Question

For \(A > 0\), 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 \([-\infty, \infty]\).

Short answer

The area is \(\pi\).

Step-by-step solution

Understanding the Integral

The given integral is \(I(A)=\int_{-A}^{A} \frac{d t}{1+t^{2}}\). This is an integral of the function \(\frac{1}{1+t^{2}}\), which resembles the derivative of the arctangent function, \(\arctan(t)\). We will use this relationship to compute the integral.

Integration Using the Antiderivative

The antiderivative of \(\frac{1}{1+t^{2}}\) is \(\arctan(t)\). Therefore, the integral \(I(A)\) can be calculated as follows: \[ I(A) = \int_{-A}^{A} \frac{1}{1+t^{2}} \, dt = \left[ \arctan(t) \right]_{-A}^{A} \].

Evaluating the Definite Integral

To evaluate \(\left[ \arctan(t) \right]_{-A}^{A}\), we need to compute \(\arctan(A) - \arctan(-A)\). We know that \(\arctan(-A) = -\arctan(A)\), thus: \[ I(A) = \arctan(A) - (-\arctan(A)) = 2\arctan(A) \].

Taking the Limit as \(A\) Approaches Infinity

We need to evaluate \(\lim_{A \to \infty} I(A) = \lim_{A \to \infty} 2\arctan(A)\). Since \(\arctan(A)\) approaches \(\frac{\pi}{2}\) as \(A\) approaches infinity, it follows that \[ \lim_{A \to \infty} 2\arctan(A) = 2\left(\frac{\pi}{2}\right) = \pi \].

Key concepts

Definite Integral

A definite integral represents the exact area under a curve within a specified interval. In our example, we evaluate the definite integral \(I(A)=\int_{-A}^{A} \frac{1}{1+t^{2}}\,dt\), which gives us the area under the graph of \(\frac{1}{1+t^{2}}\) from \(-A\) to \(A\).
This integral is determined by using the antiderivative. By calculating the difference between the antiderivative values at the bounds, \(-A\) and \(A\), we find the net area between the curve and the x-axis. Here's how:

• Identify the antiderivative of the function inside the integral.
• Evaluate this antiderivative at both the upper and lower limits.
• Subtract the value at the lower limit from the value at the upper limit.

This process provides the total accumulated value, emphasizing the importance of limits in defining where the calculation occurs.

Arctangent Function

The arctangent function, denoted as \(\arctan(t)\), is a crucial element in calculus, specifically when dealing with integrals of functions like \(\frac{1}{1+t^{2}}\). This is because the derivative of the arctangent function is \(\frac{1}{1+t^{2}}\), making it an ideal match for integration problems involving such expressions.
A few important properties of the arctangent function include:

• \(\arctan(x)\) will return values ranging from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
• It is an odd function, meaning \(\arctan(-x) = -\arctan(x)\).
• As \(x\) approaches infinity, \(\arctan(x)\) tends toward \(\frac{\pi}{2}\).

Understanding these characteristics allows students to decipher the integral results swiftly and apply limits efficiently when needed.

Limit of a Function

The limit of a function describes the behavior of the function as the input approaches a particular point. In calculus, limits are fundamental for evaluating scenarios where direct substitution might not provide an answer. Our task involves finding the limit \(\lim_{A \to \infty} I(A)\).
Here, as \(A\) becomes very large, you need to understand that the limit gives us an insight into the function's behavior toward infinity. By knowing that \(\arctan(A)\) approaches \(\frac{\pi}{2}\) when \(A\) grows indefinitely, the computation

• Executes \(\lim_{A \to \infty} 2\arctan(A) = 2\left(\frac{\pi}{2}\right) = \pi\).

This limit's evaluation illustrates how the integral converges to a specific value, providing insight on how the area under the curve behaves on an infinite domain.

Improper Integral

An improper integral involves integrating a function over an unbounded interval or where the integrand becomes unbounded. In our exercise, we deal with the integral over \([-\infty, \infty]\), making it improper.
Handling improper integrals requires taking the limit of the definite integral as the bounds approach infinity. This often uses the concept of limits to verify whether the integral converges to a finite value or diverges.
Our solution to \(\lim _{a \rightarrow \infty} I(A)\) illustrates this approach:

• First, solve the definite integral within a finite boundary \([-A, A]\).
• Extend the concept by calculating the limit as these bounds head to infinity.
• Confirm convergence by checking if the limit results in a finite number, such as \(\pi\) in this instance.

Recognizing when integrals are improper and understanding how to compute them is fundamental for exploring calculus at advanced levels.