Chapter 6.5, Problem 9E

To determine

To calculate:The volume of the water in the pool using the trapezoidal rule with n =10.

Answer to Problem 9E

  ${\displaystyle \underset{0}{\overset{50}{\int }}30h\left(x\right)}dx=15,990$

Explanation of Solution

Given information:

A rectangularswimming pool is 30 ft wide and 50 ft long. The table below shows the depth h(x) of the water at 5-fl intervals from one end of the pool to the other. The volume of water in the pool using the Trapezoidal Rule with n z 10‘applied to the integral

  $V={\displaystyle {\int }_0^{50}30h\left(x\right)}dx$

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

  $\begin{array}{l}n=10\\ \left[a,b\right]=\left[0,50\right]\end{array}$

So,

  $\begin{array}{l}h=\frac{b-a}{n}\\ =\frac{20-0}{10}\\ h=5\end{array}$

  

Now, substitute the values from given table in this equation

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{50}{\int }}30h\left(x\right)}dx=\frac{30h}{2}\left(h_0+2h_1+2h_2+2h_3+2h_4+2h_5+2h_6+2h_7+2h_8+2h_9+h_{10}\right)\\ =\frac{150}{2}\left(\begin{array}{l}6.0+2\left(8.2\right)+2\left(9.1\right)+2\left(9.9\right)+2\left(10.5\right)+2\left(11.0\right)+2\left(11.5\right)+2\left(11.9\right)\\ +2\left(12.3\right)+2\left(12.7\right)+13\end{array}\right)\\ =\frac{150}{2}\left(6+16.4+18.2+19.8+21+22+23+23.8+24.6+25.4+13\right)\\ {\displaystyle \underset{0}{\overset{50}{\int }}30h\left(x\right)}dx=15,990\end{array}$

Conclusion:

The value of ${\displaystyle \underset{0}{\overset{50}{\int }}30h\left(x\right)}dx=15,990$