Problem 19 - Calculus Concepts

Question

For Activities 17 through \(22,\) use algebraic manipulation or integration by substitution as well as limits to evaluate the improper integral. $$ \int_{2}^{\infty} \frac{x^{3}}{x^{4}+1} d x $$

Step-by-Step Solution

Answer

The integral diverges to infinity.

Step 1: Identify the Type of Improper Integral

The integral \( \int_{2}^{\infty} \frac{x^{3}}{x^{4}+1} dx \) is improper because the upper limit is infinite. We'll need to evaluate this integral by taking a limit.

Step 2: Set Up the Integral with a Limit

Replace the upper limit of infinity with a variable, say \( b \), and then consider the limit as \( b \) approaches infinity. \[ \lim_{b \to \infty} \int_{2}^{b} \frac{x^{3}}{x^{4}+1} dx \]

Step 3: Use Substitution for Integration

Use the substitution \( u = x^4 + 1 \), thus \( du = 4x^3 dx \). Rearrange this to find \( x^3 dx = \frac{1}{4} du \). This simplifies the integral: \[ \int \frac{x^3}{x^4+1} dx = \frac{1}{4} \int \frac{1}{u} du \]

Step 4: Integrate and Apply Limits

Integrate to find:\[ \frac{1}{4} \ln |u| + C \]Substitute back \( u = x^4 + 1 \) and evaluate between \( 2 \) and \( b \):\[ \frac{1}{4} \left( \ln |b^4 + 1| - \ln |2^4 + 1| \right) \]

Step 5: Evaluate the Limit as b Approaches Infinity

Evaluate the limit:\[ \lim_{b \to \infty} \frac{1}{4} \ln |b^4 + 1| = \infty \]Since \( \ln |b^4 + 1| \to \infty \) as \( b \to \infty \), the whole expression diverges.

Key Concepts

Limit Evaluation

In calculus, evaluating an improper integral often involves taking a limit. This is especially the case when the integral has an infinite boundary or includes an unbounded function within its limits. For our specific problem, the integral extends from 2 to infinity. Thus, the first step is to reformulate the problem with a variable boundary, replacing the infinite limit with a variable, say \( b \). We then re-evaluate the integral as \( b \) approaches infinity. This changes our original expression to:

• \( \lim_{b \to \infty} \int_{2}^{b} \frac{x^3}{x^4+1} \, dx \)

To solve this, you need to calculate the integral for a variable upper boundary \( b \), and then examine how the result behaves as \( b \) tends towards infinity. If the limit exists and is finite, the integral is said to converge and evaluate to this limit. Conversely, if the limit does not exist or is infinite, the integral diverges. As we will later see, this particular integral turns out to be divergent.

Integration by Substitution

Integration by substitution is a powerful technique used to simplify integrals. It transforms the integral into a simpler form, which can be integrated easily. In this problem, the substitution \( u = x^4 + 1 \) is used. From this, we find that \( du = 4x^3 \, dx \), and rearranging gives us:

• \( x^3 \, dx = \frac{1}{4} \, du \)

Using these substitutions, the original integral changes to a much simpler form:

• \( \frac{1}{4} \int \frac{1}{u} \, du \)

This simplification is crucial as it transforms a complex algebraic expression into a straightforward logarithmic integral. The integral of \( \frac{1}{u} \) is \( \ln |u| \), which gives:

• \( \frac{1}{4} \ln |u| + C \)

Once integrated, we substitute back \( u = x^4 + 1 \) to express the integral in terms of the original variable \( x \). This back-substitution allows us to fully evaluate and contextualize the result of the integral.

Divergent Integrals

An integral is termed divergent if it does not settle to a finite number. When evaluating our example integral, even after successfully integrating the expression and substituting back the original limits, we encounter:

• \( \lim_{b \to \infty} \frac{1}{4} \left( \ln |b^4 + 1| - \ln |2^4 + 1| \right) \)

Upon further evaluation of \( \ln |b^4 + 1| \) as \( b \) approaches infinity, it becomes evident that \( \ln |b^4 + 1| \) also approaches infinity. This leads to the conclusion that the entire expression is divergent as it does not converge to a finite limit. In practical terms, divergent integrals are considered undefined in a standard calculus sense because they do not represent a bounded area under a curve or a finite quantity. Thus, the result of our evaluated improper integral reaches infinity, verifying its divergence.