anto Posted May 12, 2007 Report Posted May 12, 2007 Hi, well I'm having some trouble on some integration by parts homework. I can't figure out how to integrate e^(root x), it will be of great help if you guys could give me some hints here, like what u or v are etc. Quote
Jay-qu Posted May 13, 2007 Report Posted May 13, 2007 are you told to do it by parts or can you use any method? Quote
anto Posted May 13, 2007 Author Report Posted May 13, 2007 well this is for a presentation i need to make in which I need to create 4 problems with their respective answers (and procedure to get the answer). This problem is straight from the integration by parts section of the book though so it should be solvable that way Quote
Nootropic Posted May 13, 2007 Report Posted May 13, 2007 what do you mean by "e^(root x)", that's relatively general and could have a lot of interpretations. Quote
Jay-qu Posted May 13, 2007 Report Posted May 13, 2007 he means [math] \int{e^{\sqrt{x}}}dx[/math] Quote
ronthepon Posted May 13, 2007 Report Posted May 13, 2007 It's possile that you've got to express sqrt(x) at 't' or something, and since that develops another constant nearby, you can integerate by parts. Try taking [math]\sqrt{x}[/math] as t. I think that'll do it. Quote
anto Posted May 13, 2007 Author Report Posted May 13, 2007 yeah it should work, I'm going to try and post some steps here so you guys can correct me if i make a mistake.t = root xdt = 1/2root x dxwhen I replace dt for dx my integral would now look like: e^t (2 root x) dt then I need to replace root x for te^t (2t) dtIntegration by Parts = uv - (integral of )vduu = 2t dv = e^t dtdu = 2dt v = e^t 2t (e^t) - 2e^t2e^t ( t - 1) Substituting:2e^root x ( rootx-1) Please post some comments/corrections :) Quote
max4236 Posted May 15, 2007 Report Posted May 15, 2007 differentiate (2e^rootx)*(rootx-1) [2(rootx-1)*e^rootx]'2(rootx-1)*(e^rootx)' + (2(rootx-1))'*e^rootx[ 2(rootx-1)*e^rootx*1/(2rootx) ]+[2/(2rootx) * e^rootx]e^rootx * [(rootx-1)/rootx+ 1/rootx]e^rootx * [1-1/rootx+1/rootx]e^rootx yep, looks good Quote
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