Problem 13 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

It is a fact that $$ [\arctan x]^{\prime}=\frac{1}{1+x^{2}} $$ Compute $$ \int_{0}^{\infty} \frac{1}{1+x^{2}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{2} \).

Step 1: Recognize the Integrand's Derivative

We are given that the derivative of \( \arctan(x) \) is \( \frac{1}{1+x^2} \). This means that the function inside the integral, \( \frac{1}{1+x^2} \), is the derivative of \( \arctan(x) \).

Step 2: Set Up the Indefinite Integral

Since \( \frac{1}{1+x^2} = [\arctan(x)]' \), we can write the indefinite integral as \( \int \frac{1}{1+x^2} \, dx = \arctan(x) + C \), where \( C \) is the constant of integration.

Step 3: Evaluate the Definite Integral

We are asked to evaluate \( \int_0^{\infty} \frac{1}{1+x^2} \, dx \). Using the result from Step 2, we have \( \left[ \arctan(x) \right]_0^{\infty} = \lim_{b \to \infty} \arctan(b) - \arctan(0) \).

Step 4: Calculate the Limits

The limit as \( x \) approaches \( \infty \) for \( \arctan(x) \) is \( \frac{\pi}{2} \). Meanwhile, \( \arctan(0) = 0 \). Therefore, \( \lim_{b \to \infty} \arctan(b) - \arctan(0) = \frac{\pi}{2} - 0 \).

Step 5: Finish the Calculation

The evaluation of the definite integral thus simplifies to \( \frac{\pi}{2} \). Hence, \( \int_0^{\infty} \frac{1}{1+x^2} \, dx = \frac{\pi}{2} \).

Key Concepts

Definite Integral

A definite integral represents the area under the curve of a function between two specified points. This is a key concept in integral calculus as it allows us to calculate total quantities from rates of change.
In the exercise provided, we are calculating the definite integral of \( \frac{1}{1+x^2} \) from 0 to infinity. Here's a step-by-step explanation of how this works:

• First, recognize the integrand's derivative. We know that the derivative of \( \arctan(x) \) is \( \frac{1}{1+x^2} \), which suggests that \( \arctan(x) \) is the antiderivative of the function within our integral.
• Once you identify this, you can set up the indefinite integral, yet to calculate between specific boundaries. We symbolize this calculation as \( \left[ \arctan(x) \right]_0^{\infty} \), meaning we will find the value of \( \arctan(x) \) at the upper bound and subtract its value at the lower bound.
• Finally, by plugging in infinity and zero into the \( \arctan(x) \), you get its respective values. The limit of \( \arctan(x) \) as \( x \to \infty \) is \( \frac{\pi}{2} \) and at 0, it's 0. Thus, the definite integral evaluates to \( \frac{\pi}{2} \).

This process shows how powerful the definite integral is for finding exact areas even over infinite intervals.

Indefinite Integral

An indefinite integral, unlike a definite integral, does not involve specific limits for calculation. Instead, it gives a family of functions, or an antiderivative of the function we are integrating. This results in a general form with an added constant \( C \) because removing a derivative from a function loses any constant value originally present.
In the exercise, we observe the identification of the indefinite integral of \( \frac{1}{1+x^2} \), which translates to \( \int \frac{1}{1+x^2} \, dx = \arctan(x) + C \). Here's a detailed breakdown:

• Recognizing that the derivative of \( \arctan(x) \) is \( \frac{1}{1+x^2} \), we can conclude our indefinite integral yields the function \( \arctan(x) \).
• The \( + C \) represents any potential constant since integration loses the constant value when it differentiates.

The indefinite integral provides a broad understanding of all possible antiderivatives and sets the stage for simplifying the evaluation with bounds in the definite integral.

Limits in Calculus

Limits are a fundamental concept in calculus that measure the value a function approaches as the input approaches some value. They are critical for functioning in scenarios involving definites and barring undefined situations or calculation over infinities.
In step 4 of the provided solution, limits play a key role, especially where we consider \( \lim_{b \to \infty} \arctan(b) \). Here's how limits help here:

• The limit of \( \arctan(x) \) as \( x \to \infty \) is \( \frac{\pi}{2} \), because \( \arctan(x) \) plateaus at this horizontal asymptote on the graph of the function as \( x \) increases indefinitely.
• For \( \arctan(0) \), the limit presents a straightforward evaluation of 0 because there's no tendency or approach happening at zero.
• Applying these limits allows us to precisely conclude the overall value of the area of interest, resulting in \( \frac{\pi}{2} \) in this exercise.

With limits, problems involving boundless intervals or elusive points become manageable, revealing important insights and results in calculus.