Determine whether the improper integral converges or diverges.
-$$\int _ { 0 } ^ { 2 } \frac { d x } { | x - 1 | } d x$$
A) Diverges
B) Converges

Question

Quizplus · Accepted Answer

The integrand $f(x) = \frac{1}{|x-1|}$ is not integrable on any interval containing $x=1$ due to the vertical asymptote at $x=1$. Thus, we need to consider the limits as $x$ approaches 1 from the left and from the right.

Approaching 1 from the left: 
$$\int_0^{1-\epsilon} \frac{dx}{|x-1|} = \int_0^{1-\epsilon} \frac{dx}{1-x} = \ln|1-x| \Big|_0^{1-\epsilon} = \ln (\epsilon) ightarrow \infty 	ext{ as } \epsilon ightarrow 0^+$$

Approaching 1 from the right: 
$$\int_{1+\epsilon}^2 \frac{dx}{|x-1|} = \int_{1+\epsilon}^2 \frac{dx}{x-1} = \ln|x-1| \Big|_{1+\epsilon}^2 = \ln(1+\epsilon) ightarrow \infty 	ext{ as } \epsilon ightarrow 0^+$$

Since both limits diverge to infinity, the improper integral diverges. Therefore, the answer is (A) Diverges.

Determine Whether the Improper Integral Converges or Diverges ∫02dx∣x−1∣dx\int _ { 0 } ^ { 2 } \frac { d x } { | x - 1 | } d x∫02​∣x−1∣dx​dx

Determine Whether the Improper Integral Converges or Diverges $\int _ { 0 } ^ { 2 } \frac { d x } { | x - 1 | } d x$