Expert answers to "Compute ∫021x−1dx"

vestirme4

Answered question

2021-03-09

Compute

\int_{0}^{2} \frac{1}{x - 1} d x

Answer & Explanation

Adnaan Franks

Skilled2021-03-10Added 92 answers

$\int_{0}^{2} \frac{1}{x - 1} d x is indeterminate since y = \frac{1}{x - 1} has a vertical asymptote at x = 1$ , which lies between the upper and lower integration bounds.
To determine if the integral is convergent, you must then split the integral at x=1, and then use limits to evaluate:
$\int_{0}^{2} \frac{1}{x - 1} d x$
$= \int_{0}^{1} \frac{1}{x - 1} d x + \int_{1}^{2} \frac{1}{x - 1} d x$ (Rewrite as a sum)
$lim_{m \to 1^{-}} \int_{0}^{m} \frac{1}{x - 1} d x + lim_{m \to 1^{-}} \int_{n}^{2} \frac{1}{x - 1} d x$ (Rewrite using limits)
$lim_{m \to 1^{-}} (\ln | x - 1 |) (∣_{0}^{m}) + lim_{m \to 1^{-}} (\ln | x - 1 |) ∣_{n}^{2}$ (Integrate)
$lim_{m \to 1^{-}} (\ln | m - 1 | - \ln | 0 - 1 |) + lim_{m \to 1^{-}} (\ln | 2 - 1 | - \ln | n - 1 |)$ (Evaluate)
$lim_{m \to 1^{-}} (\ln | m - 1 | - \ln 1) + lim_{m \to 1^{-}} (\ln 1 - \ln | n - 1 |)$ (Simplify)
$lim_{m \to 1^{-}} (\ln | m - 1 |) + lim_{m \to 1^{-}} (\ln | n - 1 |)$ (Simplify using $\ln 1 = 0$ )
$lim_{m \to 1^{-}} (\ln | m - 1 |) and lim_{m \to 1^{-}} (\ln | n - 1 |)$ aren't finite since $\ln 0$ is not defined.Therefore, the integral is divergent.

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