation:

For upper estimate during that week take the value of oil consumption rates from Monday to Sunday, then add the rates and multiplies with 24 because length of subinterval is 1 day and 1 day is equal to 24 hour.

  $\begin{array}{l}\left(\begin{array}{l}\text{upper}\text{estimate}\\ \text{for}\text{oil}\text{consumed}\end{array}\right)=24\left[0.020+0.021+0.023+0.025+0.028+0.031+0.035\right]\\ \left(\begin{array}{l}\text{upper}\text{estimate}\\ \text{for}\text{oil}\text{consumed}\end{array}\right)=4.392\text{L}\end{array}$

For lower estimate during that week take the value of oil consumption rates from Sunday to Saturday, then add the rates and multiplies with 24 because length of subinterval is 1 day and 1 day is equal to 24 hour.

  $\begin{array}{l}\left(\begin{array}{l}\text{lower}\text{estimate}\\ \text{for}\text{oil}\text{consumed}\end{array}\right)=24\left[0.019+0.020+0.021+0.023+0.025+0.028+0.031\right]\\ \left(\begin{array}{l}\text{lower}\text{estimate}\\ \text{for}\text{oil}\text{consumed}\end{array}\right)=4.008\text{L}\end{array}$

Conclusion:

The lower estimate is 4.392 L and the upper estimate is 4.008 L

(b)

To determine

To find: The amount of oil consumed by the generator during that week.

Answer to Problem 33RE

The value of the integral is ${\displaystyle \underset{0}{\overset{7}{\int }}f\left(x\right)}dx=4.2\text{L}$

Explanation of Solution

Given information:

Oil Consumption on Pathfinder Island A diesel generator runs continuously, consuming oil at a gradually increasing rate until it must be temporarily shut down to have the filters replaced.

  

Formula used:

  $h=\frac{b-a}{n}$

Calculation:

In the trapezoidal rule, if [a, b] is partitioned into n subintervals of equal length$h=\frac{b-a}{n}$, the graph of f on [a, b] can be approximated by a straight line segment at each subinterval. The region between the curve and the x-axis is then approximated by the trapezoids, the area of each trapezoid being the length of its horizontal “altitude" times the average of two vertical “bases".

That is,

  $\begin{array}{l}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=h\frac{y_0+y_1}{2}+h\frac{y_1+y_2}{2}+\mathrm{.....}+h\frac{y_{n-1}+y_n}{2}\\ =h\left(\frac{y_0}{2}+y_1+y_2+\mathrm{....}+y_{n-1}+\frac{y_n}{2}\right)\\ {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=\frac{h}{2}\left(y_0+2y_1+2y_2+\mathrm{.....}+2y_{n-1}+y_n\right)\end{array}$

Where,

  $\begin{array}{l}{y}_0=f\left(a\right)\\ y_1=f\left(x_1\right)\\ .\\ .\\ y_{n-1}=f\left(x_{n-1}\right)\\ y_n=f\left(b\right)\end{array}$

Now, apply the trapezoidal rule with

  $\begin{array}{l}n=7\\ \left[a,b\right]=\left[0,7\right]\end{array}$

So,

  $\begin{array}{l}h=\frac{b-a}{n}\\ =\frac{7-0}{7}\\ h=1\end{array}$

The given table is shown below,

  

To apply the trapezoidal rule,

To take ‘sun’ as $x_0=0$ and the corresponding oil consumption as$y_0=0.019$ similarly by taking the remaining tabular values as shown in the following table,

  

Now, substitute the values from given table in this equation, equation has a multiplication factor of 24 because rates are given in liters/hour.

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{7}{\int }}f\left(x\right)}dx=\text{amount of oil consumed}\\ =\frac{24h}{2}\left(y_0+2y_1+2y_2+2y_3+2y_4+2y_5+2y_6+y_7\right)\\ =\frac{24}{2}\left(0.019+2\left(0.021\right)+2\left(0.023\right)+2\left(0.025\right)+2\left(0.028\right)+2\left(0.031\right)+0.035\right)\\ {\displaystyle \underset{0}{\overset{7}{\int }}f\left(x\right)}dx=4.2\text{L}\end{array}$

Conclusion:

The value of the integral is ${\displaystyle \underset{0}{\overset{7}{\int }}f\left(x\right)}dx=4.2\text{L}$