Question

a rectangular swimming pool is 42 ft wide by 75 ft long. the accompanying table shows the depth ( h(x) ) of the water in 5 - ft intervals from one end of the pool to the other. estimate the volume of water in the pool using the trapezoidal rule with ( n = 15 ) applied to the integral ( v=int_{0}^{75}42cdot h(x)dx ).

the volume of the pool is ( square mathrm{ft}^{3} ).
(round to the nearest integer as needed.)

Explanation:

Step1: Recall the Trapezoidal Rule formula

The Trapezoidal Rule formula for \( \int_{a}^{b}f(x)dx\) with \(n\) sub - intervals is \(T=\frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+\cdots+2f(x_{n - 1})+f(x_n)]\), where \(\Delta x=\frac{b - a}{n}\). Here, \(a = 0\), \(b = 75\), \(n = 15\), so \(\Delta x=\frac{75-0}{15}=5\), and \(f(x)=42h(x)\).

Step2: Calculate the sum inside the Trapezoidal Rule formula

Let \(S = h(x_0)+2h(x_1)+2h(x_2)+\cdots+2h(x_{14})+h(x_{15})\).
\(h(x_0)=4\), \(h(x_1) = 6.2\), \(h(x_2)=7.2\), \(h(x_3)=7.9\), \(h(x_4)=8.5\), \(h(x_5)=9\), \(h(x_6)=9.5\), \(h(x_7)=9.9\), \(h(x_8)=10.3\), \(h(x_9)=10.7\), \(h(x_{10})=11.1\), \(h(x_{11})=11.4\), \(h(x_{12})=11.7\), \(h(x_{13})=12.1\), \(h(x_{14})=12.4\), \(h(x_{15})=12.7\).

$$
LATEXBLOCK0
$$

Step3: Calculate the volume \(V\)

Since \(V=\frac{\Delta x}{2}\times42\times S\) (because \(V=\int_{0}^{75}42h(x)dx\) and using the Trapezoidal Rule). Substitute \(\Delta x = 5\) and \(S = 298.5\) into the formula.
\(V=\frac{5}{2}\times42\times298.5\)
First, \(\frac{5}{2}\times42=105\). Then \(V = 105\times298.5=31342.5\)

Answer:

\(31343\)