Problem 49 - Calculus: Early Transcendental Functions

Question

A function \(f\) is said to be even if \(f(-x)=f(x)\) for all \(x\) A function \(f\) is said to be odd if \(f(-x)=-f(x) .\) Suppose that \(f\) is continuous for all \(x\). Show that if \(f\) is even, then \(\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x .\) Also, if \(f\) is odd, show that \(\int_{-a}^{a} f(x) d x=0\)

Step-by-Step Solution

Answer

For an even function, the integral from -a to a equals to two times the integral from 0 to a. For an odd function, the integral from -a to a equals to zero.

Step 1: Recognizing the Properties of Even and Odd Functions

An even function satisfies the condition \(f(-x) = f(x)\) for all \(x\), and an odd function satisfies the condition \(f(-x) = -f(x)\) for all \(x\). This means that even functions are symmetrical about the y-axis, while odd functions are symmetrical about the origin.

Step 2: Proving the First Statement for Even Functions

If the function \(f\) is even, then we have \(\int_{-a}^{a} f(x) dx = \int_{-a}^{0} f(x) dx + \int_{0}^{a} f(x) dx\). Due to the function being even, \(\int_{-a}^{0} f(x) dx = \int_{0}^{a} f(x) dx\), hence the integral from -a to a of f(x) dx equals to two times the integral from 0 to a, as \(\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx\).

Step 3: Proving the Second Statement for Odd Functions

If the function \(f\) is odd, then the integral \(\int_{-a}^{a} f(x) dx = \int_{-a}^{0} f(x) dx + \int_{0}^{a} f(x) dx\). But because the function is odd, \(\int_{-a}^{0} f(x) dx = -\int_{0}^{a} f(x) dx\). Hence when added, they sum to zero as \(\int_{-a}^{a} f(x) dx = 0\).

Key Concepts

Symmetry of Functions

When we talk about the symmetry of functions, we explore how a function behaves when inputs change, especially when considering negative values of the input. If a function is even, then it maintains a specific symmetry about the y-axis. This means that for every point on the graph of the function, there is a mirror image opposite to the y-axis. Think of it like a reflection in a vertical mirror placed along the y-axis. This symmetry is captured by the condition \(f(-x) = f(x)\) for all \(x\).
Odd functions have a different type of symmetry, one that originates from the origin itself. This can be thought of as a combination of both a vertical and a horizontal reflection. In formula terms, this symmetry is expressed as \(f(-x) = -f(x)\). This essentially means that rotating the graph 180 degrees around the origin results in the same graph again. Both symmetries simplify problems involving integration over symmetric intervals, as we'll see in the integral section.

Integral Properties

Integrals are powerful tools in calculus used to find areas under curves. When functions are even or odd, these integrals can be simplified significantly, using their symmetry properties.

• **Even Functions:** Given an even function where \(f(-x) = f(x)\), the integral from \(-a\) to \(a\) of \(f(x)\) is \(\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx\). This essentially indicates that the areas on either side of the y-axis are equal, enabling us to calculate the integral over just one side and then double it.
• **Odd Functions:** For odd functions with \(f(-x) = -f(x)\), we find that \(\int_{-a}^{a} f(x) \, dx = 0\). This occurs because the positive area from \(-a\) to 0 precisely cancels with the negative area from 0 to \(a\), resulting in a net zero area.

Recognizing these properties allows for faster computation and a deeper understanding of the behavior of integrals involving symmetric functions.

Continuity of Functions

Continuity of a function means there are no abrupt jumps, breaks, or holes in the graph of the function over a particular interval. Geometrically, you can imagine that a continuous function allows you to trace the curve without lifting your pen.
Continuity plays a crucial role in calculating integrals, especially when leveraging the symmetry properties of even and odd functions. For the integrals of even and odd functions over symmetric limits to hold, the function must be continuous over that interval (including endpoints). This ensures that any symmetry observed on a small scale applies uniformly across the interval.
Moreover, continuous functions guarantee that small changes in the input result in small changes in output, which is an important principle that upholds the truth of the integration properties. This smooth behavior assists in evaluating integrals and proving fundamental properties of even and odd functions on continuous intervals.