Problem 44 - Calculus Single Variable

Question

Calculate \(A=\int_{-1}^{1}|x| d x\). Let \(N\) be an even positive integer that is not divisible by \(4 .\) Show that if a uniform partition of order \(N\) is used, then the Midpoint Rule approximation of \(A\) is exact, but the Simpson's Rule approximation is not.

Step-by-Step Solution

Answer

Midpoint Rule gives an exact approximation; Simpson's Rule does not.

Step 1: Express the Integral

The given integral is \( A = \int_{-1}^{1} |x| \, dx \). Since \(|x|\) is an even function, we can split the integral into two parts: one over \([-1, 0]\) where \(|x| = -x\), and the other over \([0, 1]\) where \(|x| = x\). Thus, the integral becomes:\[ A = \int_{-1}^{0} -x \, dx + \int_{0}^{1} x \, dx. \]

Step 2: Solve the Integrals

Evaluate each integral separately. For \([-1, 0]\), the integral is:\[ \int_{-1}^{0} -x \, dx = \left[-\frac{x^2}{2}\right]_{-1}^{0} = 0 - \left(-\frac{1}{2}\right) = \frac{1}{2}. \]For \([0, 1]\), the integral becomes:\[ \int_{0}^{1} x \, dx = \left[\frac{x^2}{2}\right]_{0}^{1} = \frac{1}{2} - 0 = \frac{1}{2}. \]Thus, the total integral value is \( A = \frac{1}{2} + \frac{1}{2} = 1. \)

Step 3: Calculate Midpoint Rule

The Midpoint Rule formula for approximation with uniform partitions of \( n \) subintervals is given by:\[ M = \sum_{i=0}^{n-1} f\left(\frac{x_i + x_{i+1}}{2}\right) \Delta x, \]where \( \Delta x = \frac{b-a}{n} \) and \( x_i = a + i \Delta x. \)Since \(|x|\) is symmetric around 0, each subinterval contributes equally positive and negative values, and their sum along the interval \([-1, 0]\) and \([0, 1]\) cancels other contributions exactly, leading \( M \) to be precise when evaluated over \([-1, 1]\).

Step 4: Verify Simpson's Rule

Simpson's Rule for numerical approximation for an even \( n \) is given by:\[ S = \frac{\Delta x}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + \cdots + 4f(x_{n-1}) + f(x_n) \right], \]where \( x_i = a + i \Delta x \) and \( \Delta x = \frac{b-a}{n} \). In this case, since \( |x| \) is not a polynomial function of degree \leq 3 (required for Simpson's Rule to be exact), although \(|x|\) is linear in parts, across the interval as a whole, the rule does not apply exactly when \( n \) is even but \( n/2 \) is odd (as it isn't divisible by 4).

Step 5: Conclusion

For \( n \) not divisible by 4, the exactness condition fails since Simpson's Rule relies on symmetry and polynomial fitting in pieces where \(|x|\) breaks the polynomial requirement within the combined interval.Therefore, the Midpoint Rule gives an exact approximation for \( A \) while Simpson's Rule does not under this condition.

Key Concepts

Midpoint Rule

In numerical integration, the Midpoint Rule is a simple yet effective method for approximating the integral of a function over a specified interval. The key idea is to replace the function with its value at the midpoint of each subinterval, then multiply by the width of the subinterval. This method is particularly useful for functions that are symmetric or have equal contributions in either direction of the midpoint.

The formula for the Midpoint Rule with uniform partitions is given by:

• \[ M = \sum_{i=0}^{n-1} f\left(\frac{x_i + x_{i+1}}{2}\right) \Delta x \]
• where \( \Delta x = \frac{b-a}{n}, \) and \( x_i = a + i \Delta x. \)

In the context of the absolute value function \(|x|\), which is symmetric around zero, the Midpoint Rule excels in finding the correct integral value over \([-1, 1]\).

Because \(|x|\) is an even function, each positive contribution cancels out its corresponding negative contribution over symmetric intervals, leading to an exact calculation of the integral.

Simpson's Rule

Simpson's Rule is another technique for numerical integration, which is particularly useful for functions that are relatively smooth or that can be well-approximated by polynomials. The rule combines the areas under parabolic segments of the curve, assuming the function behaves like a quadratic polynomial over each interval.

Simpson's Rule is expressed as:

• \[ S = \frac{\Delta x}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + \cdots + 4f(x_{n-1}) + f(x_n) \right] \]
• where \( x_i = a + i \Delta x \) and \( \Delta x = \frac{b-a}{n}. \)

For Simpson's Rule to be exact, the function should ideally be a polynomial of degree three or less over the interval. However, the absolute value function \(|x|\) does not fit this criterion as it changes linearity at zero. Consequently, if \( n \) is even but \( n/2 \) isn't divisible by four, the symmetry Simpson's Rule relies on is not present. Therefore, this rule cannot deliver an exact result in such cases.

Absolute Value Function

The absolute value function, denoted as \(|x|\), plays a crucial role in various areas of mathematics as it represents the distance of a number from zero on the real number line, without accounting for direction.

The mathematical definition is:

• \( |x| = x, \) if \( x \geq 0 \)
• \( |x| = -x, \) if \( x < 0 \)

For numerical integration purposes, \(|x|\) poses an interesting scenario because of its piecewise nature. It can be viewed as linear in separate intervals - specifically, it is linear and decreasing over \([-1, 0]\) and linear and increasing over \([0, 1]\).

This characteristic introduces a unique aspect when applying integration rules such as the Midpoint Rule and Simpson's Rule, and necessitates careful consideration to ensure accurate approximations over symmetric intervals.

Even and Odd Functions

Understanding whether a function is even or odd is fundamental in simplifying integration processes. An even function, like \(|x|\), satisfies the condition \(f(-x) = f(x)\) for all \(x\). This symmetry about the y-axis allows certain simplifications in calculations.

An even function possesses the following properties:

• The integral over symmetric limits around zero equals twice the integral from zero to the bound: \[ \int_{-a}^{a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx \]

In contrast, an odd function satisfies \(f(-x) = -f(x)\) and its integral over symmetric limits is zero. The even nature of \(|x|\) uniquely affects the application of numerical integration rules, as it results in exact outcomes with approaches like the Midpoint Rule over symmetric intervals, while challenging polynomial-based approximations like Simpson's Rule.