Problem 18 - Calculus for Biology and Medicine

Question

Determine whether each integral is convergent. If the integral is convergent, compute its value. $$ \int_{1}^{\infty} \frac{1}{x^{1 / 3}} d x $$

Step-by-Step Solution

Answer

The integral diverges, so it is not convergent.

Step 1: Identify the Type of Integral

The integral given is from 1 to infinity of the function \( \frac{1}{x^{1/3}} \). This is an improper integral because its upper limit is infinite.

Step 2: Rewrite the Integral with a Limit

To solve an improper integral, we first rewrite it using a limit. The integral becomes: \[ \lim_{t \to \infty} \int_{1}^{t} \frac{1}{x^{1/3}} \, dx \]

Step 3: Integrate the Function

Now, find the indefinite integral of \( \frac{1}{x^{1/3}} \). Rewrite the integrand as \( x^{-1/3} \) and use the power rule for integration: \[ \int x^{-1/3} \, dx = \frac{x^{2/3}}{2/3} + C = \frac{3}{2} x^{2/3} + C \]

Step 4: Evaluate the Definite Integral

Apply the limits of integration from 1 to \( t \): \[ \int_{1}^{t} \frac{1}{x^{1/3}} \, dx = \left[ \frac{3}{2} x^{2/3} \right]_{1}^{t} = \frac{3}{2} t^{2/3} - \frac{3}{2} \cdot 1^{2/3} \] Simplifying gives: \[ \int_{1}^{t} \frac{1}{x^{1/3}} \, dx = \frac{3}{2} t^{2/3} - \frac{3}{2} \]

Step 5: Compute the Limit

Now, take the limit as \( t \to \infty \): \[ \lim_{t \to \infty} \left( \frac{3}{2} t^{2/3} - \frac{3}{2} \right) = \lim_{t \to \infty} \frac{3}{2} t^{2/3} - \frac{3}{2} \]Since \( t^{2/3} \to \infty \) as \( t \to \infty \), the integral diverges.

Key Concepts

Limits of Integration

When we talk about the limits of integration, we are discussing the bounds of the integral. This tells us where the integration process starts and ends. In the context of the given problem, we are dealing with an improper integral, as indicated by one of the limits, which extends to infinity. An improper integral is one where at least one limit of integration is either infinite or the integrand becomes infinite within the interval of integration.

To handle this, we must rewrite the integral using a limit. For example, instead of integrating from 1 to infinity directly, we replace infinity with a variable, say \( t \), which we let approach infinity. This turns the problem into a limit problem:

• Original integral: \( \int_{1}^{\infty} \frac{1}{x^{1/3}} \, dx \)
• Rewrite with limit: \( \lim_{t \to \infty} \int_{1}^{t} \frac{1}{x^{1/3}} \, dx \)

The integration process is then carried out over a finite interval \([1, t]\), allowing us to evaluate it and later compute what happens as \( t \) becomes very large.

Power Rule for Integration

The power rule for integration is a handy tool in calculus, giving us a straightforward way to find antiderivatives of power functions. To use this rule, the integrand must be a power of \( x \), like \( x^n \), where \( n \) is any real number that is not \(-1\). The formula is:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

Here, \( C \) is the constant of integration, essential in indefinite integrals.

For our improper integral example, we rewrite \( \frac{1}{x^{1/3}} \) as \( x^{-1/3} \). Applying the power rule, we find:

• \( \int x^{-1/3} \, dx = \frac{x^{2/3}}{2/3} + C = \frac{3}{2} x^{2/3} + C \)

This step shows the power rule's simplicity and elegance, allowing us to transition smoothly to the next stage of solving the integral. Once we've determined the indefinite integral, we can apply the limits of integration to evaluate the definite integral.

Divergence of Integrals

The divergence of integrals is an important concept, particularly with improper integrals. It tells us whether an integral produces a finite result or not. Divergence indicates that the integral "blows up" or becomes infinite.

In the original problem, after applying the limits of integration, the integral we need to evaluate becomes:

• \( \int_{1}^{t} \frac{1}{x^{1/3}} \, dx = \frac{3}{2} t^{2/3} - \frac{3}{2} \)

To see if it converges or diverges, we look at the behavior as \( t \to \infty \). Specifically, as \( t^{2/3} \to \infty \), \( \frac{3}{2} t^{2/3} \) also goes to infinity. Consequently, the original integral \( \int_{1}^{\infty} \frac{1}{x^{1/3}} \, dx \) diverges.

Divergence in this context means there is no bounded area under the curve \( \frac{1}{x^{1/3}} \) from 1 to infinity, which is crucial for determining the behavior of the function over infinite intervals.