Question

Evaluate the integrals. If the integral diverges, answer "diverges."\(\int_{0}^{1} \frac{d x}{1-x}\)

Short answer

The integral diverges.

Step-by-step solution

Identify the Type of Integral

The integral given is \( \int_{0}^{1} \frac{dx}{1-x} \). This is an improper integral because the integrand has a discontinuity at \( x = 1 \), within the interval of integration.

Set Up the Limit for the Improper Integral

Since the integrand becomes undefined at \( x = 1 \), we must approach it using a limit. We change the upper limit \( 1 \) to \( b \), where \( b \to 1^- \), and evaluate \( \int_{0}^{b} \frac{dx}{1-x} \).

Evaluate the Integral

The integral of \( \frac{1}{1-x} \) is \( -\ln|1-x| + C \). Evaluate from 0 to \( b \).

Apply the Fundamental Theorem of Calculus

Substitute the bounds into the antiderivative: \(-\ln|1-b| + \ln|1-0| = -\ln(1-b) + \ln(1) \). \( \ln(1) = 0 \), so the expression simplifies to \(-\ln(1-b) \).

Evaluate the Limit

Now take the limit as \( b \to 1^- \) for \( -\ln(1-b) \). As \( b \to 1 \), \( 1-b \to 0^+ \), and \( \ln(1-b) \to -\infty \). Thus, \(-\ln(1-b) \to \infty \).

Determine Convergence or Divergence

Since the limit evaluates to \( \infty \), the integral diverges.

Key concepts

Limit of Integration

In calculus, the limit of integration refers to the bounds between which an integral is evaluated. When evaluating definite integrals, these limits specify where the calculation starts and ends. For the integral \( \int_{0}^{1} \frac{dx}{1-x} \), the limits of integration are 0 and 1. However, there's a special consideration needed here because the function \( \frac{1}{1-x} \) becomes undefined as \( x \) approaches 1. This is why we modify the upper limit from 1 to \( b \), where \( b \to 1^- \), guaranteeing the calculation approaches the discontinuity without actually touching it.
This adjustment is crucial in dealing with improper integrals, where one or both limits are at the boundaries of where the function is undefined. The proper handling of limits of integration ensures accurate evaluation of the integral even in the presence of discontinuities.

Divergence

In calculus, especially when dealing with improper integrals, divergence refers to the failure of the integral to converge, meaning it doesn't approach a finite number. In the given exercise \( \int_{0}^{1} \frac{dx}{1-x} \), the integral is said to diverge.
This is discovered through the step of evaluating the limit: as \( b \to 1^- \), the value of \(-\ln(1-b) \) tends towards \( \infty \).
Since the final limit is not a finite value, but instead infinitely large, this signifies that the integral does not have a finite solution. Thus, it is declared that the integral diverges.

• Divergence often indicates an infinite area under a curve in a given interval, suggesting the sum just keeps growing.
• It’s an essential conclusion in calculus to identify when certain integrals do not yield results within the scope of real numbers.

Discontinuity

Discontinuity in a function occurs where the function is not continuous, and calculations become challenging at that point. In the integral \( \int_{0}^{1} \frac{dx}{1-x} \), the integrand \( \frac{1}{1-x} \) shows discontinuity at \( x = 1 \). This point is recognized as a vertical asymptote of the function, where the function tends towards infinity, and the value is undefined.
The presence of this discontinuity necessitates treating the integral as improper, thereby requiring limits to handle the evaluation accurately.

• Understanding where discontinuities occur is vital for correctly setting up and solving integrals.
• The handling of these points through limits leads to solutions that reflect the true behavior of the function near these discontinuities.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, providing an efficient way to evaluate definite integrals. It states that if a function is continuous over an interval, the integral of its derivative over that interval will give the change in the function’s values over that interval.
For the problem \( \int_{0}^{1} \frac{dx}{1-x} \), the theorem helps transition from the indefinite integral \( \int \frac{dx}{1-x} = -\ln|1-x| + C \) to evaluating definite bounds over our adjusted limits from \( 0 \) to \( b \) (where \( b \to 1^- \)).
Applying the theorem allows us to nd:

• Substitute bounds to compute definite integrals.
• Efficiently resolve complex integrals by studying antiderivatives.

The theorem essentially provides the fundamental link that exploits the relationship between antiderivatives and the area under curves.