Question

Explain why the integral is improper and determine whether it diverges or converges. Evaluate the integral if it converses. $$ \int_{0}^{2} \frac{1}{(x-1)^{2}} d x $$

Short answer

This improper integral is divergent.

Step-by-step solution

Identifying the nature of the integral

The function under the integral is \( \frac{1}{(x-1)^2} \). For x=1, the denominator becomes zero, rendering the function undefined. Because of this discontinuity in the interval of integration, the integral is classified as improper.

Converting the improper integral to a limit form

To handle the issue of the undefined point within the range, partition the improper integral into two proper integrals that include this problematic point as a limit. Therefore, \(\int_{0}^{2} \frac{1}{(x-1)^{2}} d x\) becomes \(\lim_{t \to 1^-} {\int_{0}^{t} \frac{1}{(x-1)^{2}} dx }\) + \(\lim_{t \to 1^+} {\int_{t}^{2} \frac{1}{(x-1)^{2}} dx }\) .

Evaluating the integrals separately

Now each of these integrals can be tackled separately and the result will be the desired full integral value. Using the fact that the antiderivative of \( \frac{1}{(x-1)^2} \) is \( -\frac{1}{(x-1)}\), you find \(\lim_{t \to 1^-} [- \frac{1}{(x-1)} ]_{0}^{t}\) = \(\lim_{t \to 1^-} -1/(t-1)-[(-1)/(0-1)]\) and \(\lim_{t \to 1^+} [- \frac{1}{(x-1)} ]_{t}^{2}\) = \(\lim_{t \to 1^+} -1/(2-1)-[(-1)/(t-1)]\) .

Evaluating the limits

Evaluate the limits: \(\lim_{t \to 1^-} -1/(t-1) = -\infty\), and \(\lim_{t \to 1^+} -1/(t-1) = +\infty\). Summing these up yields an indeterminate form, so the limits diverge.

Key concepts

Convergence and Divergence

When discussing improper integrals, we often need to determine if they converge or diverge. Simply put, an improper integral is one where the function becomes undefined or approaches infinity at some point within the integration interval.
Convergence occurs when the integral results in a finite number. If this is the case, you can evaluate it like a regular definite integral. Conversely, divergence happens when the integral's value grows without bound, leading to an infinite result.In the original problem, the integrand \( \frac{1}{(x-1)^2} \) becomes undefined at \( x = 1 \). This makes the integral improper. By breaking the integral into two parts around this undefined point, we check if each converges. In this case, both limits evaluated around \( x = 1 \) approach infinity, indicating that neither part converges, hence the entire integral diverges.
This conclusion is crucial because it indicates that the integral cannot be evaluated further to give a finite number.

Limit Evaluation

To evaluate improper integrals, we rely on the concept of limits. Since direct integration of undefined points is not possible, limits help us approach these points while calculating the integral.
The integral in the exercise splits into two proper integrals with the limits \( \lim_{t \to 1^-} \int_{0}^{t} \frac{1}{(x-1)^2} dx \) and \( \lim_{t \to 1^+} \int_{t}^{2} \frac{1}{(x-1)^2} dx \).The first limit evaluates the behavior of the integral as \( x \) approaches 1 from the left, and the second as \( x \) approaches 1 from the right.

• For \( \lim_{t \to 1^-} [-\frac{1}{x-1}]_{0}^{t} \), this results in \( -\infty \) as \( t \to 1^- \)
• For \( \lim_{t \to 1^+} [-\frac{1}{x-1}]_{t}^{2} \), this results in \( +\infty \) as \( t \to 1^+ \)

These results indicate divergence because both limits lead to infinite values. Understanding limits in improper integrals is key as it allows handling tricky undefined points and determining if the integral has a finite result or not.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are crucial for solving integrals. They represent the reverse operation of differentiation and are used to find the original function from its derivative.
In the context of improper integrals, finding the antiderivative helps us evaluate the integrals once the limits are set.For instance, the antiderivative of \( \frac{1}{(x-1)^{2}} \) is \( -\frac{1}{(x-1)} \). This makes it crucial for solving the integral. When evaluating each section of the improper integral:

• The antiderivative leads to expressions like \( -\frac{1}{t-1} \) for the limit \( \lim_{t \to 1^-} \int_{0}^{t} \), and \( -\frac{1}{2-1} = -1 \) for \( \int_{t}^{2} \).

Subtracting these values from each other based on their limits gives us insight into whether the integral converges. In this exercise, despite finding the antiderivative, the issues with the limits still caused the integral to diverge. Knowing antiderivatives simplifies evaluating definite integrals, making them essential in calculus.