Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use symmetry to help you evaluate the given integral. $$ \int_{-\pi / 2}^{\pi / 2} z \sin ^{2}\left(z^{3}\right) \cos \left(z^{3}\right) d z $$

Step-by-Step Solution

Answer

The integral evaluates to 0 due to the odd nature of the function across a symmetric interval.

Step 1: Identify Symmetry of the Function

Examine the function inside the integral. We have \(z \sin^2(z^3) \cos(z^3)\). Notice that the function comprises an odd component \(z\) and an even function \( \sin^2(z^3) \cos(z^3) \). Overall, the function \(f(z) = z \sin^2(z^3) \cos(z^3)\) is an odd function because multiplying an odd function by another function gives an odd function.

Step 2: Apply Properties of Odd Functions

Recall that the integral of an odd function over a symmetric interval about zero, like \([-a, a]\), is zero. The given integral is over the interval \([-\pi/2, \pi/2]\), which is symmetric about zero. Therefore, the integral of any odd function over this interval is zero.

Key Concepts

Symmetry in Functions

Understanding the concept of symmetry in functions can greatly simplify the evaluation of integrals. Symmetry helps identify specific properties of functions, such as whether they are odd or even. This can often lead to significant shortcuts in solving definite integrals.

When we talk about symmetry in mathematical functions, we're generally referring to two main types: even and odd function symmetry. Each of these has distinct properties that relate to how the function behaves across the y-axis in a Cartesian coordinate system.

• Even functions have mirror symmetry about the y-axis.
• Odd functions have rotational symmetry of 180 degrees about the origin.

Recognizing these symmetries when faced with integrals over symmetric intervals, such as \([-a, a]\), allows us to use these function properties to simplify our calculations, potentially even eliminating the need for complex integrations.

In our exercise, noticing the symmetry of the function over \([-\pi/2, \pi/2]\) can immediately reveal results without performing any integration calculations.

Odd and Even Functions

A crucial aspect of definite integrals is identifying whether the function in the integral is odd or even. This identification simplifies the computation of integrals over symmetric intervals.

Characteristics of Even Functions

An even function is symmetric about the y-axis. Mathematically, this can be expressed as \(f(-x) = f(x)\).

• If you test an even function over a symmetric interval \([-a, a]\), the integral will be \2 \times \int_{0}^{a} f(x)\, dx\.
• Example: \( \int_{-a}^{a} x^2 \, dx = 2 \int_{0}^{a} x^2 \, dx \).

Characteristics of Odd Functions

On the other hand, an odd function has rotational symmetry around the origin. This means \(f(-x) = -f(x)\).

• For odd functions, the integral over a symmetric interval \([-a, a]\) is zero, effectively simplifying evaluation.
• Example: \( \int_{-a}^{a} x^3 \, dx = 0 \).

In the context of the given exercise, the function \(z \, \sin^2(z^3) \, \cos(z^3)\) exhibits the properties of an odd function, leading directly to a solution of zero for the integral over \([-\pi/2, \pi/2]\).

Properties of Integrals

The properties of integrals are fundamental in simplifying the process of solving integrals, particularly definite integrals over symmetric intervals. Understanding these properties can save a lot of time and effort.

The Integral of an Odd Function

One of the most powerful properties in definite integrals is that if a function is odd, then the integral over a symmetric interval \([-a, a]\) equals zero. This applies to our function \(f(z) = z \, \sin^2(z^3) \, \cos(z^3)\).

• This property arises because the contributions from each point on the positive side of the x-axis are canceled out by the corresponding point on the negative side.

The Integral of an Even Function

Conversely, if a function is even, the definite integral from \([-a, a]\) can be computed as twice the integral from 0 to \a\.

• This results because of the mirror symmetry about the y-axis, meaning both halves of the integral contribute equally.

Using these properties efficiently reduces the need for longer calculations and provides clear insights into the behavior of the integrals involved. In solving the original exercise, identifying the function as odd directly led to applying the zero integral property, demonstrating the power of leveraging integral properties.