Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use symmetry to help you evaluate the given integral. $$ \int_{-\sqrt[3]{\pi}}^{\sqrt[3]{\pi}} x^{2} \cos \left(x^{3}\right) d x $$

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Understand the Symmetry

The integral is from \(-\sqrt[3]{\pi}\) to \(\sqrt[3]{\pi}\). We need to determine if the function \(f(x) = x^2 \cos(x^3)\) is even or odd. A function \(f(x)\) is even if \(f(-x) = f(x)\). In this case, \((-x)^2 = x^2\), and \(\cos((-x)^3) = \cos(-(x^3)) = \cos(x^3)\) because the cosine function is even. Therefore, \(f(-x) = (-x)^2 \cos((-x)^3) = x^2 \cos(x^3) = f(x)\). Thus, \(f(x)\) is an even function.

Step 2: Apply Symmetry Property

For an even function, the integral over a symmetric interval around zero, \([-a, a]\), simplifies to twice the integral from \(0\) to \(a\). Therefore, we can use the symmetry property: \[ \int_{-\sqrt[3]{\pi}}^{\sqrt[3]{\pi}} x^2 \cos(x^3)\, dx = 2 \int_0^{\sqrt[3]{\pi}} x^2 \cos(x^3)\, dx. \]

Step 3: Solve the Symmetric Integral

Now, compute the integral \( \int_0^{\sqrt[3]{\pi}} x^2 \cos(x^3)\, dx \). Use the substitution method, letting \( u = x^3 \) which implies \( du = 3x^2 \, dx \) or \( x^2 \, dx = \frac{1}{3} \, du \). Change the limits of integration accordingly: when \( x = 0 \), \( u = 0^3 = 0 \), and when \( x = \sqrt[3]{\pi} \), \( u = (\sqrt[3]{\pi})^3 = \pi \). The integral becomes \[ \int_0^{\pi} \cos(u) \cdot \frac{1}{3} \, du = \frac{1}{3} \int_0^{\pi} \cos(u) \, du. \]

Step 4: Evaluate the Integral

Evaluate the integral \( \int_0^{\pi} \cos(u) \, du \). The antiderivative of \( \cos(u) \) is \( \sin(u) \). Therefore, \[ \frac{1}{3} \int_0^{\pi} \cos(u) \, du = \frac{1}{3} [\sin(u)]_0^{\pi} = \frac{1}{3} (\sin(\pi) - \sin(0)) = \frac{1}{3} (0 - 0) = 0. \]

Step 5: Conclusion

Refer back to Step 2, where the integral over the symmetric interval was doubled: \( \int_{-\sqrt[3]{\pi}}^{\sqrt[3]{\pi}} x^2 \cos(x^3) \, dx = 2 \times 0 = 0 \). Thus, the value of the original integral is 0.

Key Concepts

Symmetry in Calculus

Symmetry plays a crucial role in simplifying calculus problems, especially when computing definite integrals. In particular, functions can exhibit symmetry that makes integration easier. There are two main types of symmetry in calculus: even symmetry and odd symmetry.
An even function is symmetric about the y-axis, meaning if you fold the graph along the y-axis, the two halves match perfectly.
Odd functions, on the other hand, are symmetric about the origin. This means if you rotate the graph 180° around the origin, it looks the same.
Recognizing these symmetries can be beneficial because it allows us to apply special properties to evaluate integrals more efficiently. This aligns with the broader goal of using calculus to solve problems in a straightforward manner, leveraging inherent patterns and structures in mathematical functions.

Even and Odd Functions

Understanding whether a function is even or odd helps immensely in calculus problem solving. A function is even if for every value of x, the function's value at -x is identical to that at x, mathematically expressed as:

• Even function: \(f(-x) = f(x)\)
• Odd function: \(f(-x) = -f(x)\)

For any integral \(\int_{-a}^{a} f(x)\, dx\):

• If \(f(x)\) is an even function, then \(\int_{-a}^{a} f(x)\, dx = 2 \int_{0}^{a} f(x)\, dx\).
• If \(f(x)\) is odd, then \(\int_{-a}^{a} f(x)\, dx = 0\).

In the given exercise, the function \(f(x) = x^2 \cos(x^3)\) is even, making it much easier to evaluate the integral. Recognizing the symmetry and whether a function is even or odd can thus lead to significant simplifications.

Substitution Method

The substitution method, often a go-to technique in calculus, involves changing the variables in an integral to simplify its evaluation. When direct integration is tricky, substitution can be a lifesaver.
The process involves a few key steps:

• Select a new variable to replace a function inside the integral. In our exercise, we set \(u = x^3\).
• Determine the differential \(du\) in terms of the original variable and replace the original variable in the integral with this new differential. In this case, \(du = 3x^2\, dx\) translates to \(x^2\, dx = \frac{1}{3} du\).
• Adjust the limits of integration based on the new variable. Here, the limits change from \(x = 0\) to \(u = 0\) and from \(x = \sqrt[3]{\pi}\) to \(u = \pi\).

Ultimately, simplifying the integral using substitution can make integration accessible and often necessary to solve more complex calculus problems efficiently.

Calculus Problem Solving

To effectively tackle calculus problems, it's essential to adopt a structured approach. Calculus problem solving often requires multiple techniques, including recognizing symmetry and employing substitution.
Approaching a calculus problem typically involves:

• Analyzing the function and its properties (e.g., symmetry, domain).
• Determining the best method for integration, whether it's direct integration, substitution, or parts.
• Simplifying and breaking down the integral into manageable parts, as was done in the given exercise by applying symmetry and a strategic substitution.

Understanding the underlying principles and methods enables you to not only solve the problem at hand but also apply these strategies to varied calculus challenges. Familiarity with common functions and integration techniques is a strong asset in efficiently solving calculus problems.