How do you find the definite integral of #1/x^2# from 0 to 1?

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Answer 1 · 2015-04-24T14:27:52

As this function is not continuous in the in the interval [0,1], it is not integrable in the interval [0,1].

Answer 2 · 2015-04-24T14:39:32

This is an improper integral because the function is not defined at one of the limits of integration.

To (attempt to) evaluate the integral use the replacement:

#int_0^1 1/x^2 dx = lim_(ararr0^+) int_a^1 1/x^2 dx #

(The "#=#" is "equals if the limit exists")

#int_0^1 1/x^2 dx = lim_(ararr0^+) int_a^1 1/x^2 dx #

#color(white)"ssssssssss"## = lim_(ararr0^+) (-1)/x ]_a^1 #

#color(white)"ssssssssss"## = lim_(ararr0^+) (-1+1/a) = oo#

The integral diverges. (Does not exist.)