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

1 Answer
Apr 17, 2016

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.)