Question Statement

Evaluate ${\int }_{-1}^{7}f(x)dx$ using the Trapezoidal rule and Simpson's $\frac{1}{3}$ rule.

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Background and Explanation

This problem involves numerical integration techniques used to approximate definite integrals when the function is given as discrete data points rather than a continuous formula. The Trapezoidal rule approximates the area using linear segments between points, while Simpson's $\frac{1}{3}$ rule fits quadratic curves through triplets of points for greater accuracy (requiring an even number of intervals).

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Solution

From the given table, we observe that the $x$-values are equally spaced with a difference of $1$. Thus, the step size is: $h=1$

Given that $n=8$ (even number of intervals), we can apply both methods. The function values are:

$\begin{array}{rl}f(x_0)& =f(-1)=0.9848\\ f(x_1)& =f(0)=1\\ f(x_2)& =f(1)=0.9848\\ f(x_3)& =f(2)=0.9397\\ f(x_4)& =f(3)=0.8660\\ f(x_5)& =f(4)=0.7660\\ f(x_6)& =f(5)=0.6428\\ f(x_7)& =f(6)=0.500\\ f(x_8)& =f(7)=0.3420\end{array}$

Trapezoidal Rule

By the Trapezoidal rule: ${\int }_{-1}^{7}f(x)dx\approx \frac{h}{2}\left[f(x_0)+2\left(f(x_1)+f(x_2)+f(x_3)+f(x_4)+f(x_5)+f(x_6)+f(x_7)\right)+f(x_8)\right]$

Putting the values in the above formula:

$\begin{array}{rl}{\int }_{-1}^{7}f(x)dx& \approx \frac{1}{2}[0.9848+2(1+0.9848+0.9397+0.8660\\ & +0.7660+0.6428+0.500)+0.3420]\\ & \approx 0.5[0.9848+2(5.6993)+0.3420]\\ & \approx 0.5[0.9848+11.3986+0.3420]\\ & \approx 0.5(12.7254)\\ & \approx 6.3627\end{array}$

Simpson's $\frac{1}{3}$ Rule

By Simpson's $\frac{1}{3}$ rule (valid since we have an even number of intervals): ${\int }_{-1}^{7}f(x)dx\approx \frac{h}{3}\left[f(x_0)+4\left(f(x_1)+f(x_3)+f(x_5)+f(x_7)\right)+2\left(f(x_2)+f(x_4)+f(x_6)\right)+f(x_8)\right]$

Putting the values in the above formula:

$\begin{array}{rl}{\int }_{-1}^{7}f(x)dx& \approx \frac{1}{3}[0.9848+4(1+0.9397+0.7660+0.500)\\ & +2(0.9848+0.8660+0.6428)+0.3420]\\ & \approx \frac{1}{3}[0.9848+4(3.2057)+2(2.4936)+0.3420]\\ & \approx \frac{1}{3}[0.9848+12.8228+4.9872+0.3420]\\ & \approx \frac{1}{3}(19.1368)\\ & \approx 6.3789\end{array}$

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Key Formulas or Methods Used

• Step Size: $h=\frac{b-a}{n}$ (determined from tabular spacing as $h=1$)
• Trapezoidal Rule: ${\int }_a^{b}f(x)dx\approx \frac{h}{2}\left[y_0+2\sum _{i=1}^{n-1}{y}_i+y_n\right]$
• Simpson's $\frac{1}{3}$ Rule: ${\int }_a^{b}f(x)dx\approx \frac{h}{3}\left[y_0+4\sum _{\text{odd }i}{y}_i+2\sum _{\text{even }i}{y}_i+y_n\right]$ (requires even number of intervals)

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Summary of Steps

1. Identify parameters: Determine $h=1$ from the equally spaced $x$-values and verify $n=8$ intervals (even, so Simpson's rule applies).
2. List function values: Assign $f(x_0)$ through $f(x_8)$ from the given table.
3. Apply Trapezoidal Rule: Calculate $\frac{h}{2}[f(x_0)+2(\text{sum of interior points})+f(x_8)]$ to get approximately $6.3627$.
4. Apply Simpson's $\frac{1}{3}$ Rule: Calculate $\frac{h}{3}[f(x_0)+4(\text{odd-indexed points})+2(\text{even-indexed points})+f(x_8)]$ to get approximately $6.3789$.
5. Compare results: Note that Simpson's rule generally provides better accuracy due to its quadratic approximation.