**Problem Statement**: A student solves the following integral:\[\int_{0}^{2} \frac{dx}{(x-1)^{2}} = \left[ -\frac{1}{x-1} \right]_{0}^{2} = -1 - 1 = -2\]**Explanation of the Mistake**:The student evaluated the indefinite integral correctly when they found that the antiderivative of \(\frac{1}{(x-1)^2}\) is \(-\frac{1}{x-1}\). However, the mistake occurred during the evaluation of the definite integral's limits from 0 to 2:When calculating \(\left[ -\frac{1}{x-1} \right]_{0}^{2}\), the student should have recognized that \(x = 1\) is a point of discontinuity. Specifically, \(-\frac{1}{x-1}\) becomes undefined when \(x = 1\). The student incorrectly assessed both the limit evaluations at \(x=0\) and \(x=2\) without addressing this discontinuity, leading to an incorrect solution.**Correct Solution**:1. **Evaluate Limit as \(x\) Approaches 1**: - Recognize that the integral is improper due to the discontinuity at \(x = 1\). 2. **Split the Integral**: \[ \int_{0}^{2} \frac{dx}{(x-1)^{2}} = \int_{0}^{1} \frac{dx}{(x-1)^{2}} + \int_{1}^{2} \frac{dx}{(x-1)^{2}} \]3. **Evaluate Each Integral Separately**: - For \(\int_{0}^{1} \frac{dx}{(x-1)^{2}}\), calculate the limit as \(x\) approaches 1 from the left. - For \(\int_{1}^{2} \frac{dx}{(x-1)^{2}}\), calculate the limit as \(x\) approaches 1 from the right.4. **Note Divergence**: - Both limits diverge as they approach infinity.Thus, the integral diverges to \(\infty\) due to the discontinuity, indicating that the improper integral is undefined over

Name: **Problem Statement**: A student solves the following integral:\[\int
Uploaded: 2024-03-20T06:47:08.000Z
Channel: Bartleby
Description: **Problem Statement**: A student solves the following integral:\[\int_{0}^{2} \frac{dx}{(x-1)^{2}} = \left[ -\frac{1}{x-1} \right]_{0}^{2} = -1 - 1 = -2\]**Explanation of the Mistake**:The student evaluated the indefinite integral correctly when the