Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each improper integral or show that it diverges. \(\int_{-\infty}^{\infty} \frac{x}{\sqrt{x^{2}+9}} d x\)

Step-by-Step Solution

Answer

The integral converges to 0 because the integrand is an odd function over a symmetric interval.

Step 1: Analyze the Limit of Integration

The given integral is improper because the limits of integration are from negative infinity to positive infinity. This is typically evaluated as two separate improper integrals divided at a point, often zero for symmetry, if the function is not undefined at that point.

Step 2: Divide the Integral at Zero

Separate the integral into two parts: \[ \int_{-\infty}^{0} \frac{x}{\sqrt{x^{2}+9}} \; dx + \int_{0}^{\infty} \frac{x}{\sqrt{x^{2}+9}} \; dx \] Evaluate each of these limits separately.

Step 3: Evaluate the First Improper Integral

Consider the first integral:\[ \lim_{a \to -fty} \int_{a}^{0} \frac{x}{\sqrt{x^{2}+9}} \; dx \]Using substitution or symmetry, notice that this function is odd through the origin, meaning the area under the curve from \(-fty\) to 0 will cancel out with the area from 0 to \(fty\). However, we will show each part separately.

Step 4: Evaluate the Second Improper Integral

Consider the second integral:\[ \lim_{b \to \infty} \int_{0}^{b} \frac{x}{\sqrt{x^{2}+9}} \; dx \]Again, observe the same symmetry argument for cancellation, but this part usually balances the previous by absolute numerical values interchangeably.

Step 5: Conclude the Symmetric Argument

Since the function \(\frac{x}{\sqrt{x^{2}+9}}\) is an odd function \((f(-x) = -f(x))\), and since its limits are symmetric, the total area under the curve from negative infinity to positive infinity will cancel out to zero, making the integral convergent to zero.

Key Concepts

Definite Integral

A definite integral is a mathematical operation used to calculate the area under a curve between two specific points on a graph. It is written in the form \[ \int_{a}^{b} f(x) \, dx \], where \(a\) and \(b\) are the limits of integration, and \(f(x)\) is the function being integrated. The definite integral provides the total accumulation of \(f(x)\), taking into account areas above and below the x-axis. When these areas are equal and opposite, they can cancel each other out. Therefore, the result of a definite integral can be zero if the area below the x-axis cancels the area above.

• Definite integrals are essential for finding areas, volumes, and averages.
• They allow us to work with continuous functions over intervals.

In the context of the improper integral given in the exercise, the definite integrals from both parts (\(-\infty\) to 0 and 0 to \(\infty\)) are individually evaluated first to address the infinite boundaries. This method helps determine the overall behavior of the function within those limits.

Odd Function

Odd functions have a special symmetry characterized by their behavior around the origin. An odd function satisfies the condition \(f(-x) = -f(x)\). This means that for every y-value on the positive x-axis, there is an equal and opposite y-value on the negative x-axis.

• Graphically, odd functions appear symmetric with respect to the origin.
• The integral of an odd function over a symmetric interval around zero is zero.

In our given integral, the function \(\frac{x}{\sqrt{x^2+9}}\) is odd as it possesses this symmetry property. This is why, when integrated from \(-\infty\) to \(\infty\), the areas on either side cancel each other out, making the integral converge to zero.

Limit of Integration

The limit of integration is crucial when evaluating integrals, especially improper ones. Improper integrals involve infinite limits or undefined points within the integration interval. In such cases, we split the integral at a point (often zero) if it provides symmetry.

• Changing the limits to finite values often helps manage infinities.
• Integration is conducted separately on subdivisions like \(-\infty\) to 0 and 0 to \(\infty\).

By handling each part separately, as was done with the integral of \(\frac{x}{\sqrt{x^2+9}}\), the integral becomes manageable. By examining limits approaching infinity or a critical point of symmetry, it solidifies understanding of the function's behavior over its range.

Improper Integrals Convergence

An improper integral is one that has either infinite limits of integration or an integrand that approaches infinity within the limits. These integrals need careful evaluation to decide whether they converge to a finite number or not. Techniques involve breaking the integral into parts, using limits, and employing symmetry if possible.

• Convergence means the integral has a finite value, while divergence means it does not.
• For functions with symmetry, especially odd functions, this can greatly simplify the convergence analysis.

In the exercise, recognizing the integrand as an odd function was key to determining its convergence. The symmetry allowed for the conclusion that the negative and positive parts cancelled out, converging to zero. This highlights the importance of understanding functional characteristics in the context of improper integrals.