Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use symmetry to help you evaluate the given integral. $$ \int_{-1}^{1} \frac{x^{3}}{\left(1+x^{2}\right)^{4}} d x $$

Step-by-Step Solution

Answer

The integral is 0, since the function is odd and symmetric about the y-axis.

Step 1: Identify the Symmetric Property

First, we need to check whether the function inside the integral is symmetric. A function is considered odd if \( f(-x) = -f(x) \) for all \( x \). Substitute \( -x \) into the function inside the integral, \( f(x) = \frac{x^3}{(1+x^2)^4} \), to get \( f(-x) = \frac{(-x)^3}{(1+(-x)^2)^4} = \frac{-x^3}{(1+x^2)^4} = -f(x) \). Therefore, the function is odd.

Step 2: Use Properties of Odd Functions

If a function is odd, then its integral over a symmetric interval around zero is zero. Since \( f(x) \) is odd and we are integrating from \(-1\) to \(1\), the integral \( \int_{-1}^{1} f(x) \, dx = 0 \).

Key Concepts

Symmetric Property

When dealing with integrals, the symmetric property can greatly simplify your work if approached right. Essentially, it helps us understand how certain functions behave around the y-axis or any vertical line. For a function to be symmetric, it must possess either even or odd symmetry.

- **Even Symmetry**: This occurs when the function satisfies the condition \( f(-x) = f(x) \). Graphically, this means the left side of the function mirrors the right side across the y-axis.
- **Odd Symmetry**: This happens when \( f(-x) = -f(x) \). In terms of graphs, each point \( (x, y) \) is reflected as \((-x, -y) \), giving the function a rotatable symmetry about the origin.

Translating these properties into the realm of definite integrals, these symmetries have specific ramifications that can significantly condense our calculations.

Odd Function

An odd function is fascinating in the context of definite integrals, especially when integrated over symmetric intervals. As stated before, the defining characteristic of an odd function is \( f(-x) = -f(x) \). This property is a game changer when it comes to calculating integrals.

When evaluating an integral over a symmetric interval, such as \([-a, a]\), the integral of an odd function becomes zero:

• This happens because the areas under the curve on the positive side of the y-axis cancel out the areas on the negative side.
• Mathematically, it simplifies as \( \int_{-a}^{a} f(x) \, dx = 0 \) for any odd function.

In the example provided, the integral from \(-1\) to \(+1\) of the function \( \frac{x^3}{(1+x^2)^4} \) confirms this principle, as we've already shown it is an odd function.

By recognizing the inherently symmetric property of odd functions, we drastically reduce the complexity of solving certain definite integrals.

Definite Integrals

Definite integrals are a cornerstone concept in calculus, crucial for calculating the actual numerical value of the integral of a function between two limits. Unlike indefinite integrals, which involve a constant of integration, definite integrals give us a specific bounded value.

When evaluating definite integrals, especially on symmetric intervals, understanding the properties of the function you are integrating, like symmetry, can be exceptionally advantageous. With definite integrals, particularly when dealing with functions that exhibit odd or even symmetry, simplifications emerge.

• If the function is odd and is being integrated over a symmetric interval, the integral evaluates to zero.
• If the function is even, integrate across each half and multiply by two, or simply integrate over the entire interval as usual.

The definite integral in our exercise is simplified elegantly by recognizing the symmetry properties and relocating focus towards the evaluation over the given limits. This is why calculus often challenges us to see beyond the numbers and recognize underlying patterns and properties.