Question

Prove that if \(f\) is an odd function, then $$ \int_{-L}^{L} f(x) d x=0 $$

Short answer

Question: Prove that the integral of an odd function over a symmetric interval is equal to 0. Answer: To prove that the integral of an odd function over a symmetric interval is equal to 0, we first split the integral into two integrals over separate intervals from -L to 0 and from 0 to L. Next, we apply the property of odd functions to the integral over the interval [-L, 0] and use substitution to rewrite the first integral. Then, we reverse the limits of integration for the first integral and combine the two integrals into one. Finally, we observe that the integrand is the sum of an odd function with itself, which is always equal to zero. Therefore, the integral evaluates to zero, proving that the integral of an odd function over a symmetric interval is equal to 0.

Step-by-step solution

Write down the integral of an odd function over a symmetric interval.

Given an odd function \(f(x)\), we want to find the value of the integral over the symmetric interval \([-L, L]\): $$ \int_{-L}^{L} f(x) dx $$

Split the integral into two integrals over separate intervals from \(-L\) to \(0\) and from \(0\) to \(L\) respectively.

To simplify the problem, we can split the integral into two parts: one for the interval \([-L, 0]\) and another one for the interval \([0, L]\): $$ \int_{-L}^{L} f(x) dx = \int_{-L}^{0} f(x) dx + \int_{0}^{L} f(x) dx $$

Apply the property of odd functions to the integral over the interval \([-L, 0]\).

For the interval \([-L, 0]\), let us perform the substitution \(u = -x\). Then, \(du = - dx\). When \(x = -L\), \(u = L\), and when \(x = 0\), \(u = 0\). Using this substitution and the fact that \(f(-x) = -f(x)\), we can rewrite the first integral as: $$ \int_{-L}^{0} f(x) dx = -\int_{L}^{0} f(-u) du $$

Rewrite the integrals.

Now, reverse the limits of integration for the first integral and combine the two integrals into one: $$ -\int_{L}^{0} f(-u) du + \int_{0}^{L} f(x) dx = \int_{0}^{L}(-f(-u)+f(x))dx $$

Observe that the integrand is zero and evaluate the integral.

Now we can see that the integrand is just the sum of an odd function with itself, which is always equal to zero: $$ (-f(-u)+f(x)) = 0 $$ Since the integrand is zero, the integral evaluates to zero: $$ \int_{0}^{L}(-f(-u)+f(x))dx = 0 $$ Therefore, we have proven that if \(f\) is an odd function, then $$ \int_{-L}^{L} f(x) dx = 0 $$

Key concepts

Symmetric Interval

A symmetric interval is a range on the real number line that is equidistant from zero, extending equally in both the negative and positive directions. For instance, the interval \([-L, L]\) is symmetric about zero. This means that the interval extends from \(-L\) on the negative side to \(+L\) on the positive side.
This concept is especially important when dealing with odd functions, as it allows us to make significant simplifications.
An important property of symmetric intervals is that when integrating odd functions over them, the contributions from the negative and positive sides cancel each other out, leading to an integral of zero. This makes problems involving symmetric intervals and odd functions much easier to solve, with the symmetry playing a crucial role in simplifying calculations.

Properties of Integrals

Integrals have several important properties that make them a key tool in calculus. Understanding these properties helps in evaluating complex functions.
The linearity property states that the integral of a sum of functions is the sum of their integrals:

• \( \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \)

This allows us to split or combine integrals as needed.
The property of odd functions states that if a function \(f(x)\) is odd, then \(f(-x) = -f(x)\).
This characteristic means that when you integrate an odd function over a symmetric interval \([-L, L]\), the result is zero:

• \( \int_{-L}^{L} f(x) \, dx = 0 \)

With symmetric intervals, each negative contribution cancels out with its positive counterpart.

Substitution Method

The substitution method is a powerful tool for evaluating integrals and transforming them into simpler forms. It involves changing the variable of integration to another variable, typically to take advantage of a function's symmetry or other properties.
The basic idea is to define a substitution variable \(u = g(x)\), which simplifies the original integral. You'll also have to change the differential, using \( du = g'(x) \, dx \), and adjust the limits if the integral is definite.
For example, in an integral involving an odd function over the interval \([-L, 0]\), using the substitution \(u = -x\) allows you to express the integral in a new form:

• When \( x = -L \), \( u = L \)
• When \( x = 0 \), \( u = 0 \)

Using this substitution, the original integral over a symmetric interval can be transformed and often simplified, ultimately confirming the zero result for symmetric integrals of odd functions.