Problem 54 - 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. $$\int_{-a}^{a} f(p(x)) d x$$

Step-by-Step Solution

Answer

Answer: The simplified form of the integral is \(2 \int_{0}^{a} f(p(x)) dx\).

Step 1: Determine if the integrand is even or odd

To check if the integrand is even or odd, we will evaluate the integrand function, \(f(p(x))\), at \(-x\) : $$ f(p(-x))$$ Since \(f\) is an even function: $$f(p(-x))=f(-p(x))$$ And since \(p(x)\) is an odd function: $$f(-p(x))=f(p(x))$$ Since \(f(p(-x))=f(p(x))\), the integrand is an even function.

Step 2: Evaluate the integral or simplify it

Since the integrand is even, we can use the property of even functions to simplify the integral: $$\int_{-a}^{a} f(p(x)) dx = 2 \int_{0}^{a} f(p(x)) dx$$ Now, we simplify the integral without evaluating it. The final simplified integral is: $$2 \int_{0}^{a} f(p(x)) dx$$

Key Concepts

Properties of Even and Odd Functions

Understanding the properties of even and odd functions is fundamental when working with integrals. An even function is characterized by symmetry about the y-axis, which means that for any point on the function, its reflection across the y-axis will also be on the function. Mathematically, a function f(x) is even if for all x in the function's domain, f(x) = f(-x). This type of symmetry enables several simplifications in calculus, particularly in integration.

On the other hand, an odd function exhibits rotational symmetry around the origin. For such functions, f(-x) = -f(x) holds true for all x. Odd functions have the intriguing property that their integral over a symmetric interval around the origin is zero. This is because the areas bounded by the function and the x-axis on either side of the y-axis cancel each other out. Recognizing these properties aids in anticipating the behavior of integrals involving even and odd functions.

Simplifying Integrals

When dealing with integrals, particularly definite integrals, the goal is often to simplify the expression to make it easier to evaluate. This can involve a variety of techniques such as u-substitution, integration by parts, or taking advantage of function properties, as with even and odd functions.

The simplification process is greatly aided by recognizing when an integrand falls into one of these categories. As seen in the provided step by step solution, knowing that f and p were even and odd functions respectively allowed for the simplification of the integral without actually evaluating it. Simplifying an integral can turn a seemingly daunting task into a straightforward calculation and is a key skill in the study of calculus.

Symmetry in Integration

Symmetry plays a crucial role in the integration of functions, especially when the domain is symmetric about the origin. For an even function, the integral over a symmetric interval can be simplified because the area under the curve from zero to a is exactly the same as the area from zero to -a. This property enables us to multiply the integral from zero to a by two, thus reducing the computational effort.

In contrast, as mentioned earlier, the integral of an odd function over a symmetric interval is zero due to the cancellation of areas. This property can lead to significant simplifications, especially in complex integral problems. For example, if an integrand consists of several terms and some are odd functions, those terms can be disregarded in the computation of the definite integral over a symmetric interval, as their contribution will be zero.