Question

How do you show whether the improper integral `int (1/3x-6) dx` converges or diverges from negative infinity to zero?

Answer 1

Evaluating the integral shows that it diverges

Explanation:

`int_(-oo)^0(1/3x-6)dx=lim_(t->-oo)int_t^0(1/3x-6)dx`

`=lim_(t->-oo)[x^2/6-6x]_t^0`

`=lim_(t->-oo)(0-t^2/6+6t)`

`=-oo !inRR`

Therefore this integral diverges