Question

Evaluate each improper integral or show that it diverges. \(\int_{10}^{\infty} \frac{x}{1+x^{2}} d x\)

Short answer

The improper integral diverges.

Step-by-step solution

Identify the Type of Integral

The integral \( \int_{10}^{\infty} \frac{x}{1+x^{2}} dx \) is an improper integral because one of its limits is infinity. To solve it, we need to take the limit as \( b \to \infty \) where \( b \) is the upper limit of integration.

Set Up the Limit for the Improper Integral

Rewrite the integral with a finite upper limit \( b \), and take the limit as \( b \to \infty \):\[\lim_{{b \to \infty}} \int_{10}^{b} \frac{x}{1+x^{2}} dx\]

Determine the Antiderivative

Find the antiderivative of the function \( \frac{x}{1+x^{2}} \). Let \( u = 1 + x^{2} \), then \( du = 2x \, dx \). This gives:\[\int \frac{x}{1+x^{2}} dx = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|u| + C\]Replace \( u \) back to get: \( \frac{1}{2} \ln(1+x^{2}) + C \).

Evaluate the Definite Integral

Evaluate \( \int_{10}^{b} \frac{x}{1+x^{2}} dx \) using the antiderivative:\[\left[ \frac{1}{2} \ln(1+x^{2}) \right]_{10}^{b} = \frac{1}{2} \ln(1+b^{2}) - \frac{1}{2} \ln(101)\]

Take the Limit as $b$ Approaches Infinity

Compute the limit:\[\lim_{{b \to \infty}} \left( \frac{1}{2} \ln(1+b^{2}) - \frac{1}{2} \ln(101) \right)\]As \( b \to \infty \), \( \ln(1+b^{2}) \to \infty \) so the limit is \( \infty \).

Conclude

Since the limit is infinite, the improper integral diverges. Thus, \( \int_{10}^{\infty} \frac{x}{1+x^{2}} dx \) diverges.

Key concepts

Convergence and Divergence

Improper integrals involve limits at one or both ends of their interval of integration. Understanding if they converge or diverge is crucial. - **Convergence** occurs if the integral results in a finite number. - **Divergence** happens if it results in an infinite value or does not exist.
In the problem, we have the improper integral \[ \int_{10}^{\infty} \frac{x}{1+x^{2}} dx \] One endpoint is at infinity. We evaluated this by taking limits to find out whether it converges or diverges.
After computing, the limit approaches infinity, indicating divergence. Therefore, the integral doesn't have a finite solution.
When dealing with improper integrals, always look at the behavior of the integral's function as it approaches infinity.

Antiderivative Calculation

Finding the antiderivative is central for solving many calculus problems, especially integrals. It involves determining a function whose derivative is the given function.
In solving the integral \( \int \frac{x}{1+x^{2}} dx \), we use substitution. Let \( u = 1+x^2 \) with \( du = 2x \, dx \). This simplifies the process.
The problem becomes evaluating the integral of \( \frac{1}{u} du \), which is a classic integral.- The antiderivative is \( \frac{1}{2} \ln |u| + C \). - Substitute back \( u \) to get \( \frac{1}{2} \ln (1+x^2) + C \).
It’s essential to properly apply substitution techniques when calculating antiderivatives. This allows for straightforward integration of otherwise complex functions.

Limits at Infinity

Limits at infinity are used to evaluate the behavior of functions as they approach infinity. They are particularly useful in the context of improper integrals.
For the integral function \( \frac{x}{1+x^2} \), as \( x \) approaches infinity, it impacts the behavior and result of the entire calculation.
To compute the proper limit, we transformed the original integral:- The expression \( \lim_{{b \to \infty}} \left( \frac{1}{2} \ln(1+b^2) - \frac{1}{2} \ln(101) \right) \) shows this process. - As \( b \to \infty \), \( \ln(1+b^2) \) increases indefinitely.
The result is that the limit is infinite, providing a clear reason why the integral diverges.
Understanding limits is vital for analyzing the long-term behavior of functions in calculus.