Question

How do you find the definite integral for: `1/x^2` for the intervals `[0, 1]`?

Answer 1

It is

#int_0^1 1/x^2 dx=lim_(t->0^+) int_t^1 (-1/x)'dx=lim_(t->0^+) [-1/x]_t^1= -1+lim_(t->0^+) 1/t=oo#

Hence the integral does not converge