gutswarrior Posted November 23, 2008 Report Share Posted November 23, 2008 What should i do to be able to solve this? ∫e^2 dx upper limit is 2lower limit is 0 The answer is 2e^2.I don't how to get it.Answer this immediately.Thank you... Link to comment Share on other sites More sharing options...
Death Metal Posted November 23, 2008 Report Share Posted November 23, 2008 You should give up. Simple as that. Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 i don't know how to give up... i don't know how to do it...please help me... Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 oh the answer to that is simple. Once the integration is complete (but before the limits are put into place), you'd have x*e^2. Now you go from the values of 0 to 2. That means you take x=2 for x*e^2 minus x=- for x*e^2. The first part gives you 2*e^2, the second gives you 0*e^2. Therefore, it is 2*e^2 - 0*e^2 = 2e^2 Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 how can i integrate e^2, their both constants and why i have to place x. Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 you don't integrate e^2, e is just a constant, remember? And you have to place an x there because of the dx at the end of the integration Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 so i bring e^2 outside the integral sign and then i integrate? oh yes, thank you... Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 exactly my point. Remember, e^2 is a constant, not a variable. Anything that's a constant you can bring outside the integration because it can't be changed Link to comment Share on other sites More sharing options...
Bakura Vessal Posted November 23, 2008 Report Share Posted November 23, 2008 um 2 + 2 is 4 Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 don't listen to SOS, gutswarrior. He doesn't understand this process Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 this time now. ∫ [4^ln(1/x)] / x dx u= ln (1/x)du= x the answer the i've got is + something but the answer is -something.please help again Link to comment Share on other sites More sharing options...
Bakura Vessal Posted November 23, 2008 Report Share Posted November 23, 2008 don't listen to SOS' date=' gutswarrior. He doesn't understand this process[/quote'] lol, i was joking. could have been worse i could have said 1 + 1 Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 are you trying to do this via substitution or by parts? Link to comment Share on other sites More sharing options...
Death Metal Posted November 23, 2008 Report Share Posted November 23, 2008 um 2 + 2 is 4 Funniest thing i've seen all day. =D Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 i think i've do it by parts.. Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 It's been a LONG time since I've done that. First tell me what you chose for u and what you chose for dv Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 i remember, my answer is 4^ln (1/x) / ln 4 but the answer is negative of thatmy u is ln (1/x)and its du is x because ln u = 1/u (du) Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 I'm a little unsure. From what I see you're doing it by substitution, not by parts because substitution only requires u, parts requires u and dv Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 so what should i do now? my positive answer and that negative answer... Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 first I need to know if you're sure this has to be done by parts or not Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 i don't know what is by parts? my professor only uses u and du that's it.. Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 then that's substitution. And your du is wrong. You said u = ln(1/x). If that's true then du = -1/x Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 why is it negative? Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 think 1/x as the same thing as x^-1 Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 oh yes... i got it now... thanks for your help... Link to comment Share on other sites More sharing options...
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