Chapter 25: Problem 3
(a) approximate the value of each of the given integrals by use of Simpson's rule, using the given value of \(n,\) and \((b)\) check by direct integration. $$\int_{0}^{2}\left(1+x^{3}\right) d x, n=2$$
Short Answer
Expert verified
The integral \( \int_{0}^{2}(1+x^{3}) dx \) is approximately 6, verified by both Simpson's Rule and direct integration.
Step by step solution
01
Understand Simpson's Rule
Simpson's Rule is a numerical method to approximate the value of a definite integral. For an integral \( \int_{a}^{b} f(x) \, dx \) and an even integer \( n \), the approximation is:\[\text{Approximation} = \frac{\Delta x}{3} \left[ f(x_0) + 4\sum_{\text{odd } i} f(x_i) + 2\sum_{\text{even } i} f(x_i) + f(x_n) \right]\]where \( x_i = a + i\Delta x \) and \( \Delta x = \frac{b-a}{n} \).
02
Compute \( \Delta x \) and \( x_i \) Values
For the integral \( \int_{0}^{2} (1+x^3) \, dx \) with \( n = 2 \), first calculate \( \Delta x = \frac{2 - 0}{2} = 1 \). Next, determine the \( x_i \) values: \( x_0 = 0 \), \( x_1 = 1 \), and \( x_2 = 2 \).
03
Evaluate the Function at \( x_i \)
Compute the function values: \( f(x_0) = 1+0^3 = 1 \), \( f(x_1) = 1+1^3 = 2 \), and \( f(x_2) = 1+2^3 = 9 \).
04
Apply Simpson's Rule Formula
Substitute values into Simpson's Rule formula:\[\text{Approximation} = \frac{1}{3} [ 1 + 4 \times 2 + 9 ] = \frac{1}{3} [ 1 + 8 + 9 ] = \frac{1}{3} [ 18 ] = 6\]
05
Direct Integration
To verify the approximation, integrate directly:\[ \int_0^2 (1 + x^3) \, dx = \left. \left(x + \frac{x^4}{4} \right) \right|_0^2 = \left( 2 + \frac{2^4}{4} \right) - \left( 0 + \frac{0^4}{4} \right) = 2 + 4 = 6\]
06
Conclusion
Both the approximation using Simpson's Rule and the actual value from direct integration result in 6, confirming that the approximation is accurate.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Simpson's Rule
Simpson's Rule is a powerful technique used in numerical integration to approximate the definite integral of a function. It is particularly useful when we cannot easily find an explicit solution through direct calculations. The technique is based on the idea of approximating segments of the curve with a series of parabolic arcs drawn across subintervals of the range of integration.
Simpson's Rule is applied when the number of subintervals \( n \) is even.
This approximation formula is given by:\[\text{Approximation} = \frac{\Delta x}{3} \left[ f(x_0) + 4\sum_{\text{odd } i} f(x_i) + 2\sum_{\text{even } i} f(x_i) + f(x_n) \right]\]where:
Simpson's Rule is applied when the number of subintervals \( n \) is even.
This approximation formula is given by:\[\text{Approximation} = \frac{\Delta x}{3} \left[ f(x_0) + 4\sum_{\text{odd } i} f(x_i) + 2\sum_{\text{even } i} f(x_i) + f(x_n) \right]\]where:
- \( \Delta x = \frac{b-a}{n} \) – the width of each subinterval.
- \( x_i = a + i\Delta x \) – the endpoints and midpoints of subintervals.
Definite Integral
A definite integral symbolically represents the area under a curve described by a function over a specific interval \([a, b]\). It's written as:\[ \int_{a}^{b} f(x) \, dx \]where \( f(x) \) is the function being integrated. The main focus is to find the total accumulation of quantities, like area or volume.
- In comparison to an indefinite integral, which includes a constant of integration, a definite integral provides a numeric value.
- The limits of integration, \(a\) and \(b\), define the interval over which we perform the integration.
Direct Integration
Direct integration is the classical method of finding an exact solution for a definite integral. It involves finding the antiderivative (or integral) of the function, and then evaluating this antiderivative at the upper and lower bounds of the interval.To directly integrate a function like \(1 + x^3\) over the interval [0, 2], we first find the antiderivative:\[ \int (1 + x^3) \, dx = x + \frac{x^4}{4} \]After finding the antiderivative, we apply the limits of integration using the Fundamental Theorem of Calculus:\[ \left.\left(x + \frac{x^4}{4} \right) \right|_0^2 \]Evaluating this expression at \(x = 2\) and \(x = 0\), we verify that the exact value of the integral is 6.
This technique not only affirms the accuracy of approximate methods like Simpson's Rule but also provides a clear and definitive result for the area under the curve.
This technique not only affirms the accuracy of approximate methods like Simpson's Rule but also provides a clear and definitive result for the area under the curve.