Find out the answer to "∫021(x−1)2dx"

Trent Carpenter

Answered question

2021-02-18

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

Answer & Explanation

Nola Robson

Skilled2021-02-19Added 94 answers

Here the integral does not converge. In order for a function to be integrable, it must be continuous on its domain. Here, notice that the function

\frac{1}{{(x - 1)}^{2}}

has a discontinuity at

x = 1 \in [0, 2]

. If we try to integrate beside the discontinuity, we also encounter problems with the improper integrals:

\int_{0}^{2} \frac{1}{{(x - 1)}^{2}} d x = \int_{0}^{1} \frac{1}{{(x - 1)}^{2}} d x + \int_{1}^{2} \frac{1}{{(x - 1)}^{2}} d x

Notice that:

\int_{0}^{1} \frac{1}{{(x - 1)}^{2}} d x = lim t \to 1 \int_{0}^{t} \frac{1}{{(x - 1)}^{2}} d x = lim t \to 1 (\frac{- t}{t - 1})

which does not exist since we have this limit approach ∞ and −∞ from the left and right, respectively.
Therefore, we cannot integrate by this method either.

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