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Xavier
 one year ago
Best ResponseYou've already chosen the best response.0With ln(x) in the denominator like that that's not one of our preatty integrals.

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2you will not find a nice closed form for this one

dan815
 one year ago
Best ResponseYou've already chosen the best response.0is it one of those repeating integrating ones

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2you mean like if you integrate by parts will you go around in a circle? i don't think so

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2i 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 \(\frac{x}{\ln(x)}\)

Xavier
 one year ago
Best ResponseYou've already chosen the best response.0It'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.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0this is what im trying to evaultate dw:1363059332628:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.0is there another method to equate integral like i know this isnt conservative so

Xavier
 one year ago
Best ResponseYou've already chosen the best response.0That'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.

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.0Wolfram says it's not expressible in terms of elementary functions  http://www.wolframalpha.com/input/?i=integrate+x%2Fln+x

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.0imo you'd best be off solving it numerically

dan815
 one year ago
Best ResponseYou've already chosen the best response.0oh hold on i think i got it xD i forgot the lnx cancels if i do dy first

dan815
 one year ago
Best ResponseYou've already chosen the best response.0xavier 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

Xavier
 one year ago
Best ResponseYou've already chosen the best response.0I'm here on and off. I'll try to help. Sure other people will jump onboard if I take too long

dan815
 one year ago
Best ResponseYou've already chosen the best response.0just 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.0look at this one its about setting the right bounds dw:1363060518856:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i 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

Xavier
 one year ago
Best ResponseYou've already chosen the best response.0What is correct there? The diagram or the bounds on your integrals?

dan815
 one year ago
Best ResponseYou've already chosen the best response.0the whole thing is correct i just need to covert to an another double integration where im doing it in reverse order dy dx

dan815
 one year ago
Best ResponseYou've already chosen the best response.0but before i do that i wanna know why they set up bounds like that

Xavier
 one year ago
Best ResponseYou've already chosen the best response.0Either 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.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0the shaded region is what im integrating over its in the bounds the function is just like a force equation

Xavier
 one year ago
Best ResponseYou've already chosen the best response.0Then your bounds on x should be \[0 \le x \le y^2\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.0oh nvm i see what u mean the diagram is wrong

dan815
 one year ago
Best ResponseYou've already chosen the best response.0so the bottom part shud be the shaded region right

Xavier
 one year ago
Best ResponseYou've already chosen the best response.0Ah 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.0so if i wanted to switch order of integration to dy dx then

Xavier
 one year ago
Best ResponseYou've already chosen the best response.0Just use the bounds on y on the inner integral. x bounds go on outer integral \[0 \le y \le \sqrt{x}\ ; 0 \le x \le 4\]
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