Calculate the integral, if it converges. You may calculate the limit by appealing to the dominance of one function over another, or by l'Hopital's rule.
dx/(4-x)^2 from 3 to 6
I have the integral as 1/(4-x) but im not sure what to do next because the hint tells me "The integral is improper at the interior of the interval; split the integral there. Determine the convergence of each part separately."

Question

Calculate the integral, if it converges. You may calculate the limit by appealing to the dominance of one function over another, or by l'Hopital's rule.
dx/(4-x)^2 from 3 to 6
I have the integral as 1/(4-x) but im not sure what to do next because the hint tells me "The integral is improper at the interior of the interval; split the integral there. Determine the convergence of each part separately."

anonymous · Answer

$$\int_3^4\frac{dx}{(4-x)^2}$$ and $$\int_4^6\frac{dx}{(4-x)^2}$$

fibonaccichick666 · Answer

On your domain, is there any point that causes your function not to exist?

anonymous · Answer

$$\int_3^4\frac{dx}{(4-x)^2}=\lim_{l\to 4}\int_3^l\frac{dx}{(4-x)^2}$$ is a start

anonymous · Answer

alright, i got -1 for that integral. but dont I also need to take it from 4 to 6? and what do i do after I get the integral for that?

anonymous · Answer

oh wait, just figured it out. it doesn't converge. haha thank you both @satellite73 and @FibonacciChick666.