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Chapter 6: Problem 7

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

Short Answer

Expert verified
This improper integral is convergent and evaluates to 2.

Step by step solution

01

Identifying the Improper Integral and Break into Proper Integral Forms

The given improper integral \(\int_{0}^{2} \frac{1}{(x-1)^{2}} dx\) is improper at \(x=1\), which is within the bounds of integration. Therefore, this integral is split into two proper integrals around \(x=1\), resulting in: \(\int_{0}^{1} \frac{1}{(x-1)^{2}} dx + \int_{1}^{2} \frac{1}{(x-1)^{2}} dx\).
02

Evaluating the Integral

The integrals can be evaluated individually as: - For the first integral, we make a substitution \(u = x - 1\), \(du = dx\). Then the integral becomes \(- \int_{-1}^{0} \frac{1}{u^2} du\), which can be solved as \(-[u^{-1}]_{-1}^{0} = -[0 - -1] = 1\).- Similarly, for the second integral, we make the substitution \(v = x - 1\), \(dv= dx\). The integral then becomes \(\int_{0}^{1} \frac{1}{v^2} dv\), which evaluates to \([v^{-1}]_{0}^{1} = 1 - 0 = 1\).
03

Sum the Results

By adding the solutions of both integrals together, we get \(1 + 1 = 2\).

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