Calculus
posted by Hannah .
Determine whether the integral is divergent or convergent. If it is convergent, evaluate it and enter that value as your answer. If it diverges to infinity, state your answer as "INF" (without the quotation marks). If it diverges to negative infinity, state your answer as "MINF". If it diverges without being infinity or negative infinity, state your answer as "DIV".
The integral from 0 to 7 of 1/((x5)^2)dx
I keep getting 0.7 but apparently I am wrong.

Apparently it diverges to negative infinity. Can someone please explain to me why?

I do not know how you got 0.7, negative infinity seems to me correct because there is an asymptote at x=5.
For improper integrals where the vertical asymptote is between the integration limites, we have to be careful not just evaluate the integral at the two limits.
Instead, we will subdivide the integral into two, from 0 to 5 and from 5 to 7.
If both integrals converge, then you have convergence and the integral is defined.
If there are more vertical asymptotes, you will need to subdivide accordingly.