Question

For \(1

Short answer

The area under the curve is \(\frac{\pi}{2}\).

Step-by-step solution

Select a Suitable Substitution

To simplify the integration, use the substitution \( t = \sec(x) \). This gives \( dt = \sec(x)\tan(x)\, dx \). Notice that when \( t = 1 \), \( x = 0 \), and when \( t = B \), \( x = \sec^{-1}(B) \).

Transform the Integral Limits

The integral \( I(B) \) now becomes \( \int_{0}^{\sec^{-1}(B)} \frac{\sec(x)\tan(x)\, dx}{\sec(x)\sqrt{\sec^2(x)-1}} \). Recall that \( \sqrt{\sec^2(x) - 1} = \tan(x) \).

Simplify the Integral

After substitution, the expression inside the integral simplifies to \( \int_{0}^{\sec^{-1}(B)} \frac{\sec(x)\tan(x)\, dx}{\sec(x)\tan(x)} = \int_{0}^{\sec^{-1}(B)} dx = x \bigg|_{0}^{\sec^{-1}(B)} \).

Evaluate the Definite Integral

Evaluate the integral to get \( I(B) = \sec^{-1}(B) - 0 = \sec^{-1}(B) \).

Compute the Limit as \( B \to \infty \)

Consider \( \lim_{B \to \infty} I(B) = \lim_{B \to \infty} \sec^{-1}(B) \). The arcsecant function, \( \sec^{-1}(B) \), approaches \( \frac{\pi}{2} \) as \( B \to \infty \).

Interpret the Result

The result \( \lim_{B \to \infty} I(B) = \frac{\pi}{2} \) represents the area under the graph from \( t=1 \) to \( t=\infty \).

Key concepts

Trigonometric Substitution

Trigonometric substitution is a method used in calculus to simplify integrals involving square roots. It is particularly useful when dealing with integrals that involve expressions like \( \sqrt{a^2 + x^2} \), \( \sqrt{a^2 - x^2} \), or \( \sqrt{x^2 - a^2} \).

In the given exercise, we substitute \( t = \sec(x) \), which is effective because \( \sec(x) = \frac{1}{\cos(x)} \) and helps us manage the expression \( \sqrt{t^2-1} \). This substitution exploits the trigonometric identity \( \sec^2(x) - 1 = \tan^2(x) \), simplifying the square root to \( \tan(x) \).

This substitution transforms the integral into one with easier limits and integrands, enabling us to solve it straightforwardly. Trigonometric substitutions can initially seem daunting, but they are a powerful technique that often make challenging integrals much more tractable.

Definite Integrals

Definite integrals represent the area under a curve between two points. Specifically, they calculate the accumulated value of a function over a certain interval. In this exercise, we evaluate \( I(B) = \int_{1}^{B} \frac{d t}{t \, \sqrt{t^2-1}} \), computing the total area under the function \( \frac{1}{t \sqrt{t^2-1}} \) from 1 to \( B \).

Through the trigonometric substitution \( t = \sec(x) \), the limits of integration change from \( t = 1 \) to \( t = B \) into \( x = 0 \) to \( x = \sec^{-1}(B) \).

The beauty of definite integrals lies in their ability to transform complex problems into computable areas and can often give insights into the behavior of functions across specific domains. It is important to remember that these integrals yield a numerical result, unlike indefinite integrals which provide a function.

Limits in Calculus

The concept of limits is fundamental in calculus, providing a foundation for derivatives, integrals, and continuity. In our specific problem, we examine the behavior of the integral \( I(B) \) as \( B \to \infty \).

Limits help us understand the end behavior of functions. In this problem, \( \lim_{B \to \infty} \sec^{-1}(B) \) approaches \( \frac{\pi}{2} \) as \( B \to \infty \). This result tells us that the area under the curve from 1 to infinity is finite, and equals \( \frac{\pi}{2} \).

Understanding limits is crucial because it allows us to analyze situations where variables approach a certain point — whether finite or infinite. In essence, limits provide a "snapshot" of what happens at extremes of a function and help cement the link between different branches of calculus.