Question

Compute int_0^2 1/(x-1)dx

Integrals
ANSWERED
asked 2021-03-09
Compute \(\displaystyle{\int_{{0}}^{{2}}}\frac{{1}}{{{x}-{1}}}{\left.{d}{x}\right.}\)

Answers (1)

2021-03-10

\(\displaystyle{\int_{{0}}^{{2}}}\frac{1}{{{x}-{1}}}{\left.{d}{x}\right.}\ \text{ is indeterminate since }\ {y}=\frac{1}{{{x}-{1}}}\ \text{ has a vertical asymptote at }\ {x}={1}\), 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:
\(\displaystyle{\int_{{0}}^{{2}}}\frac{{1}}{{{x}-{1}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{1}}}\frac{{1}}{{{x}-{1}}}{\left.{d}{x}\right.}+{\int_{{1}}^{{2}}}\frac{{1}}{{{x}-{1}}}{\left.{d}{x}\right.}\) (Rewrite as a sum)
\(\displaystyle\lim_{{{m}\rightarrow{1}^{{-}}}}{\int_{{0}}^{{m}}}\frac{{1}}{{{x}-{1}}}{\left.{d}{x}\right.}+\lim_{{{m}\rightarrow{1}^{{-}}}}{\int_{{n}}^{{2}}}\frac{{1}}{{{x}-{1}}}{\left.{d}{x}\right.}\) (Rewrite using limits)
\(\displaystyle\lim_{{{m}\rightarrow{1}^{{-}}}}{\left({\ln{{\left|{{x}-{1}}\right|}}}\right)}{\left({{\mid}_{{0}}^{{m}}}\right)}+\lim_{{{m}\rightarrow{1}^{{-}}}}{\left({\ln{{\left|{{x}-{1}}\right|}}}\right)}{{\mid}_{{n}}^{{2}}}\) (Integrate)
\(\displaystyle\lim_{{{m}\rightarrow{1}^{{-}}}}{\left({\ln{{\left|{{m}-{1}}\right|}}}-{\ln{{\left|{{0}-{1}}\right|}}}\right)}+\lim_{{{m}\rightarrow{1}^{{-}}}}{\left({\ln{{\left|{{2}-{1}}\right|}}}-{\ln{{\left|{{n}-{1}}\right|}}}\right)}\) (Evaluate)
\(\displaystyle\lim_{{{m}\rightarrow{1}^{{-}}}}{\left({\ln{{\left|{{m}-{1}}\right|}}}-{\ln{{1}}}\right)}+\lim_{{{m}\rightarrow{1}^{{-}}}}{\left({\ln{{1}}}-{\ln{{\left|{{n}-{1}}\right|}}}\right)}\) (Simplify)
\(\displaystyle\lim_{{{m}\rightarrow{1}^{{-}}}}{\left({\ln{{\left|{{m}-{1}}\right|}}}\right)}+\lim_{{{m}\rightarrow{1}^{{-}}}}{\left({\ln{{\left|{{n}-{1}}\right|}}}\right)}\) (Simplify using \(\ln 1=0\))
\(\displaystyle\lim_{{{m}\rightarrow{1}^{-}}}{\left( \ln{{\left|{{m}-{1}}\right|}}\right)}{\quad\text{and}\quad}\lim_{{{m}\rightarrow{1}^{-}}}{\left( \ln{{\left|{{n}-{1}}\right|}}\right)}\) aren't finite since \(\ln 0\) is not defined.Therefore, the integral is divergent.

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