Problem 84 - Calculus Single Variable

Question

Let \(f:[-a, a] \rightarrow \mathbb{R}\) be continuous and odd \((f(-x)=\) \(-f(x))\). Show that $$ \int_{-a}^{a} f(x) d x=0 $$

Step-by-Step Solution

Answer

The integral of an odd function over a symmetric interval around zero is zero.

Step 1: Understand the definition of an odd function

A function is called odd if it satisfies the property \( f(-x) = -f(x) \) for all \( x \). In our case, this function is given as continuous and odd over the interval \([-a, a]\).

Step 2: Evaluate the integral by splitting the domain

We will split the integral from \(-a\) to \(a\) into two parts: \( \int_{-a}^{0} f(x)\,dx \) and \( \int_{0}^{a} f(x)\,dx \). Therefore, the integral becomes:\[\int_{-a}^{a} f(x)\,dx = \int_{-a}^{0} f(x)\,dx + \int_{0}^{a} f(x)\,dx\]

Step 3: Substitute using the odd function property

Apply the property of odd functions to the first part of the integral. For a function \(f\) that is odd, \( f(-x) = -f(x) \).\[\int_{-a}^{0} f(x)\,dx = \int_{a}^{0} f(-u) \, du = \int_{a}^{0} -f(u) \, du\]

Step 4: Change the limits of integration

Since we are changing variable substitution from \(x\) to \(u\), change the limits of the integral from \(-a\) to 0 to 0 to \(a\): \[\int_{0}^{a} -f(u)\, du = -\int_{0}^{a} f(u)\, du\]

Step 5: Add the modified integrals

Combine the two parts of the integrals. The first is from Step 3 and the second is an unchanged integral from Step 2:\[\int_{-a}^{a} f(x)\,dx = \int_{0}^{a} f(x)\,dx + \int_{0}^{a} -f(x)\,dx\]This simplifies to:\[\int_{0}^{a} f(x)\,dx - \int_{0}^{a} f(x)\,dx = 0\]

Step 6: Conclude

The two integrals from Step 5 cancel each other because they are equal in magnitude but opposite in sign. Hence, \[\int_{-a}^{a} f(x)\,dx = 0.\] This concludes the proof that the integral of an odd function over a symmetric interval around zero is zero.

Key Concepts

Continuous Function

A continuous function is a fundamental concept in calculus and analysis. It means that the function doesn't have any breaks, jumps, or holes in its domain. For a function to be continuous, you should be able to draw its graph without lifting your pencil from the paper. This idea is crucial for integral calculus because it ensures that the function behaves predictably as you move from one point to another.

• Intuitively: If every tiny movement along the x-axis results in a small change in the function's output, the function is continuous.
• Mathematically: A function \( f(x) \) is continuous at a point \( c \) if \( \lim_{x \to c} f(x) = f(c) \).

When we deal with integrals, like the definite integral we have in the problem, assuming the function is continuous over the interval helps in evaluating the area under the curve without any unexpected complications.

Definite Integral

The definite integral is a powerful tool in calculus for calculating the accumulation of quantities, such as areas under curves over an interval from \(a\) to \(b\). In simple terms, it measures the total accumulation of a quantity, considering both magnitude and direction, over a specified interval.

• Notation: \( \int_{a}^{b} f(x)\, dx \) where \(a\) and \(b\) are the limits of integration and \(f(x)\) is the function being integrated.
• Geometric Interpretation: It represents the net area between the graph of \(f(x)\) and the x-axis, considering parts below the x-axis as negative contributions.

For our particular problem, the definite integral of the continuous and odd function over a symmetric interval results in zero. This occurs because the positive and negative areas cancel each other out, due to the symmetrical properties of odd functions.

Properties of Odd Functions

Understanding odd functions is key to solving problems involving symmetry in calculus. A function \( f(x) \) is defined as odd if for every point \(x\) in its domain, the function satisfies the condition \( f(-x) = -f(x) \). This results in a symmetrical graph about the origin.

• Visual Representation: If you flip the graph of the function over the y-axis and then over the x-axis, it looks the same.
• Effect on Integration: For odd functions integrated over symmetrical limits \([-a, a]\), the areas from \(-a\) to 0 and from 0 to \(a\) are equal in size but have opposite signs, leading to the integral's net value being zero.

These properties make odd functions particularly interesting in integral calculus because they allow simplifications of otherwise complex calculations.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with integration, two core ideas in calculus. It serves as a bridge connecting the evaluation of an integral with antiderivatives.This theorem has two main parts:

• First Part: If \(f\) is continuous over \([a, b]\) and \(F\) is an antiderivative of \(f\) (meaning \(F'(x) = f(x)\)), then \(\int_{a}^{b} f(x)\, dx = F(b) - F(a)\).
• Second Part: It states that if \(f\) is a continuous function on an interval, \( x \mapsto \int_{a}^{x} f(t)\, dt \) is an antiderivative of \(f\).

In relation to our problem, while the direct application of the Fundamental Theorem of Calculus was not used explicitly, understanding how integrals and antiderivatives are connected helps deepen your comprehension. When you found that the integral of our odd function over a symmetric interval results in zero, it's a practical confirmation of these concepts in action.