note the antiderivative is ...
converges
Hey guys, I have a question about improper integrals.
For example lets say the question is 1/(x-2)^3/2 dx and the interval is between infinity and 3.
How would I be able to determine if its convergent or divergent and if its convergent solve.
So I tried the problem out myself and what I did was I took the (x-2)^3/2 to the top so it becomes (x-2)^-3/2 and from there i just did the anti derivative of that which i thought was 1/(-3/2+1/1) giving me 1/(-1/4) which i then brought the 4 to the top, meaning the equation is now -4(x-2)^(-3/2+1/1) which turns out to be -4(x-2)^(-1/2) between the interval of infinity and 3. Is what i did so far correct or completely off? If it's wrong can you tell me why I can't do that and what the correct way of doing it is?
Sorry I had to make it long, i just wanted you guys to understand the problem I had.
oh now I get it, but like for the last line where it says 1/square root of b-2 then -1 right howd u get that to =2? Like how did you manage to get the answer 2? Since its infinity aren't you able to sub in anything meaning lets say 1000. so 1/square root of 1000-2 then -1 doesn't give you 2. So how'd you get that value?
Are you able to quickly summarize it for me? I could've swore it was just subbing in a really high number but if it's not how do you solve for it?
Can anyone please tell me how to evaluate using the infinity limit?
If you're doing improper integrals, you should have already been taught how to evaluate limits at infinity. If you need a review, check your book or another student or your teacher or one of the many fine online tutorials such as this.
Thread closed.