Problem 434 - Calculus Volume 1

Question

For \(1 < B < \infty,\) compute \(I(B)=\int_{1}^{B} \frac{d t}{t \sqrt{t^{2}-1}}\) and evaluate \(\lim _{B \rightarrow \infty} I(B), \quad\) the area under the graph of \(\frac{1}{t \sqrt{t^{2}-1}}\) over \([1, \infty)\).

Step-by-Step Solution

Answer

The limit is \( \pi/2 \).

Step 1: Understand the Integral

The integral we're trying to solve is \( I(B) = \int_{1}^{B} \frac{d t}{t \sqrt{t^{2}-1}} \). We need to find an antiderivative for the integrand and evaluate it from 1 to B.

Step 2: Variable Substitution

To handle the term \( \sqrt{t^2 - 1} \), we use trigonometric substitution. Let \( t = \sec(\theta) \). Then, \( dt = \sec(\theta) \tan(\theta) \, d\theta \) and \( \sqrt{t^2 - 1} = \sqrt{\sec^2(\theta) - 1} = \tan(\theta) \).

Step 3: Substitute and Simplify

Replace \( t \) and \( dt \) in the integral: \( I(B) = \int \frac{\sec(\theta) \tan(\theta) \, d\theta}{\sec(\theta) \tan(\theta)} \). This simplifies to \( \int d\theta \).

Step 4: Compute the Primitive Function

The integral \( \int d\theta \) equals \( \theta + C \). Thus, the antiderivative of the original function is \( \theta = \arccos(1/t) \), because \( t = \sec(\theta) \) implies \( \theta = \arccos(1/t) \).

Step 5: Evaluate the Integral from 1 to B

Substitute the bounds into the antiderivative: \( I(B) = \arccos(1/B) - \arccos(1/1) = \arccos(1/B) - 0 = \arccos(1/B) \).

Step 6: Calculate the Limit

Evaluate \( \lim_{B \to \infty} \arccos(1/B) \). As \( B \to \infty \), \( 1/B \to 0 \), and \( \arccos(0) = \pi/2 \). Thus, \( \lim_{B \to \infty} I(B) = \pi/2 \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a clever technique used to evaluate integrals that involve expressions like \( \sqrt{t^2 - 1} \). The idea is to replace the variable \( t \) with a trigonometric expression that simplifies the square root.In this particular exercise, we use the substitution \( t = \sec(\theta) \). This choice is motivated by the identity \( \sec^2(\theta) - 1 = \tan^2(\theta) \), which helps simplify the square root \( \sqrt{t^2 - 1} \) to \( \tan(\theta) \). Here's how it works step-by-step:

• Start with the substitution \( t = \sec(\theta) \).
• Calculate \( dt = \sec(\theta) \tan(\theta) \, d\theta \) using differentiation.
• Simplify the square root: \( \sqrt{t^2 - 1} = \tan(\theta) \).

This transformation allows the original integral \( I(B) \) to be expressed in terms of \( \theta \), simplifying the integration process. It effectively turns a potentially complex integral into a more manageable form.

Antiderivative

Finding the antiderivative of a function is akin to determining the original function whose derivative gives the specified function. In the context of integrals, this process is known as finding the primitive function.For the integral \( I(B) = \int \frac{d t}{t \sqrt{t^2 - 1}} \), the substitution \( t = \sec(\theta) \) simplifies the integrand to a basic form: \( \int d\theta \). Calculating this integral is straightforward:

• The antiderivative of \( d\theta \) is simply \( \theta + C \), where \( C \) is the constant of integration.
• However, we need this antiderivative in terms of the original variable \( t \). Using the relation \( \theta = \arccos(1/t) \), we derive that the antiderivative in terms of \( t \) is \( \arccos(1/t) \).

Therefore, computing the definite integral involves evaluating \( \theta \) over the given range, effectively utilizing the antiderivative.

Limit of a Function

The concept of the limit is crucial in evaluating integrals that extend over infinite intervals, like this improper integral. The goal is to determine the behavior of a function as the variable approaches a certain value.In our problem, after computing the antiderivative as \( \arccos(1/B) \), we need to assess what happens as \( B \) approaches infinity.Here's a breakdown:

• As \( B \rightarrow \infty \), the term \( 1/B \) tends to zero.
• The key function here is \( \arccos(x) \), which approaches its limit as \( x \to 0 \), resulting in \( \arccos(0) = \pi/2 \).

Thus, the limit of \( I(B) \) as \( B \rightarrow \infty \) is \( \pi/2 \), signifying the area under the curve from 1 to infinity for our original function. This limiting process is a vital tool in dealing with integrals over unbounded domains, providing a meaningful and finite result for what might first appear as an indeterminate expression.