Question

Evaluate the improper integral `int_3^4 1/(x-3)^3 dx` ?

`int_3^4 1/(x-3)^3 dx`

How do you know if it's divergent?

Answer 1

The entire solution blows up to `\infty`, which means the integral is divergent.

Explanation:

We declare a variable for the bound that makes the integral improper. Let's use `t = 3`. We now want to integrate and assess the bound as a limit as the integral approaches it.

We now perform the integration with the limit definition:
`lim_(t->3)\int_t^4 (1)/(x-3)^3 dx`

`= lim_(t->3)[-1/(2(x-3)^2)]_(x=t)^(x=4)`

`=lim_(t->3){-1/(2(1)^2)+1/(2(t-3)^2)}`

`=lim_(t->3){-1/2 + 1/(2(t-3)^2)}`

If `t->3`, then that fraction blows up to `\infty`.

This means that the entire solution blows up to `\infty`, since `-1/2 + (\infty)` in essence is still just `\infty`. This means that the integral does not have a finite solution and is divergent.