Problem 35 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(f(x)\) tends to infinity at both \(a\) and \(b\), then we define $$ \int_{a}^{b} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{b} f(x) d x $$ where \(c\) is any point between \(a\) and \(b\), provided of course that both the latter integrals converge. Otherwise, we say that the given integral diverges. Use this to evaluate \(\int_{-3}^{3} \frac{x}{\sqrt{9-x^{2}}} d x\) or show that it diverges.

Step-by-Step Solution

Answer

The integral diverges.

Step 1: Identify the Nature of the Integral Limits

Examine the integral \( \int_{-3}^{3} \frac{x}{\sqrt{9-x^{2}}} d x \) to determine if there are points where the function becomes undefined or infinite. Notice that at \( x = 3 \) and \( x = -3 \), the denominator \( \sqrt{9-x^{2}} \) becomes zero, leading to potential points of infinity.

Step 2: Choose a Splitting Point

Since the function becomes infinite at both \( x = -3 \) and \( x = 3 \), select a splitting point \( c \), for example \( c = 0 \), within the interval to separate the integral into two parts: \( \int_{-3}^{0} \frac{x}{\sqrt{9-x^{2}}} dx + \int_{0}^{3} \frac{x}{\sqrt{9-x^{2}}} dx \).

Step 3: Evaluate Each Sub-Integral Separately

Evaluate the two sub-integrals \( \int_{-3}^{0} \frac{x}{\sqrt{9-x^{2}}} dx \) and \( \int_{0}^{3} \frac{x}{\sqrt{9-x^{2}}} dx \) to check for convergence. Using substitution where \( x = 3\sin(\theta) \), we find that both integrals simplify. However, symmetry and integral properties imply \( \int_{-3}^{0} ... = -\int_{0}^{3} ... \).

Step 4: Show the Integral Diverges

After calculating, since \( \int_{-3}^{0} \frac{x}{\sqrt{9-x^{2}}} dx + \int_{0}^{3} \frac{x}{\sqrt{9-x^{2}}} dx = 0 \), and if each sub-integral were separately convergent, their sum should also converge to a single finite value. The cancellation to zero, considering separate infinite limits at the boundaries, shows an issue with convergence at these points indicating divergence of the initial integral.

Key Concepts

Convergence

Convergence is a fundamental concept when handling improper integrals. An integral converges if, as we approach the bounds, the area under the curve approaches a finite value. When evaluating improper integrals, one tests for convergence by separating the integral into manageable parts, especially when the function approaches infinity.
In the given problem, the integral is split into two parts using a mid-point where the function remains finite, simplifying the assessment of convergence.

• For each sub-integral, check whether the limit, as the variable goes towards infinity or a point of discontinuity, exists and is finite.
• If each split integral results in a finite value, the overall integral is considered convergent.
• Otherwise, if it results in an infinite area under the curve, the integral diverges.

Limits of Integration

In the context of improper integrals, the limits of integration can make the function undefined. This occurs when the function exhibits infinite or discontinuous behavior at the limits.
In our original problem, the integral \( \int_{-3}^{3} \frac{x}{\sqrt{9-x^{2}}} d x \) shows that the limits -3 and 3 cause the denominator to become zero. This results in an undefined and infinite value, signifying improper limits of integration.

• Identify points within the interval that make the function undefined to determine improper limits.
• Choose a point within the interval that avoids such issues to separate the integral safely into parts.

Understanding and managing the limits correctly is key to tackling impropriety and determining whether the integral converges or diverges.

Trigonometric Substitution

Trigonometric substitution is a useful method in calculus for evaluating integrals with specific types of rational functions. This technique often helps in simplifying expressions involving square roots, such as the one in our problem.
Using trigonometric identities can transform a troublesome integral into a more standard form.

• For expressions involving terms like \( \sqrt{a^2 - x^2} \), substitute \( x = a \sin(\theta) \).
• This simplifies the radial expression and increases ease when evaluating the integral.

In this exercise, by substituting \( x = 3 \sin(\theta) \), the integral becomes simpler and avoids undefined areas in the original limits effectively. This step is crucial for revealing symmetry or solving more cumbersome integrals.