Question

How do you evaluate `int 1/ x^2` from 0 to 1?

Answer 1

You evaluate `int 1/x^2 dx` on `[a,1]`, then find the limit as `ararr0^+` (if the limit exists).

Explanation:

Because `0` is not in the domain of the integrand, this is what is called an improper integral.
To (attempt to) evaluate the intergral, find `lim_(ararr0^+) int_a^1 1/x^2 dx` (if that limit exists).

`int_a^1 1/x^2 dx = int_a^1 x^-2 dx = -x^-1]_a^1 = -1/x]_a^1`

So, `int_a^1 1/x^2 dx = [-1/(1)] - [-1/(a)] = 1/a-1`

`int_a^1 1/x^2 dx = lim_(ararr0^+) int_a^1 1/x^2 dx` `" "` if the limit exits.

But, `lim_(ararr0^+) int_a^1 1/x^2 dx=lim_(ararr0^+) [1/a-1]=oo`.

So, the limit does not exist and neither does the integral. (We say the integral diverges.)