Problem 55 - Calculus Early Transcendentals

Question

Prove that the integrand is either even or odd. Then give the value of the integral or show how it can be simplified. Assume that \(f\) and \(g\) are even functions and \(p\) and \(q\) are odd functions. $$\int_{-a}^{a} p(g(x)) d x$$

Step-by-Step Solution

Answer

Answer: The given integrand, \(p(g(x))\), is an even function. The integral can be simplified using the property of even functions: \(\int_{-a}^{a} p(g(x)) \,dx = 2\int_{0}^{a} p(g(x)) \,dx\). Without more information about the functions \(p\) and \(g\), we cannot evaluate the integral further.

Step 1: Understanding even and odd functions

An even function is a function \(f(x)\) that satisfies the condition \(f(-x) = f(x)\) for all \(x\) in its domain. An odd function is a function \(g(x)\) that satisfies the condition \(g(-x) = -g(x)\) for all \(x\) in its domain. It's given that \(f\) and \(g\) are even functions, so they satisfy: $$f(-x) = f(x)$$ $$g(-x) = g(x)$$ And \(p\) and \(q\) are odd functions, so they satisfy: $$p(-x) = -p(x)$$ $$q(-x) = -q(x)$$

Step 2: Check if the given integrand is even or odd

Let's evaluate the integrand, \(p(g(x))\), at \(-x\): $$p(g(-x)) = p(g(x))$$ Since \(g\) is even, we can see that the function \(p(g(x))\) is also an even function, as it satisfies the condition for even functions.

Step 3: Find the value of the integral or simplify it

Since the integrand is even, we can use the property of even functions to rewrite the integral as: $$\int_{-a}^{a} p(g(x)) \,dx = 2\int_{0}^{a} p(g(x)) \,dx$$ This integral has been simplified. Based on the given information and the analysis, we cannot evaluate this integral further without knowing more about the functions \(p\) and \(g\).

Key Concepts

Integration

Integration is a fundamental concept in calculus, representing the process of finding the area under a curve. Imagine you have a crumpled piece of paper; you can smooth it out and measure its area. That's similar to what integration does with functions and curves.
When dealing with integration of a function like \( \int_{-a}^{a} p(g(x)) \, dx \), you want to determine the total area between the curve \( p(g(x)) \) and the axis from \(-a\) to \(a\).
This process might simplify depending on the nature of the function, such as whether it's even or odd.

• An even function, being symmetric about the y-axis, allows you to simplify the integral by calculating only from 0 to \(a\) and then doubling the result.
• For odd functions, the integral from \(-a\) to \(a\) equals zero because the areas on both sides of the y-axis cancel each other out.

Symmetry of Functions

The symmetry of functions plays a significant role in determining how we can simplify integrals. Symmetry can be visualized as the mirroring of a function, making it very powerful when working with definite integrals over symmetric intervals like \([-a, a]\).

• Even functions: An even function satisfies \(f(-x) = f(x)\). This symmetry about the y-axis means that we only need the data from one side (either positive or negative x-values) to understand the whole curve.
  For example, if a function \(g\) is even, then \(g(-x) = g(x)\) for all \(x\), meaning that \(g\)'s output does not change if you input opposite x-values.
• Odd functions: An odd function satisfies \(f(-x) = -f(x)\), resulting in rotational symmetry about the origin (180 degrees). Here, if \(p\) is an odd function, its values for \(-x\) are simply the negative of its values for \(x\).

Properties of Definite Integrals

The properties of definite integrals provide the framework through which integrals can be manipulated or simplified, especially when dealing with functions that exhibit symmetry. These properties are essential tools that help in breaking down complex integrals.

• Even Function Property: If a function \(f(x)\) is even, then the integral from \(-a\) to \(a\) can be rewritten as \(2 \cdot \int_{0}^{a} f(x) \, dx\). This saves time and effort by reducing the interval over which you must compute the integral.
• Odd Function Property: Likewise, if \(p(x)\) is an odd function, then \(\int_{-a}^{a} p(x) \, dx = 0\). Any area calculated on the negative side of the y-axis will perfectly cancel out with the positive side.

In the given exercise, we utilized the even function property to simplify \(\int_{-a}^{a} p(g(x)) \, dx\) into \(2 \int_{0}^{a} p(g(x)) \, dx\). Without specific details about \(p\) and \(g\), we can't reach further numerical result, but understanding these properties aids in simplifying integrals substantially.