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

2 Answers
Apr 24, 2015

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

Apr 24, 2015

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.)