Question

a. Prove that \(\int_{-\pi}^{\pi} f(x) d x=0\) if \(f\) is odd and that \(\int_{-\pi}^{\pi} f(x) d x=2 \int_{0}^{\pi} f(x) d x\) if \(f\) is even. b. Prove that \(\frac{1}{2}[f(x)+f(-x)]\) is even and that \(\frac{1}{2}[f(x)-f(-x)]\) is odd for any function \(f .\) Note that they sum to \(f(x)\).

Short answer

a. Odd: integral is zero; Even: integral is twice from 0 to \(\pi\). b. Averages create even and odd functions.

Step-by-step solution

Define Odd and Even Functions

A function \( f(x) \) is called odd if \( f(-x) = -f(x) \) for all \( x \). It is called even if \( f(-x) = f(x) \) for all \( x \). These properties will be crucial in the proof.

Integrate Odd Function from \(-a\) to \(a\)

Consider the integral \( \int_{-a}^{a} f(x) \, dx \) for an odd function. Substitute \( u = -x \), then \( du = -dx \). The limits of integration change from \(-a\) to \(a\) to \(a\) to \(-a\), and the integral becomes \( \int_{a}^{-a} f(-u) \, (-du) = \int_{-a}^{a} f(-u) \, du = \int_{-a}^{a} -f(u) \, du = -\int_{-a}^{a} f(x) \, dx \). This leads to \( \int_{-a}^{a} f(x) \, dx = 0 \). Therefore, \( \int_{-\pi}^{\pi} f(x) \, dx = 0 \) if \( f \) is odd.

Integrate Even Function from \(-a\) to \(a\)

For an even function, substitute \( u = -x \), \( du = -dx \). The integral from \(-a\) to \(a\) becomes \( \int_{a}^{-a} f(-u) \, (-du) = \int_{-a}^{a} f(u) \, du \). Since \( f(x) = f(-x) \), \( \int_{-a}^{a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx \). Hence, for an even function, \( \int_{-\pi}^{\pi} f(x) \, dx = 2\int_{0}^{\pi} f(x) \, dx \).

Define \(g(x) = \frac{1}{2}[f(x) + f(-x)]\) and Prove It's Even

To show \( g(x) \) is even, compute \( g(-x) = \frac{1}{2}[f(-x) + f(x)] = \frac{1}{2}[f(x) + f(-x)] = g(x) \). Thus, \( g(x) \) is even.

Define \(h(x) = \frac{1}{2}[f(x) - f(-x)]\) and Prove It's Odd

For \( h(x) \), compute \( h(-x) = \frac{1}{2}[f(-x) - f(x)] = -\frac{1}{2}[f(x) - f(-x)] = -h(x) \). Therefore, \( h(x) \) is odd.

Show That Functions Add Up to Original Function

By definition, \( h(x) + g(x) = \frac{1}{2}[f(x) - f(-x)] + \frac{1}{2}[f(x) + f(-x)] = f(x) \), which confirms that these components sum to the original function \( f(x) \).

Key concepts

Odd and Even Functions

In mathematics, functions can be classified as odd or even based on how they behave with respect to negation of the variable. This distinction is important because it helps simplify integrals and understand function properties better.
For odd functions, the definition is straightforward: a function \( f(x) \) is odd if, for every \( x \), the equation \( f(-x) = -f(x) \) holds true. Graphically, this means that the function is symmetric about the origin.
An example of an odd function is \( f(x) = x^3 \). As you can verify, \( f(-x) = (-x)^3 = -x^3 = -f(x) \).

• Odd functions have this symmetry around the origin, so their graph reflects over the origin.
• When integrated over symmetric limits such as from \(-a\) to \(a\), the areas above and below the x-axis cancel out, resulting in a zero integral.

For even functions, the scenario is different: a function \( f(x) \) is even if \( f(-x) = f(x) \) for all \( x \). This results in symmetry across the y-axis.
An example of an even function is \( f(x) = x^2 \). It satisfies \( f(-x) = (-x)^2 = x^2 = f(x) \).

• Even functions exhibit symmetry across the vertical axis, meaning the curve on one side of the y-axis is a mirror image of the curve on the other side.
• This property allows the integral over symmetric limits \(-a\) to \(a\) to be calculated as \(2\int_{0}^{a} f(x) \, dx\), simplifying computations.

Integrals

Integrals are fundamental in calculus and they measure the area under a curve between two points. In the context of odd and even functions, they reveal interesting patterns when taken over symmetric limits.
For odd functions, an integral from \(-a\) to \(a\) can be shown to always equal zero. Why? Because the negative parts of the function cancel out the positive parts due to their symmetry around the origin. Essentially, this result arises from the property that if \( f \) is odd, then \( \int_{-a}^{a} f(x) \, dx = 0 \).
Even functions, conversely, allow us to exploit their symmetry around the y-axis to ease calculations. If \( f \) is even, calculating \( \int_{-a}^{a} f(x) \, dx \) can be reduced to doubling the integral from \(0\) to \(a\); thus, \( \int_{-a}^{a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx \).
This simplification is possible because the function’s behavior on the positive side is mirrored on the negative side, and the areas add up.

• Integrals help determine properties like area, displacement, and more by measuring the cumulative effect of a function over an interval.
• For symmetric limits, using properties of odd and even functions can greatly simplify computations.

Function Properties

Understanding the properties of functions is crucial in calculus as it helps break down complex problems into manageable parts. One way to do this is by decomposing functions into even and odd components.
Consider any function \( f(x) \). We can express it as a sum of an even function \( g(x) = \frac{1}{2}[f(x) + f(-x)] \) and an odd function \( h(x) = \frac{1}{2}[f(x) - f(-x)] \). This allows us to analyze each component separately with the properties we know.
The even part \( g(x) \) satisfies \( g(-x) = g(x) \), showing it’s symmetric along the y-axis.
The odd part \( h(x) \) satisfies \( h(-x) = -h(x) \), confirming its symmetry around the origin.
These decomposed functions have significant importance:

• They illustrate how any function can be viewed as having symmetries inherent to its structure.
• Working with these components can simplify integration and other operations on \( f(x) \).

Recognizing and applying these function properties can streamline the process of solving integrals and understanding the overall behavior of complex functions. By breaking functions into even and odd parts, we can often solve problems more efficiently.