a rectangular swimming pool is 40 ft wide by 75 ft long. the accompanying table shows the depth ( h(x) ) of…

a rectangular swimming pool is 40 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}40cdot h(x)dx ).

a rectangular swimming pool is 40 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}40cdot h(x)dx ).

Answer

Explanation:

Step1: Recall the Trapezoidal Rule formula

The Trapezoidal Rule for $\int_{a}^{b}f(x)dx$ is $T_n=\frac{\Delta x}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+\cdots+2f(x_{n - 1})+f(x_n)\right]$, where $\Delta x=\frac{b - a}{n}$. Here, $a = 0$, $b = 75$, $n = 15$, so $\Delta x=\frac{75-0}{15}=5$, and $f(x)=40h(x)$.

Step2: Substitute the values into the formula

[ \begin{align*} T_{15}&=\frac{5}{2}\left[40h(0)+2\times40h(5)+2\times40h(10)+\cdots+2\times40h(70)+40h(75)\right]\ &= 100\left[h(0)+2h(5)+2h(10)+2h(15)+2h(20)+2h(25)+2h(30)+2h(35)+2h(40)+2h(45)+2h(50)+2h(55)+2h(60)+2h(65)+2h(70)+h(75)\right] \end{align*} ] Substitute $h(0) = 4.4$, $h(5)=6.6$, $h(10)=7.6$, $h(15)=8.3$, $h(20)=8.9$, $h(25)=9.4$, $h(30)=9.9$, $h(35)=10.3$, $h(40)=10.7$, $h(45)=11.1$, $h(50)=11.5$, $h(55)=11.8$, $h(60)=12.1$, $h(65)=12.5$, $h(70)=12.8$, $h(75)=13.1$ into the above formula: [ \begin{align*} &100\left[4.4+2\times(6.6 + 7.6+8.3+8.9+9.4+9.9+10.3+10.7+11.1+11.5+11.8+12.1+12.5+12.8)+13.1\right]\ &=100\left[4.4 + 2\times(6.6+7.6+8.3+8.9+9.4+9.9+10.3+10.7+11.1+11.5+11.8+12.1+12.5+12.8)+13.1\right]\ &=100\left[4.4+2\times(15.2+8.3+8.9+9.4+9.9+10.3+10.7+11.1+11.5+11.8+12.1+12.5+12.8)+13.1\right]\ &=100\left[4.4+2\times(23.5+8.9+9.4+9.9+10.3+10.7+11.1+11.5+11.8+12.1+12.5+12.8)+13.1\right]\ &=100\left[4.4+2\times(32.4+9.4+9.9+10.3+10.7+11.1+11.5+11.8+12.1+12.5+12.8)+13.1\right]\ &=100\left[4.4+2\times(41.8+9.9+10.3+10.7+11.1+11.5+11.8+12.1+12.5+12.8)+13.1\right]\ &=100\left[4.4+2\times(51.7+10.3+10.7+11.1+11.5+11.8+12.1+12.5+12.8)+13.1\right]\ &=100\left[4.4+2\times(62+10.7+11.1+11.5+11.8+12.1+12.5+12.8)+13.1\right]\ &=100\left[4.4+2\times(72.7+11.1+11.5+11.8+12.1+12.5+12.8)+13.1\right]\ &=100\left[4.4+2\times(83.8+11.5+11.8+12.1+12.5+12.8)+13.1\right]\ &=100\left[4.4+2\times(95.3+11.8+12.1+12.5+12.8)+13.1\right]\ &=100\left[4.4+2\times(107.1+12.1+12.5+12.8)+13.1\right]\ &=100\left[4.4+2\times(119.2+12.5+12.8)+13.1\right]\ &=100\left[4.4+2\times(131.7+12.8)+13.1\right]\ &=100\left[4.4+2\times144.5+13.1\right]\ &=100\left[4.4 + 289+13.1\right]\ &=100\times306.5 \end{align*} ]

Answer:

$30650$ cubic - feet.