Question

Volume of water in a swimming poolA rectangular swimming pool is 30 ft wide and 50 ft long. The accompanying table shows the depth h(x) of the water at 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 = 10 applied to the integralV = ∫ from 0 to 50 of 30 · h(x) dx.

Accepted Answer

Identify the integral to estimate the volume of water in the pool: $$V = \$$int_0$$^{50} 30 \cdot h(x) \, dx$$, where 30 ft is the width of the pool and $$h(x) is $$the depth at position $$x. $$Note that the depth values $$h(x)$$ are given at 5-ft intervals from $$x=0 to$$ $$x=50$$, so the interval width is $$\Delta x = 5 ft$$, and the number of subintervals is $$n=10. $$Apply the Trapezoidal Rule formula for $$n=10$$ subintervals:

$$\int_a^b f(x) \, dx \approx \frac{\Delta x}{2} \left[f(x_0$) + 2f(x_1$) + 2f(x_2$) + \cdots + 2f(x_{n-1}) + f(x_n)ight]$$,

where $$f(x) = 30 \cdot h(x) in $$this problem. Calculate $$f(x_i) = 30 	imes h(x_i)$$ for each depth value $$h(x_i)$$ from the table, then substitute these values into the trapezoidal sum expression. Finally, multiply the sum by $$\frac{\Delta x}{2} = \frac{5}{2} to $$estimate the volume $$V. $$This will give the approximate volume of water in the pool.

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Key Concepts

Definite Integral as Volume Calculation

Trapezoidal Rule for Numerical Integration

Using Discrete Data Points for Approximation