Question

Evaluate \(\int_{-1}^{1} \frac{x d x}{x^{2}+1}\)

Short answer

The integral evaluates to 0.

Step-by-step solution

Set Up the Integral

We start by rewriting the integral we need to evaluate: \[ \int_{-1}^{1} \frac{x \, dx}{x^{2} + 1} \] We will evaluate this definite integral by finding an antiderivative and then using the Fundamental Theorem of Calculus.

Recognize the Integrand

Notice that the integrand \( \frac{x}{x^2 + 1} \) suggests a substitution method could be useful since the numerator is the derivative of the denominator except for a scaling factor. Recognize that if \( u = x^2 + 1 \), then \( du = 2x \, dx \).

Apply Substitution

In order to apply substitution, let \( u = x^2 + 1 \), then \( du = 2x \, dx \). Therefore, \( \frac{1}{2} du = x \, dx \). Rewrite the integral in terms of \( u \): \[ \int \frac{x \, dx}{x^{2} + 1} = \frac{1}{2} \int \frac{du}{u} \]

Find the Antiderivative

The antiderivative of \( \frac{1}{u} \) with respect to \( u \) is \( \ln|u| + C \). Thus, \[ \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| + C \] Substitute back \( u = x^2 + 1 \) to get the antiderivative: \[ \frac{1}{2} \ln|x^2 + 1| + C \]

Evaluate the Definite Integral

Use the Fundamental Theorem of Calculus to evaluate the definite integral from \( -1 \) to \( 1 \). Substitute these bounds into the antiderivative:\[ \left[ \frac{1}{2} \ln|x^2 + 1| \right]_{-1}^{1} = \frac{1}{2} \ln|1^2 + 1| - \frac{1}{2} \ln|(-1)^2 + 1| \] Calculate:\[ \frac{1}{2} \ln|2| - \frac{1}{2} \ln|2| = 0 \]

Conclude the Evaluation

Since the result of the evaluation of the integral from \(-1\) to \(1\) of the function \( \frac{x}{x^2 + 1} \) is zero, it confirms the symmetry of the function across the y-axis canceling contributions from both sides equally.

Key concepts

Substitution Method

The substitution method is a powerful tool for solving integrals, especially when the integral's format suggests a simpler form once a substitution is made. In our problem, we needed to evaluate \[ \int_{-1}^{1} \frac{x \, dx}{x^{2}+1} \]. The integrand, \( \frac{x}{x^2 + 1} \), indicates that a substitution might simplify the expression. - Choose a substitution that transforms the integrand into a simpler form.- In this exercise, we set \( u = x^2 + 1 \), which naturally suggested itself because the numerator \( x \) is related to the derivative of the denominator. - The corresponding differential becomes \( du = 2x \, dx \), leading to \( \frac{1}{2} du = x \, dx \). Thus, the integral is rewritten as a simpler logarithmic integral \( \frac{1}{2} \int \frac{du}{u} \). This method transforms the problem into one that's much easier to handle.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is pivotal in connecting differentiation with integration. It consists of two main parts:- The first part guarantees that if a function is continuous over an interval, then its integral function provides an antiderivative.- The second part states that if you have an antiderivative and you evaluate it over an interval, you can find the value of the definite integral.In our problem, after finding the antiderivative \( \frac{1}{2} \ln|x^2 + 1| \), we used the Fundamental Theorem to evaluate the definite integral from \(-1\) to \(1\). - Evaluate the antiderivative at the upper limit \( x = 1 \) and subtract the evaluation at the lower limit \( x = -1 \).- This straightforward process highlights the beauty and utility of having an antiderivative readily evaluated at specific bounds.

Antiderivative

An antiderivative is essentially the backwards of a derivative. Finding an antiderivative is the primary step in evaluating an indefinite integral. In our exercise, the goal was to integrate \( \frac{x}{x^2 + 1} \) by expressing it in terms of \( u \).- Given \( \frac{1}{u} \) as our integrand after substitution, its antiderivative is \( \ln|u| + C \).- Returning to our original variables involved substituting \( u = x^2 + 1 \) back into this expression.This reveals the antiderivative \( \frac{1}{2} \ln|x^2 + 1| + C \). Antiderivatives allow us to solve definite integrals by applying the Fundamental Theorem of Calculus.

Symmetric Functions

Symmetric functions across an interval possess intriguing properties. Symmetry can simplify solving definite integrals if the functions contribute equally with opposite signs.- Consider the function \( \frac{x}{x^2 + 1} \), which is odd with respect to the y-axis. This means for every positive input, there's a corresponding negative input contributing equally in magnitude but with opposite signs.- Thus, definite integrals with bounds symmetric around zero—like \(-1\) to \(1\) in our problem—result in a net area of zero due to symmetry.In evaluating \( \int_{-1}^{1} \frac{x}{x^2 + 1} \, dx \), the integral equals zero since the positive and negative areas cancel out, perfectly capturing the essence of symmetric functions.