Problem 1
Question
For each definite integral: a. Approximate it "by hand," using trapezoidal approximation with \(n=4\) trapezoids. Round calculations to three decimal places. b. Evaluate the integral exactly using antiderivatives, rounding to three decimal places. c. Find the actual error (the difference between the actual value and the approximation). d. Find the relative error (the actual error divided by the actual value, expressed as a percent). \(\int_{1}^{3} x^{2} d x\)
Step-by-Step Solution
Verified Answer
Approximate value: 8.625. Exact value: 8.667. Actual error: 0.042. Relative error: 0.485%.
1Step 1: Determine Subinterval Width
First, calculate the width of each trapezoid for the approximation. With \(n = 4\) trapezoids, the interval \([1, 3]\) is divided as follows. The width \(h\) is given by \(h = \frac{b-a}{n} = \frac{3-1}{4} = 0.5\).
2Step 2: Identify Subintervals
The interval \([1, 3]\) divided into 4 subintervals with width 0.5 gives the subinterval points: \(1, 1.5, 2, 2.5, 3\).
3Step 3: Evaluate Function at Key Points
Calculate \(f(x) = x^2\) at each key point: \(f(1) = 1^2 = 1\), \(f(1.5) = (1.5)^2 = 2.25\), \(f(2) = 2^2 = 4\), \(f(2.5) = (2.5)^2 = 6.25\), \(f(3) = 3^2 = 9\).
4Step 4: Apply the Trapezoidal Rule
Using the trapezoidal rule formula, \[ T = \frac{h}{2} \left( f(x_0) + 2 \sum_{i=1}^{n-1} f(x_i) + f(x_n) \right) \]For this problem, the approximation is \[ T = \frac{0.5}{2} \left( 1 + 2(2.25 + 4 + 6.25) + 9 \right) = 0.25 \times (1 + 24.5 + 9) = 0.25 \times 34.5 = 8.625 \] thus the approximate value is 8.625.
5Step 5: Find Exact Value Using Antiderivatives
Compute the integral using antiderivatives. The antiderivative of \(x^2\) is \(\frac{x^3}{3}\). Evaluate it from 1 to 3:\[ \frac{3^3}{3} - \frac{1^3}{3} = \frac{27}{3} - \frac{1}{3} = 9 - \frac{1}{3} = \frac{26}{3} = 8.667 \] Rounding to three decimal places gives 8.667.
6Step 6: Calculate Actual Error
The actual error is the difference between the exact value and the approximate value:\[ |8.667 - 8.625| = 0.042 \].
7Step 7: Calculate Relative Error Percent
The relative error is the actual error divided by the exact value, expressed as a percentage:\[ \left( \frac{0.042}{8.667} \right) \times 100 \approx 0.485\% \].
Key Concepts
Trapezoidal ApproximationAntiderivativesError Calculation
Trapezoidal Approximation
The trapezoidal approximation is a numerical method used to estimate the value of definite integrals. It works by dividing the region under a curve into a series of trapezoids (four, in this case) and estimating the area of each one separately. The widths of these trapezoids are equal, calculated as the difference between the upper and lower limits of integration divided by the number of trapezoids. In our exercise, this is \[ h = \frac{3-1}{4} = 0.5 \] which defines the width of each trapezoid.
Next, we identify the x-values for the endpoints of each subinterval: 1, 1.5, 2, 2.5, and 3. At these points, we evaluate the function to find its height.
The trapezoidal rule is then applied using the formula: \[ T = \frac{h}{2} \left( f(x_0) + 2 \sum_{i=1}^{n-1} f(x_i) + f(x_n) \right) \] for the sum of the areas. This formula takes into account the isolated first and last points and doubles the sum of the function values at the interior points because each interior point falls at the junction of two trapezoids. After doing the math step-by-step as shown, we come up with an approximation of 8.625.
This approximation method provides a quick insight into the integral's value, which is especially helpful when the function does not have an easy-to-find antiderivative.
Next, we identify the x-values for the endpoints of each subinterval: 1, 1.5, 2, 2.5, and 3. At these points, we evaluate the function to find its height.
The trapezoidal rule is then applied using the formula: \[ T = \frac{h}{2} \left( f(x_0) + 2 \sum_{i=1}^{n-1} f(x_i) + f(x_n) \right) \] for the sum of the areas. This formula takes into account the isolated first and last points and doubles the sum of the function values at the interior points because each interior point falls at the junction of two trapezoids. After doing the math step-by-step as shown, we come up with an approximation of 8.625.
This approximation method provides a quick insight into the integral's value, which is especially helpful when the function does not have an easy-to-find antiderivative.
Antiderivatives
Antiderivatives are the centerpiece of definite integrals. An antiderivative of a function is another function whose derivative gives back the original function. For example, an antiderivative of \( x^2 \) is \( \frac{x^3}{3} \). To calculate the definite integral from 1 to 3 of \( x^2 \), we evaluate this antiderivative at the upper limit and subtract its value at the lower limit.
Performing the calculation \[ \frac{3^3}{3} - \frac{1^3}{3} = \frac{27}{3} - \frac{1}{3} = 9 - \frac{1}{3} = \frac{26}{3} = 8.667 \] gives the exact value of the integral as 8.667.
Using antiderivatives provides the exact value of the area under a curve between two points, unlike the trapezoidal approximation, which is only an estimate. Mastering the use of antiderivatives is crucial for solving integrals accurately and analyzing areas under curves.
Performing the calculation \[ \frac{3^3}{3} - \frac{1^3}{3} = \frac{27}{3} - \frac{1}{3} = 9 - \frac{1}{3} = \frac{26}{3} = 8.667 \] gives the exact value of the integral as 8.667.
Using antiderivatives provides the exact value of the area under a curve between two points, unlike the trapezoidal approximation, which is only an estimate. Mastering the use of antiderivatives is crucial for solving integrals accurately and analyzing areas under curves.
Error Calculation
The error calculation in the context of approximations is key to understanding the accuracy of the estimation methods like the trapezoidal approximation. There are two common types of errors we look at:
For actual error, we simply calculate the absolute difference between the exact value obtained through anti-differentiation and the estimated value from the trapezoidal approximation. This gives us an error of \[ |8.667 - 8.625| = 0.042 \].
Relative error, on the other hand, offers insight into the size of the error compared to the true value. It is calculated by dividing the actual error by the exact integral value, providing a percentage. In this exercise, the relative error calculates to about \[ \left( \frac{0.042}{8.667} \right) \times 100 \approx 0.485\% \].
These error calculations are crucial in evaluating the reliability of numerical methods and in choosing the most suitable approach for approximating integrals based on the acceptable margin of error in different contexts.
- Actual Error
- Relative Error
For actual error, we simply calculate the absolute difference between the exact value obtained through anti-differentiation and the estimated value from the trapezoidal approximation. This gives us an error of \[ |8.667 - 8.625| = 0.042 \].
Relative error, on the other hand, offers insight into the size of the error compared to the true value. It is calculated by dividing the actual error by the exact integral value, providing a percentage. In this exercise, the relative error calculates to about \[ \left( \frac{0.042}{8.667} \right) \times 100 \approx 0.485\% \].
These error calculations are crucial in evaluating the reliability of numerical methods and in choosing the most suitable approach for approximating integrals based on the acceptable margin of error in different contexts.
Other exercises in this chapter
Problem 1
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For each definite integral: a. Approximate it "by hand," using trapezoidal approximation with \(n=4\) trapezoids. Round calculations to three decimal places. b.
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