Problem 434 - Calculus Volume 1

Question

For \(1

Step-by-Step Solution

Answer

The limit is \( \frac{\pi}{2} \).

Step 1: Identify the Substitution

To solve the integral, notice that the expression under the square root suggests a trigonometric substitution. Use the substitution \( t = \sec(\theta) \). This will simplify the integral.

Step 2: Execute the Substitution

If \( t = \sec(\theta) \), then \( dt = \sec(\theta)\tan(\theta) d\theta \). Also, \( t^2 - 1 = \sec^2(\theta) - 1 = \tan^2(\theta) \), so \( \sqrt{t^2 - 1} = \tan(\theta) \). Substitute these into the integral.

Step 3: Simplify the Integral

The integral \( I(B) = \int_1^B \frac{dt}{t \sqrt{t^2 - 1}} \) becomes \( \int_{\theta_1}^{\theta_B} \frac{\sec(\theta) \tan(\theta) d\theta}{\sec(\theta) \tan(\theta)} \). This simplifies to \( \int_{\theta_1}^{\theta_B} d\theta \), where \( \theta_1 = 0 \) and \( \theta_B = \arccos\left(\frac{1}{B}\right) \).

Step 4: Evaluate the Integral

The integral \( \int_{\theta_1}^{\theta_B} d\theta = \theta_B - \theta_1 = \arccos\left(\frac{1}{B}\right) - 0 = \arccos\left(\frac{1}{B}\right) \).

Step 5: Compute the Limit

To find \( \lim_{B \to \infty} I(B) \), evaluate \( \lim_{B \to \infty} \arccos\left(\frac{1}{B}\right) \). As \( B \to \infty \), \( \frac{1}{B} \to 0 \), and hence \( \arccos\left(\frac{1}{B}\right) \to \arccos(0) = \frac{\pi}{2} \).

Key Concepts

Integral Calculus

Integral Calculus is an important part of calculus and deals with the process of integration. Integration helps us find the accumulation of quantities or the area under curves.
In this exercise, we have the integral \( I(B) = \int_{1}^{B} \frac{dt}{t \sqrt{t^2 - 1}} \). This integral appears complicated, but using a technique called trigonometric substitution, it can be streamlined.
Trigonometric substitution is effective when the integrand involves expressions like \( \sqrt{t^2 - 1} \), \( \sqrt{1 - t^2} \), or \( \sqrt{t^2 + 1} \). These expressions suggest the relationships of the sides of a right triangle to a trigonometric function. In our case, substituting \( t = \sec(\theta) \) helps us convert the square root and simplify the integral drastically.

• Let's look at the substitution: if \( t = \sec(\theta) \), then \( \sqrt{t^2 - 1} = \tan(\theta) \).
• This substitution simplifies the integral to \( \int d\theta \), which is straightforward to integrate.

This method is powerful for evaluating integrals that initially seem complex without substitution.

Improper Integrals

Improper Integrals occur when an integral has one or more infinite limits or the integrand becomes infinite within the limits of integration.
In our problem, the integral \( I(B) = \int_{1}^{B} \frac{dt}{t \sqrt{t^2 - 1}} \) is improper because the upper limit of integration \( B \) approaches infinity, which means we're looking at the area under the curve from 1 to infinity.
With improper integrals, we typically cannot evaluate them directly using standard integration techniques. Instead, we evaluate the integral over a finite interval \([1, B]\) and then analyze the behavior as \( B \to \infty \).

• This means first solving for \( I(B) \) and seeing if this value approaches a finite limit or not as \( B \) increases without bound.
• Improper integrals play an essential role in physics and engineering, especially when dealing with rates of change over time or space where limits can naturally extend to infinity.

Understanding their convergence or divergence tells us if the total area under a curve can be a meaningful, finite number, which is crucial in real-world applications.

Limit Evaluation

Evaluating limits is a fundamental concept in calculus, and it's especially crucial when dealing with functions that change behavior as they extend to infinity.
In this exercise, once the integral is found, we seek to determine \( \lim_{B \to \infty} I(B) \).
This involves understanding how the expression \( \arccos\left(\frac{1}{B}\right) \) behaves as \( B \) grows larger.

• As \( B \to \infty \), the fraction \( \frac{1}{B} \) approaches 0.
• The arc cosine of 0, \( \arccos(0) \), equals \( \frac{\pi}{2} \).

Therefore, the limit comes out to be \( \frac{\pi}{2} \), indicating the area under the curve from 1 to infinity is finite and equals \( \frac{\pi}{2} \).
This result demonstrates the power of calculus in determining finite results from processes extending indefinitely. Proper understanding of limits allows us to handle these infinite behaviors and derive meaningful conclusions.