Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\int_{-\sqrt[3]{\pi}}^{\sqrt[3]{\pi}} x^{2} \cos \left(x^{3}\right) d x\)

Step-by-Step Solution

Answer

The integral evaluates to 0 due to the oddness of the integrand over a symmetric interval.

Step 1: Understanding the Problem

We are tasked with evaluating the integral \( \int_{-\sqrt[3]{\pi}}^{\sqrt[3]{\pi}} x^{2} \cos(x^{3}) \, dx \). Importantly, notice that the function involves \( x^2 \) which is even and \( \cos(x^3) \) which is odd, overall making the integrand an odd function over a symmetric interval \([-a, a]\).

Step 2: Recognize Symmetry in the Integrand

An integrand \( f(x) \) is odd if \( f(-x) = -f(x) \). For our function, substitute \( -x \) in: \((-x)^2 \cos((-x)^3) = x^2 \cos(-x^3) = x^2 (-\cos(x^3)) = -x^2 \cos(x^3)\), confirming that it is odd.

Step 3: Apply the Property of Odd Functions

For an odd function \( f(x) \) over the interval \([-a, a]\), the integral evaluates to zero: \( \int_{-a}^{a} f(x) \, dx = 0 \). Since \( x^2 \cos(x^3) \) is odd and the limits are symmetric about zero, the integral evaluates directly to zero.

Key Concepts

Odd Functions

In mathematics, an odd function refers to a function where for every value of the input, the negative of that input results in the negative of the output. Formally, if \( f(x) \) is an odd function, then \( f(-x) = -f(x) \). This property of odd functions is significant because it leads to a remarkable fact about their integrals over symmetric intervals. For example, consider a function that looks like a mirror reflection along the y-axis; this symmetry is what defines the odd function. The integrand of the integral can display this property if you substitute \(-x\) into the function and observe that it yields the negative of the original function. Hence, if you encounter such a scenario in an integral, it suggests a unique result when the limits are symmetric about zero.

Symmetric Interval

A symmetric interval is a range of integration equally spaced around zero, typically expressed as \([-a, a]\). This type of interval is particularly useful when dealing with functions that exhibit certain symmetries, such as being odd or even. When integrating over a symmetric interval, we can apply properties of these functions to simplify the calculation.
The symmetry here essentially means that if you imagine folding the interval in half at the origin, the two halves would exactly overlap. This is why, when integrating an odd function over a symmetric interval, the integral evaluates to zero. The positive and negative halves of the interval perfectly cancel each other out. This principle simplifies calculations significantly for certain functions, making symmetric intervals a handy tool in integration.

Integral Evaluation

Integral evaluation is the process used to calculate the area under the curve of a function between two limits on the x-axis. Often, this involves finding the antiderivative of the function and substituting the upper and lower bounds into it. However, when dealing with specific types of functions, like odd functions over symmetric intervals, we can use special properties to immediately determine the result.
For example, when you have an odd function over a symmetric interval, you know the integral will evaluate to zero, eliminating the need for complex calculus. This kind of insight not only saves time but helps build a deeper understanding of function properties and their practical implications in calculus.
Hence, integral evaluation isn't just about doing the calculations; it's also about recognizing when you can apply mathematical properties for an easier and more efficient solution.

Cosine Function

The cosine function, \( \cos(x) \), is a fundamental trigonometric function that is periodic and even, creating a wave-like pattern as it repeats its values in regular intervals. The property of being even means that \( \cos(-x) = \cos(x) \), which is integral to solving many calculus problems involving symmetry.
However, when combined with other functions like powers of x, the overall behavior of the function could change. In the original exercise, the problem involved \( x^3 \) within the cosine function, affecting the symmetry of the overall integrand. While \( x^2 \) is an even function, \( \cos(x^3) \) behaves like an odd function when combined as \( x^2 \cos(x^3) \). This is because when \(-x\) is substituted into the composite function, the result negates the original function, thus displaying the odd function behavior. Recognizing such intricacies is key to understanding and solving integrals efficiently.