Problem 49 - Thomas Calculus

Question

In Exercises \(35-64\) , use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$ \int_{2}^{\infty} \frac{d v}{\sqrt{v-1}} $$

Step-by-Step Solution

Answer

The integral is divergent, as it evaluates to infinity.

Step 1: Rewrite the Integral with a Substitution

To evaluate the improper integral \( \int_{2}^{\infty} \frac{dv}{\sqrt{v-1}} \), we will make a substitution to simplify the expression. Let \( u = \sqrt{v-1} \). Then \( u^2 = v - 1 \) or \( v = u^2 + 1 \). We compute the derivative: \( dv = 2u \, du \). Change the limits: when \( v = 2 \), \( u = \sqrt{2-1} = 1 \) and as \( v \to \infty \), \( u \to \infty \). The integral becomes \( \int_1^{\infty} \frac{2u \, du}{u} = \int_1^{\infty} 2 \, du \).

Step 2: Evaluate the Simplified Integral

The integral is now \( \int_1^{\infty} 2 \, du \), which is simpler. Evaluate this improper integral by taking the limit: \[ \lim_{t \to \infty} \int_1^{t} 2 \, du = \lim_{t \to \infty} \left[ 2u \right]_1^{t} = \lim_{t \to \infty} (2t - 2). \]

Step 3: Determine Convergence or Divergence

Since \( \lim_{t \to \infty} (2t - 2) = \infty \), the integral diverges. Therefore, the original integral \( \int_{2}^{\infty} \frac{dv}{\sqrt{v-1}} \) is divergent.

Key Concepts

Integration Techniques

Integration techniques are essential tools in calculus used to evaluate integrals, especially when they seem complex or cumbersome at first glance. One key technique often employed is substitution, which simplifies the integral by transforming it into a more familiar or basic form.
In the exercise provided, the integral \( \int_{2}^{\infty} \frac{d v}{\sqrt{v-1}} \) might appear initially complicated. By employing the substitution method, let \( u = \sqrt{v-1} \), which gives \( u^2 = v-1 \) and consequently \( v = u^2 + 1 \). This substitution streamlines the integral, transforming it into a simpler form:

• The limits change from \( v = 2 \) to \( u = 1 \) and as \( v \to \infty \), \( u \to \infty \).
• The integral simplifies to \( \int_1^{\infty} 2 \, du \).

The resulting integral is more straightforward, allowing for easier computation. Hence, integration techniques, particularly substitution, are valuable for tackling challenging integrals efficiently.

Direct Comparison Test

The Direct Comparison Test is a method used to determine the convergence or divergence of an improper integral by comparing it with another integral whose behavior is known.
Suppose you have an improper integral \( \int_{a}^{\infty} f(x) \, dx \). To apply the Direct Comparison Test, find a function \( g(x) \) such that:

• \( 0 \leq f(x) \leq g(x) \) for all \( x \geq a \).
• If \( \int_{a}^{\infty} g(x) \, dx \) converges, then \( \int_{a}^{\infty} f(x) \, dx \) also converges.
• If \( \int_{a}^{\infty} g(x) \, dx \) diverges, then \( \int_{a}^{\infty} f(x) \, dx \) also diverges.

This test is straightforward but requires finding an appropriate comparison function. For the example integral \( \int_{2}^{\infty} \frac{dv}{\sqrt{v-1}} \), we might compare it with simpler functions like \( \frac{1}{\sqrt{v}} \), depending on the context, to test for divergence or convergence.

Limit Comparison Test

The Limit Comparison Test is another method to assess the convergence of improper integrals and offers an advantage when direct comparison is less apparent.
To use this test with an integral \( \int_{a}^{\infty} f(x) \, dx \), you need another function \( g(x) \) such that both are positive for \( x \geq a \). The key steps are:

• Compute the limit \( L = \lim_{x \to \infty} \frac{f(x)}{g(x)} \).
• If \( L \) is a positive finite number, then both \( \int_{a}^{\infty} f(x) \, dx \) and \( \int_{a}^{\infty} g(x) \, dx \) have the same behavior; both converge or both diverge.

This test is helpful when you suspect similar asymptotic behavior between two functions but cannot use a straightforward direct comparison. For example, if examining \( \int_{2}^{\infty} \frac{dv}{\sqrt{v-1}} \), selecting a suitable \( g(x) \) to compare with can aid in deciding if the given integral converges or diverges.