eaglebtc Posted June 6, 2004 Share Posted June 6, 2004 I'm currently taking independent study classes through an out-of-state university. The courses are paper-based, which means I mail the assignments to the school after Ive completed them. The topic of this homework assignment is Integration by Parts. If you remember lessons from Calculus in high school, this involves substituting the variables u, du, v, and dv for parts of the equation in order to integrate it more easily. It's supposed to take this form: And here is my problem: I do not know what I should choose for u or dv. Please help me! Maybe it's simpler than it looks, and my brain is just tired. But I already tried "ln x" as u, but what do I do about that power of 2? setting u to "(ln x)^2" doesn't seem to help either. Link to comment Share on other sites More sharing options...
rumbleph1$h Posted June 6, 2004 Share Posted June 6, 2004 I think you want to take: u = ln(x) and dv = 2dx EDIT: whoops, didn't see that bottom part.....i'll try to work the prob. out ;) Link to comment Share on other sites More sharing options...
1337.g4M3R Posted June 6, 2004 Share Posted June 6, 2004 lol, if you had asked me about 18 months ago i would have been able to do it, having not done calculus for so long i've completely forgotton how to do it :rolleyes: sorry i can't be of help Link to comment Share on other sites More sharing options...
HellBender Posted June 6, 2004 Share Posted June 6, 2004 Oh, wow, thats painfully easy, but I can't remember for the life of me how to do it. :no: I'm going to have such a hard time taking Calculus 3 next semester... Link to comment Share on other sites More sharing options...
eaglebtc Posted June 6, 2004 Author Share Posted June 6, 2004 thanks rumbleph1sh. two heads are better than one. :) I may be posting another calculus problem a little bit later, but we'll see how the rest of the assignment goes. Link to comment Share on other sites More sharing options...
eaglebtc Posted June 6, 2004 Author Share Posted June 6, 2004 lol, if you had asked me about 18 months ago i would have been able to do it, having not done calculus for so long i've completely forgotton how to do it :rolleyes: sorry i can't be of help LOL - it's been 4 years since my high school Calculus courses. I'm in COLLEGE now, and being forced to take Calculus over again for my major. I used to be good at this stuff too :rofl: My AP tests in High School counted for Calculus 112 & 113 (for Computer Science) but since I switched majors, they have this class called "Business Calculus" -- Calc 119. Rehash of the same crap from 112, slightly easier, but more emphasis on practical applications. But man, 4 years does take its toll on the brain. Link to comment Share on other sites More sharing options...
rumbleph1$h Posted June 6, 2004 Share Posted June 6, 2004 thanks rumbleph1sh. two heads are better than one. :)I may be posting another calculus problem a little bit later, but we'll see how the rest of the assignment goes. ahh i got it: u = (lnx)^2 dv = dx :p This ones right for sure.....straight out of my Calc book ;) Link to comment Share on other sites More sharing options...
eaglebtc Posted June 6, 2004 Author Share Posted June 6, 2004 (edited) ^^^ OK, but that leaves me with du = (ln x)^2 * 1/x . How the heck do I integrate that? OOHHHH!!!! Now i get it! It looks like this: Plus, I actually bothered to go to "Table 1" in the back of the book. It contains a list of the most common integrals. PS: who (or what) is that in your sig? Edited June 6, 2004 by eaglebtc Link to comment Share on other sites More sharing options...
dduardo Posted June 6, 2004 Share Posted June 6, 2004 Look at the problem as ln(x) * ln(x) Therefore: u = ln(x) du = 1/x dv=ln(x) v=x*ln(x)-x Now plug it into the equation and you can figure it out from there Link to comment Share on other sites More sharing options...
tang_tito Posted June 6, 2004 Share Posted June 6, 2004 there you go i hope it's not wrong since I took calculus a very long time ago p.s. sorry, it's kind of unclear Link to comment Share on other sites More sharing options...
1337.g4M3R Posted June 6, 2004 Share Posted June 6, 2004 Rehash of the same crap from 112, slightly easier, but more emphasis on practical applications. surely there is no practical application of differentiation and integration...i think it's something they made up just so they can test us on it... :p Link to comment Share on other sites More sharing options...
tang_tito Posted June 6, 2004 Share Posted June 6, 2004 Look at the problem as ln(x) * ln(x)Therefore: u = ln(x) du = 1/x dv=ln(x) v=x*ln(x)-x Now plug it into the equation and you can figure it out from there yeah, you can do it like this too Link to comment Share on other sites More sharing options...
rumbleph1$h Posted June 6, 2004 Share Posted June 6, 2004 tang_tito is right you can prove this by using the integration rule: int (ln u)^2 du = u [2 - 2 (ln u) + (ln u)^2 ] + C :happy: Link to comment Share on other sites More sharing options...
eaglebtc Posted June 6, 2004 Author Share Posted June 6, 2004 So it would appear that you have to do some "nested integration" to break down the last remaining "ln x". Link to comment Share on other sites More sharing options...
tang_tito Posted June 6, 2004 Share Posted June 6, 2004 So it would appear that you have to do some "nested integration" to break down the last remaining "ln x". what do you mean? Link to comment Share on other sites More sharing options...
eaglebtc Posted June 6, 2004 Author Share Posted June 6, 2004 THe integration of "ln x" is an integral within an integral. I used to be a programmer, so sometimes I still think in terms of that. Nesting is when you layer a series of commands within another. Like nested if statements. if { blah == bleh elseif { sex == drugs elseif { rock == roll } } } That's a nested if statement, though it is very crude and does nothing. Thank you, sir. VVVV Link to comment Share on other sites More sharing options...
rumbleph1$h Posted June 6, 2004 Share Posted June 6, 2004 yeah.........its just the 'chain rule' as they say in calculus :) Link to comment Share on other sites More sharing options...
Guest Posted June 6, 2004 Share Posted June 6, 2004 The use of integration is to find the area under a graph - useful for displacement/time graphs etc The use of differentiation is to find the rate at which something changes. e.g. Differentiating VELOCITY will give you ACCELERATION Differentiating DISPLACEMENT will give you VELOCITY And Integrating these will give you the opposite (e.g. Integrating Acceleration will give you Velocity (+ a constant)) Link to comment Share on other sites More sharing options...
Recommended Posts