Problem 2 - Thomas Calculus

Question

The instructions for the integrals in Exercises \(1-10\) have two parts, one for the Trapezoidal Rule and one for Simpson's Rule. I. Using the Trapezoidal Rule a. Estimate the integral with \(n=4\) steps and find an upper bound for \(\left|E_{T}\right| .\) b. Evaluate the integral directly and find \(\left|E_{T}\right|\) c. Use the formula (( true value)) \(\times 100\) to express \(\left|E_{T}\right|\) as \right.\right. a percentage of the integral's true value. II. Using Simpson's Rule a. Estimate the integral with \(n=4\) steps and find an upper bound for \(\left|E_{S}\right| .\) b. Evaluate the integral directly and find \(\left|E_{S}\right|\) c. Use the formula ( \(\left|E_{S}\right| /(\) true value) \() \times 100\) to express \(\left|E_{S}\right|\) as a percentage of the integral's true value. $$ \int_{1}^{3}(2 x-1) d x $$

Step-by-Step Solution

Answer

Both methods provide an exact integral evaluation of 6 with 0% error.

Step 1: Set up the problem

We need to estimate the integral \( \int_{1}^{3}(2x-1) \, dx \) using the Trapezoidal Rule and Simpson's Rule with \( n = 4 \), then evaluate the error and express it as a percentage. We'll also evaluate the integral directly to find its true value.

Step 2: Compute True Value of the Integral

Evaluate the integral \( \int_{1}^{3}(2x-1) \, dx \) using basic integration techniques. The antiderivative of \( 2x - 1 \) is \( x^2 - x \). At \( x = 3 \), it evaluates to \( 3^2 - 3 = 6 \), and at \( x = 1 \), it evaluates to \( 1^2 - 1 = 0 \). The definite integral is \( 6 - 0 = 6 \). Thus, the true value is 6.

Step 3: Apply the Trapezoidal Rule

For the Trapezoidal Rule with \( n = 4 \), divide the interval \([1, 3]\) into 4 equal parts, each with a width of \( h = \frac{3-1}{4} = 0.5 \). Calculate the approximate integral: \[ I_T = \frac{h}{2}(f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4)) \] Substitute \(x_0 = 1, x_1 = 1.5, x_2 = 2, x_3 = 2.5, x_4 = 3\). Compute \(f(x_i) = 2x_i - 1\) for each point, then substitute these into the formula to obtain \( I_T = 6 \).

Step 4: Find Upper Bound for Trapezoidal Rule Error

The error bound for the Trapezoidal Rule is given by \( |E_T| \leq \frac{(b-a)^3}{12n^2} \cdot M \), where \(M\) is the maximum value of \(\left| f''(x) \right|\) on \([1, 3]\). For \(f(x) = 2x-1\), \(f''(x) = 0\), so \(M = 0\). Therefore, \( |E_T| = 0 \).

Step 5: Calculate Percentage Error for Trapezoidal Rule

The actual error \(|E_T| = 0\) because our approximate integral from the Trapezoidal Rule equals the true value. The percentage error is \( \left(\frac{|E_T|}{6}\right) \times 100 = 0\%\).

Step 6: Apply Simpson's Rule

For Simpson’s Rule with \( n = 4 \), use \( h = 0.5 \) and apply the formula: \[ I_S = \frac{h}{3}(f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + f(x_4)) \]. Substitute the earlier computed function values to find \( I_S = 6 \).

Step 7: Evaluate Simpson's Rule Error

The error bound for Simpson's Rule is \( |E_S| \leq \frac{(b-a)^5}{180n^4} \cdot M \), where \(M\) is the maximum of \(\left|f^{(4)}(x)\right|\). For our function, the fourth derivative is zero, meaning \(M = 0\), so \(|E_S| = 0\).

Step 8: Calculate Percentage Error for Simpson's Rule

The actual error \(|E_S| = 0\) because the approximation from Simpson’s Rule also matches the true integral value perfectly. Thus the percentage error is \( \left(\frac{|E_S|}{6}\right) \times 100 = 0\%\).

Key Concepts

Trapezoidal Rule

The Trapezoidal Rule is a numerical method used to approximate the definite integral of a function. It involves dividing the area under a curve into trapezoids and calculating the area of each trapezoid to estimate the total integral. This is achieved by dividing the integral's interval into small subdivisions, or segments, called \( n \), and determining the width \( h \) of each segment by calculating \( h = \frac{b-a}{n} \).
Using these segments, the formula for the Trapezoidal Rule is: \\[ I_T = \frac{h}{2}(f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)) \] \This basic approach provides a simple and often effective way to approximate integrals, especially for smooth functions.
The accuracy of the Trapezoidal Rule depends on the function being integrated and the number of subdivisions. Generally, more subdivisions result in a better approximation.
In cases where the underlying function is perfectly linear across each trapezoid, such as with our function \( 2x-1 \), this method can yield a perfect approximation.

Simpson's Rule

Simpson’s Rule is another numerical technique used to estimate definite integrals. It is generally more accurate than the Trapezoidal Rule because it approximates the underlying function with a series of parabolas, rather than straight lines.
The method requires that the number of subdivisions \( n \) be even, and calculates the width \( h \) similarly as \( h = \frac{b-a}{n} \). The formula for Simpson's Rule is: \\[ I_S = \frac{h}{3}(f(x_0) + 4f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n)) \] \This pattern of alternating coefficients of 4 and 2, with 1 at each endpoint, captures the influence of the quadratic nature of the approximation. Simpson's Rule often yields better results because it accommodates curves better.
In our exercise, by approximating the integral of \( 2x-1 \), we got an exact match, which underscores its effectiveness sometimes even with fewer subdivisions than needed by the Trapezoidal Rule to achieve the same accuracy.

Error Analysis

Error analysis in numerical integration helps us understand the accuracy of the approximation methods. For the Trapezoidal Rule, the error is primarily determined by the second derivative of the function. The error formula is given by: \\[ |E_T| \leq \frac{(b-a)^3}{12n^2} \cdot M \] \Where \( M \) is the maximum value of \( |f''(x)| \) on the interval. Essentially, if the function's second derivative is zero, as with linear functions like \( 2x-1 \), this can lead to zero error.
For Simpson's Rule, the error depends on the fourth derivative of the function, expressed as: \\[ |E_S| \leq \frac{(b-a)^5}{180n^4} \cdot M \] \Similarly, if the fourth derivative is zero, the error becomes zero, suggesting that the approximation is perfectly accurate.
The concept of percentage error further clarifies these results, enabling a comparison relative to the actual value of the integral. In our case, both methods produced a zero error when measuring against the true value of 6, yielding a 0% error mark, indicating that the numerical methods perfectly matched the direct integral evaluation.