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dan815
integrate x/lnx
With ln(x) in the denominator like that that's not one of our preatty integrals.
you will not find a nice closed form for this one
is it one of those repeating integrating ones
you mean like if you integrate by parts will you go around in a circle? i don't think so
i think you will not be able to solve in in the sense that you will not be able to find some nice function \(F\) whose derivative is \(\frac{x}{\ln(x)}\)
It's the type where our general methods of integrating get us nowhere. We have to leave it in the form of an integral and work with it that way.
this is what im trying to evaultate |dw:1363059332628:dw|
is there another method to equate integral like i know this isnt conservative so
That's completly different. If you are integrating with repect to y first then you don't need to worry about the expressions in x -- they are essentially constants.
Wolfram says it's not expressible in terms of elementary functions - http://www.wolframalpha.com/input/?i=integrate+x%2Fln+x
imo you'd best be off solving it numerically
oh hold on i think i got it xD i forgot the lnx cancels if i do dy first
xavier u got some time to help ?? im working on one of my assigments due in about 6 hours, gonna need help when i get stuck
I'm here on and off. I'll try to help. Sure other people will jump onboard if I take too long
just looking thru the questions so far i think im gonna just have some trouble in the volume integral questions ill tell ya when i get to them
look at this one its about setting the right bounds |dw:1363060518856:dw|
i get why for dx its y^2 to 4 but how much for dy just doing 0 to 2 instead is enough how come it cant be be rootX to 4
What is correct there? The diagram or the bounds on your integrals?
the whole thing is correct i just need to covert to an another double integration where im doing it in reverse order dy dx
but before i do that i wanna know why they set up bounds like that
Either the bounds are wrong or the shaded area is incorrect. Having \[y^2 \le x \le 4\] is the larger of the two areas there.
the shaded region is what im integrating over its in the bounds the function is just like a force equation
Then your bounds on x should be \[0 \le x \le y^2\]
oh nvm i see what u mean the diagram is wrong
so the bottom part shud be the shaded region right
Ah ok so you know that the curve your have is y^2=x or y=sqrt(x). So y is bounded above by sqrt(x) and below by 0. And x ranges from 0 to 4
so if i wanted to switch order of integration to dy dx then
Just use the bounds on y on the inner integral. x bounds go on outer integral \[0 \le y \le \sqrt{x}\ ; 0 \le x \le 4\]