Problem 18 - Applied Calculus

Question

17-40. Evaluate each improper integral or state that it is divergent.17-40. Evaluate each improper integral or state that it is divergent. $$ \int_{1}^{\infty} \frac{1}{\sqrt[3]{x^{4}}} d x $$

Step-by-Step Solution

Answer

The integral converges and evaluates to 3.

Step 1: Rewrite the Integrand

The integrand \( \frac{1}{\sqrt[3]{x^4}} \) can be rewritten using the properties of exponents. We know that \( \sqrt[3]{x^4} = x^{4/3} \). Thus, \( \frac{1}{\sqrt[3]{x^4}} = x^{-4/3} \).

Step 2: Set Up the Improper Integral

The integral we need to evaluate is: \[\int_{1}^{\infty} x^{-4/3} \, dx\] Since the upper limit is infinity, this is an improper integral.

Step 3: Determine Convergence or Divergence Criterion

For the integral \( \int_{1}^{\infty} x^{-p} \, dx \), it converges if \( p > 1 \) and diverges otherwise. Here, \( p = \frac{4}{3} > 1 \), so the integral converges.

Step 4: Evaluate the Integral

Evaluate: \[\int x^{-4/3} \, dx = \frac{x^{-4/3 + 1}}{-4/3 + 1} + C = \frac{x^{-1/3}}{-1/3} + C = -3x^{-1/3} + C\]

Step 5: Apply Limits Using the Limit Process

Evaluate the limit as \( b \to \infty \): \[\lim_{b \to \infty} \left( -3x^{-1/3} \right) \bigg|_{1}^{b} = \lim_{b \to \infty} \left( -3b^{-1/3} + 3(1)^{-1/3} \right)\] The term \( 3(1)^{-1/3} = 3 \).

Step 6: Calculate the Limit

As \( b \to \infty \), \( b^{-1/3} \to 0 \). Therefore: \[\lim_{b \to \infty} \left( -3b^{-1/3} + 3 \right) = -3(0) + 3 = 3\] Thus, the integral evaluates to 3.

Key Concepts

Convergence of Integrals

When evaluating improper integrals, understanding convergence is key. An improper integral occurs when the limits of integration are not finite or if the integrand becomes infinite within the limits. In this exercise, the upper limit of the integral is infinity, which makes it an improper integral.
For an improper integral of the form \( \int_{1}^{\infty} x^{-p} \, dx \), determining convergence depends on the value of \( p \).

• If \( p > 1 \), the integral converges.
• If \( p \leq 1 \), the integral diverges.

In our example, because \( p = \frac{4}{3} > 1 \), the integral converges. This means that despite integrating over an infinite interval, the integral evaluates to a finite number. Such analyses are fundamental when deciding whether an improper integral is work-worthy for further evaluation.

Exponents in Calculus

Exponents play an important role in calculus as they simplify complex expressions and integrals. In the original exercise, the expression \( \frac{1}{\sqrt[3]{x^4}} \) may look daunting at first, but using properties of exponents, it becomes much more manageable.
The cube root and the exponent on \( x \) must be combined correctly:

• The cube root of \( x^4 \) is \( x^{4/3} \).
• Therefore, \( \frac{1}{\sqrt[3]{x^4}} \) simplifies to \( x^{-4/3} \).

Converting expressions to exponential form is particularly helpful in calculus as it makes differentiation and integration straightforward using power rules. With such simplifications, you can easily apply standard integral evaluation techniques.

Limit Evaluation in Calculus

Evaluating limits is an essential skill in calculus, especially for improper integrals. Limits allow us to handle infinity within integrals by providing a finite value as a result.
When evaluating an improper integral from 1 to infinity, as in this exercise, you compute:

• First evaluate the antiderivative, here, \(-3x^{-1/3}\).
• Apply the limits of the integral: compute from 1 to \( b \), then let \( b \to \infty \).

The focus of limit evaluation is recognizing how expressions behave as they tend towards infinity. In this example, as \( b^{-1/3} \) approaches zero, the evaluating leads to a final calculated result of 3. Properly handling limits ensures accurate evaluations and conclusions, particularly in calculus problems involving infinity.