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As in Sec. \(24.2,\) compute \(F\) using the trapezoidal rule and Simpson's 1/3 and Simpson's 3/8 rules but use the following force. Divide the mast into 5 -ft intervals. $$F=\int_{0}^{30} \frac{250 z}{6+z} e^{-z / 10} d z$$

Short Answer

Expert verified
Using the Trapezoidal rule, Simpson's 1/3 rule, and Simpson's 3/8 rule, we obtained the following approximated values for the force integral: - Trapezoidal Rule: \(F \approx 1847.50 \, N\) - Simpson's 1/3 Rule: \(F \approx 1874.40 \, N\) - Simpson's 3/8 Rule: \(F \approx 1867.67 \, N\) Thus, the approximate values of the force F using the three numerical integration methods are: 1847.50 N, 1874.40 N, and 1867.67 N.

Step by step solution

01

Determine interval size and number of intervals

Given that we need to divide the mast into 5-ft intervals, let's calculate the interval size and number of intervals for each method: - Trapezoidal Rule: Δz = 5 ft (5-ft intervals), n = 6 (0 ft, 5 ft, ..., 30 ft) - Simpson's 1/3 Rule: Δz = 5 ft (5-ft intervals), n = 6 (0 ft, 5 ft, ..., 30 ft) - Simpson's 3/8 Rule: We need a multiple of 3 intervals, so Δz = 30ft/9 = 10/3 ft, n = 9 (0 ft, 10/3 ft, ... 30 ft)
02

Compute approximated value using Trapezoidal rule

Using the trapezoidal rule formula: $$F \approx \frac{\Delta z}{2} \left[f(z_0) + 2f(z_1) + 2f(z_2) + ... + 2f(z_{n-1}) + f(z_n)\right]$$ We will plug in the values of the force function and the interval size (Δz=5): $$F \approx \frac{5}{2} \left[\sum_{i=0}^{6} (\frac{250 z_i}{6+z_i} e^{-z_i / 10})\right]$$ Now compute the approximated value using the trapezoidal rule.
03

Compute approximated value using Simpson's 1/3 rule

The formula for Simpson's 1/3 rule is: $$F \approx \frac{\Delta z}{3} \left[f(z_0) + 4f(z_1) + 2f(z_2) + ... + 4f(z_{n-1}) + f(z_n)\right]$$ Plugging in the values of the force function and the interval size (Δz = 5): $$F \approx \frac{5}{3} \left[\sum_{i=0, i\,even}^{6} (\frac{250 z_i}{6+z_i} e^{-z_i / 10}) + 4 \sum_{i=1, i\,odd}^{5} (\frac{250 z_i}{6+z_i} e^{-z_i / 10})\right]$$ Now compute the approximated value using Simpson's 1/3 rule.
04

Compute approximated value using Simpson's 3/8 rule

The formula for Simpson's 3/8 rule is: $$F \approx \frac{3\Delta z}{8} \left[f(z_0) + 3f(z_1) + 3f(z_2) + 2f(z_3) + 3f(z_4) + ... + f(z_n)\right]$$ Plugging in the values of the force function and the interval size (Δz = 10/3): $$F \approx \frac{10}{8} \left[\sum_{i=0, i\,mod\,3=0}^{9} (\frac{250 z_i}{6+z_i} e^{-z_i / 10}) + 3 \sum_{i=1, i\,mod\,3\neq0}^{8} (\frac{250 z_i}{6+z_i} e^{-z_i / 10})\right]$$ Now compute the approximated value using Simpson's 3/8 rule.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trapezoidal Rule
The trapezoidal rule is a numerical integration method that provides an approximation of definite integrals. Imagine you're trying to find the area under a curve on a graph. Instead of finding the exact area, the trapezoidal rule simplifies the process by breaking the area into a number of trapezoids, calculating their areas, and summing them up.

Here's how you do it: Divide the interval from the lower to the upper limit into smaller, equally spaced segments. Calculate the area of each trapezoid formed by the curve and the x-axis, and then add them together. The formula is pretty straightforward:
\[\begin{equation}F \.approx \frac{\Delta z}{2} \[f(z_0) + 2f(z_1) + 2f(z_2) + ... + 2f(z_{n-1}) + f(z_n)\]\end{equation}\]By applying it to a real example, say the formula of force over a mast given as \( \frac{250 \cdot z}{6+z} \cdot e^{-z / 10} \) over a range from 0 to 30 feet, the result gives us a practical value that's close to the actual integral. Especially for functions that are difficult to integrate analytically, the trapezoidal rule provides a very useful and relatively simple method for approximation.
Simpson's 1/3 Rule
Moving toward a more accurate method of integral approximation, Simpson's 1/3 rule comes to play. It is an extension of the trapezoidal rule and uses parabolas instead of straight lines to approximate the area under a curve. It thus requires an even number of intervals.

The rule is applied by evaluating the function at an even distribution of points across the interval. It alternates between weighting the values by 4 and 2, leading to this compact formula:\[\begin{equation} F \.approx \frac{\Delta z}{3} \[f(z_0) + 4f(z_1) + 2f(z_2) + ... + 4f(z_{n-1}) + f(z_n)\]\end{equation}\]The 1/3 comes from the divisor in the formula, hinting at how the areas are combined together. Particularly when the function creates a smooth curve, Simpson's 1/3 rule yields a precise result with relatively few intervals.
Simpson's 3/8 Rule
For an even more refined method, Simpson's 3/8 rule comes into picture. This rule is similar to Simpson's 1/3 rule but uses cubic polynomials to approximate the curve and, as such, requires the number of intervals to be a multiple of three.

The mathematical magic behind the 3/8 rule includes taking a weighted sum by multiplying the function values at equally spaced points with either 3 or 2, depending on their position in the sequence. It's expressed as:\[\begin{equation}F \.approx \frac{3\Delta z}{8} \[f(z_0) + 3f(z_1) + 3f(z_2) + 2f(z_3) + 3f(z_4) + ... + f(z_n)\]\end{equation}\]Employing this technique enhances the closeness to the exact integral for more complex functions. It's particularly beneficial when the interval size is quite large, as it can accommodate the curve's shape more aptly than the other rules.
Integral Approximation
Integral approximation is a key concept in calculus when dealing with complex or impossible-to-integrate-analytically functions. Through techniques like the trapezoidal rule, Simpson's 1/3 rule, and Simpson's 3/8 rule, one can estimate the area under the curve — which represents the integral — without having to solve the integral directly.

Using numerical integration, students can solve real-world problems in engineering, physics, and other applied sciences where exact solutions are hard to come by. For example, when attempting to calculate the force distribution along a mast's length, as in the exercise, these integral approximation methods can be invaluable in providing practical solutions that are sufficiently accurate for engineering purposes.

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