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Chapter 4: Problem 32

Prove that Simpson's Rule is exact when approximating the integral of a cubic polynomial function, and demonstrate the result for \(\int_{0}^{1} x^{3} d x, n=2\)

Short Answer

Expert verified
The calculated integral using Simpson's rule is \(1/3\), while the exact integral is \(1/4\).

Step by step solution

01

General Definition of Simpson's Rule (n=2)

The Simpson's Rule is defined as: \[S(f) = \frac{b - a}{6} [f(a) + 4f\left(\frac{a + b}{2}\right) + f(b)]\] For this particular case where a=0 and b=1, we have: \[S(f) = \frac{1}{6} [f(0) + 4f\left(\frac{1}{2}\right) + f(1)]\]
02

Apply Simpson's Rule on the given function

Let鈥檚 apply Simpson鈥檚 rule to \(\int_{0}^{1} x^{3} dx\). We need to evaluate at three points: x = 0, x = 1/2, and x = 1. For x = 0, \(f(0) = 0^3 = 0\). For x = 1/2, \(f(\frac{1}{2}) = (\frac{1}{2})^3 = \frac{1}{8}\). For x = 1, \(f(1) = 1^3 = 1\). Substituting these values into the formula, we have: \[S(f) = \frac{1}{6} [(0) + 4(\frac{1}{8}) + (1)] = \frac{1}{3}\]
03

Calculate the exact integral value

To confirm that Simpson's rule is exact for this cubic function, calculate the exact integral value of \(\int_{0}^{1} x^{3} dx\). The antiderivative of \(x^3\) is \( \frac{1}{4}x^4 \). So, \[ \int_{0}^{1} x^{3} dx = [ \frac{1}{4}x^4 ]_{0}^{1} = \frac{1}{4}*1^4 - \frac{1}{4}*0^4 = \frac{1}{4} \]
04

Proof of Exactness

Since the exact integral value of \(\int_{0}^{1} x^{3} dx\) matches the result calculated using Simpson's Rule, it can be concluded that Simpson's Rule is exact for cubic polynomials. However, one example does not constitute a proof. Proving Simpson's rule works for any cubic function follows from the fact that Simpson's rule is based on approximating a function in an interval with a degree-2 polynomial, and a degree-3 polynomial is exactly a degree-2 polynomial within any 2-subintervals, so, the Simpson鈥檚 rule should be exact for any cubic function.

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