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Chapter 5: Problem 1

If \(f\) is an odd function, why is \(\int_{-a}^{a} f(x) d x=0 ?\)

Short Answer

Expert verified
Question: Prove that the integral of an odd function from -a to a is equal to 0. Answer: We can prove that the integral of an odd function, f(x), from -a to a is equal to 0 by using properties of definite integrals, specifically substitution and additivity. By substituting u = -x and applying the property of odd functions (f(-x) = -f(x)), we can simplify the integral to \(\int_{a}^{a} f(u) du\). Since the interval of integration is [a, a], this definite integral evaluates to 0, which proves that the integral of an odd function from -a to a is equal to 0.

Step by step solution

01

Define odd function property

An odd function is a function f(x) for which the condition f(-x) = -f(x) holds true for all x in its domain.
02

Understand the definite integral of an odd function

Our goal is to prove that \(\int_{-a}^{a} f(x) d x=0\) given that f is an odd function.
03

Use substitution in the integral

Let u = -x. Then, the differential \(du = -dx\). We will substitute "u" into the integral and make the necessary adjustments: $$ \int_{-a}^{a} f(x) d x = \int_{-(-a)}^{-(-a)} f(-u) \cdot (-du) $$
04

Simplify the integral with substitution

Now we will apply the property of odd functions and simplify the integral: $$ \int_{a}^{a} (-f(u))\cdot (-du) $$
05

Use the additivity property of definite integral

According to the additivity property of definite integrals, we have: $$ \int_{-a}^{a} f(x) d x = \int_{a}^{a} f(u) d u $$
06

Evaluate the definite integral

Now we will evaluate the definite integral over the interval [a, a]: $$ \int_{a}^{a} f(u) d u = 0 $$
07

Conclude

We have proved that the integral of an odd function from -a to a is equal to 0, as \(\int_{-a}^{a} f(x) d x=0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
The definite integral is a concept in calculus that represents the accumulation of quantities and computes the area under a curve over a given interval. It is a fundamental tool in analysis and is commonly expressed as \( \int_{a}^{b} f(x) \, dx \). Let鈥檚 break it down:
  • Limits of Integration: The terms \(a\) and \(b\) are the lower and upper limits of integration. They define the interval over which the function \(f(x)\) is being integrated.
  • Function: The function \(f(x)\) inside the integral sign represents the curve for which the area is being calculated between the specified limits.
  • Differential: The differential \(dx\) indicates that the integration is performed with respect to the variable \(x\).
The definite integral is evaluated over an interval and gives a specific number, representing the net area under the curve.
This net area can be positive or negative, depending on the region above or below the x-axis. So, you can think of it as the sum of infinitesimally small quantities over the interval \([a, b]\).
Substitution Method
The substitution method is a powerful technique for evaluating integrals. It is particularly useful when faced with a complex integral that can be simplified by substituting part of the equation with a single variable. Here's how it works:
  • Selection of Substitution: Choose a substitution, say \(u\), such that \(u = g(x)\) simplifies the integral. Often, this involves setting \(u\) as an inner function.
  • Differential Conversion: Compute \(du\) in terms of \(dx\). This step is crucial as it assists in changing the variables in the integral.
  • Integral Transformation: Rewrite the integral in terms of \(u\). Replace all \(x\)-variables and \(dx\) with \(u\) and \(du\).
  • Integration and Back-substitution: Integrate with respect to \(u\). Once the integral in \(u\) is evaluated, substitute back the original \(x\) values to find the solution in terms of \(x\).
This method is akin to reversing a complex integral into a more manageable form, transforming a complicated algebraic expression into a simpler one, and then reversing the transformation.
Properties of Integrals
The properties of integrals are helpful tools in solving integral problems more effectively and understanding the behavior of definite integrals. Here are some key properties:
  • Linearity: This property states that \(\int [cf(x) + dg(x)] \cdot dx = c\int f(x) \, dx + d\int g(x) \, dx\), where \(c\) and \(d\) are constants.
  • Reversal of Limits: If you swap the limits of integration, the integral's sign changes: \(\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx\).
  • Zero Width Interval: Integrating over an interval where the lower and upper limits are the same yields zero: \(\int_{a}^{a} f(x) \, dx = 0\).
  • Odd and Even Functions: For an odd function, \(f(-x) = -f(x)\), the integral over a symmetric interval centered at zero, i.e., \([-a, a]\), results in zero: \(\int_{-a}^{a} f(x) \, dx = 0\). This happens because areas on opposite sides of the y-axis cancel each other out.
  • Additivity on Intervals: You can break down a definite integral into sums over sub-intervals: \(\int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx\), where \(c\) is between \(a\) and \(b\).
These properties make it easier to compute integrals by providing insightful rules and shortcuts. Knowing these can greatly aid in simplifying and solving complex integral problems.

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