Problem 433 - Calculus Volume 1

Question

For \(A>0, \quad\) compute \(I(A)=\int_{-A}^{A} \frac{d t}{1+t^{2}}\) and evaluate \(\lim _{a \rightarrow \infty} I(A),\) the area under the graph of \(\frac{1}{1+t^{2}}\) on \([-\infty, \infty]\)

Step-by-Step Solution

Answer

The area is \( \pi \).

Step 1: Understanding the Functions and Limits

We need to evaluate the function \( I(A) = \int_{-A}^{A} \frac{dt}{1+t^2} \). This integral computes the area under the curve of the function \( \frac{1}{1+t^2} \) from \(-A\) to \(A\). The goal is to determine what this area approaches as \( A \) goes to infinity.

Step 2: Substitution for Integral of Inverse Trigonometric Function

The integral \( \int \frac{1}{1+t^2} \, dt \) is known to equal \( \tan^{-1}(t) + C \), where \( C \) is the constant of integration. This result utilizes the fact that the derivative of \( \tan^{-1}(t) \) is \( \frac{1}{1+t^2} \).

Step 3: Evaluation of the Definite Integral

Applying the antiderivative \( \tan^{-1}(t) \) to the limits of the definite integral from \(-A\) to \(A\), we get: \[ I(A) = \left[ \tan^{-1}(t) \right]_{-A}^{A} = \tan^{-1}(A) - \tan^{-1}(-A). \]

Step 4: Simplifying Using Properties of Arctan

Since \( \tan^{-1}(-x) = -\tan^{-1}(x) \), it follows:\[ I(A) = \tan^{-1}(A) - (-\tan^{-1}(A)) = 2\tan^{-1}(A). \]

Step 5: Finding the Limit as A Approaches Infinity

The problem asks for \( \lim_{A \to \infty} I(A) \). Using \( I(A) = 2\tan^{-1}(A) \), and knowing \( \lim_{A \to \infty} \tan^{-1}(A) = \frac{\pi}{2} \), we compute:\[ \lim_{A \to \infty} 2\tan^{-1}(A) = 2 \times \frac{\pi}{2} = \pi. \]

Key Concepts

Definite Integral

When working with a definite integral, like the one in this exercise, we aim to find the area under a curve for a specific interval. Here, the interval is from -\(A\) to \(A\). The function given is \(\frac{1}{1+t^2}\), which resembles the well-known bell-shaped curve. This integral provides the total accumulation of the function over the specified limits.

The process involves calculating the antiderivative, using it to evaluate the function at the upper and lower limits of integration, and finding their difference:

• Identify the function to integrate and its limits.
• Compute the antiderivative of the function.
• Evaluate at both limits and subtract the lower from the upper evaluation.

This simple, yet powerful, operation allows us to measure the area that the function covers within any given bounds. In this case, the symmetry about 0 simplifies the integral's evaluation, as shown in the solution steps.

Inverse Trigonometric Functions

Inverse trigonometric functions come into play in this exercise through the function \(\tan^{-1}(t)\). The integral of \(\frac{1}{1+t^2}\) is known to be \(\tan^{-1}(t) + C\), where \(C\) stands for a constant that emerges in indefinite integrals. However, since we deal with definite integrals here, constants cancel out after evaluation, streamlining our calculations.

These functions essentially reverse trigonometric operations. Whereas a tangent function gives us an angle from a ratio, the inverse or arctan function gives us the ratio's angle.

• The derivative provides the slope of tangent lines.
• The integral finds the accumulated area under the graph.
• Inverse functions switch roles from output to input.

In definite integrals, the inverse trigonometric functions serve as linear transformations that simplify calculating areas under specific curves, as we see with the crisp conclusion in the final solution step.

Limit of a Function

A limit represents the value a function approaches as the input approaches some value. It's a foundational concept in calculus, helping us understand behavior over infinite spaces.

In our problem, we face the task of finding \(\lim_{A \to \infty} I(A)\) after defining \(I(A)\) as the definite integral expression. We need this to evaluate how the function behaves as the interval expands towards infinity.The scenario considers the approaching boundaries and how the function stabilizes:

• Function \(I(A) = 2\tan^{-1}(A)\).
• Use properties of limits and inverse tangent.
• \( \lim_{A \to \infty} \tan^{-1}(A) \ = \frac{\pi}{2} \) because it asymptotically approaches this value.

Consequently, \( \lim_{A \to \infty} 2\tan^{-1}(A) = \pi \), giving us a precise numerical concept of how the function's "area" behaves over its domain indefinitely. Understanding limits provides a glimpse into the unseen boundaries of mathematics.