Problem 11 - Calculus Concepts

Question

For Activities 5 through \(16,\) evaluate the improper integral. $$ \int_{2}^{\infty} \frac{1}{\sqrt{x}} d x $$

Step-by-Step Solution

Answer

The integral diverges to infinity.

Step 1: Understand the Improper Integral

This integral is improper because it extends to infinity, so the upper limit is infinite. We will need to use limits to evaluate it.

Step 2: Set Up the Limit

Convert the integral from an improper integral to a limit: \[ \lim_{b \to \infty} \int_{2}^{b} \frac{1}{\sqrt{x}} \, dx \]

Step 3: Find the Antiderivative

The integrand is \( \frac{1}{\sqrt{x}} = x^{-1/2} \). Find the antiderivative: \[ \int x^{-1/2} \, dx = 2x^{1/2} + C \] where \( C \) is the constant of integration.

Step 4: Evaluate the Definite Integral

Evaluate the definite integral from 2 to \( b \) using the antiderivative: \[ \int_{2}^{b} \frac{1}{\sqrt{x}} \, dx = 2x^{1/2} \bigg|_{2}^{b} = 2b^{1/2} - 2(2)^{1/2} \] Simplifying, we get: \[ 2b^{1/2} - 4 \]

Step 5: Take the Limit as b Approaches Infinity

Now, evaluate the limit: \[ \lim_{b \to \infty} (2b^{1/2} - 4) \] Since \( b^{1/2} \to \infty \), \( 2b^{1/2} \to \infty \), so:\[ \lim_{b \to \infty} (2b^{1/2} - 4) = \infty \]

Step 6: Conclusion: Divergence

Since the result is \( \infty \), this improper integral diverges. The integral does not converge to a finite value.

Key Concepts

Infinite Limits

When dealing with improper integrals, one of the challenges is handling infinite limits. These integrals involve limits that extend infinitely. In the exercise provided, the integration limits stretch from a finite value (2) up to infinity. This creates a condition where traditional calculus techniques need adjustment.

To evaluate such integrals, we use a limit process.

• We reformulate the integral as a limit problem.
• Instead of integrating up to infinity directly, we integrate up to a finite boundary 'b' and then take the limit as 'b' approaches infinity.

This approach helps in systematically handling the complication of the upper limit being infinite. It breaks down the problem into manageable parts, enabling us to utilize the familiar rules of definite integration.

Antiderivatives

In calculus, finding an antiderivative is a crucial step in solving integrals. For the integral in the exercise, we're asked to find the antiderivative of the function \( \frac{1}{\sqrt{x}} \). Converting this function into some form suitable for integration is key.

The given function \( \frac{1}{\sqrt{x}} \) can be rewritten in the exponent form as \( x^{-1/2} \). To find its antiderivative, we apply basic antiderivative rules:

• We add 1 to the exponent, changing \( x^{-1/2} \) to \( x^{1/2} \).
• We then divide by the new exponent, \( \frac{1}{1/2} = 2 \).

Thus, the antiderivative of \( \frac{1}{\sqrt{x}} \) is \( 2x^{1/2} + C \), where \( C \) is the constant of integration. Understanding this step is essential in transitioning from the integrand to a usable form for evaluating limits.

Divergence of Integrals

The concept of divergence is critical when analyzing improper integrals. In our exercise, after calculating and evaluating the integral with upper limits as infinity, we encounter a result that trends towards infinity itself.

The divergence or convergence of an integral refers to whether the integral results in a finite value or grows indefinitely.

• A convergent integral safely "settles" to a finite number.
• A divergent integral, like this one, does not settle and thus tends to infinity or remains undefined.

The given integral \( \int_{2}^{\infty} \frac{1}{\sqrt{x}} \, dx \), after evaluation, leads to a limit of \( \infty \), suggesting that the integral diverges. This indicates that there isn't a finite area under the curve from 2 to infinity, shaping our understanding that some integrals, while defined in theory, span an endless value.