Ceasium Posted October 20, 2013 Share Posted October 20, 2013 Hi, I need to integrate the following function: f(x) = 1/x^{2} * e^{1/x} Where x does not equal 0. Determine a number a < 0 such that: The integral of a till 0 f(x)dx = f(a) What the question aks is on other words: Find a number a, such that the number a equals the surface under the graph and its position on the x-axis. All that remains for me is to determine the integral of the function, I have tried integration by parts, substitution with u = e^{1/x} , and improper intergrals, but none seemed to work. (I was working in loops, or making the integral even more difficult) Can someone help me in the right direction? Link to comment Share on other sites More sharing options...

Cap'n Refsmmat Posted October 21, 2013 Share Posted October 21, 2013 Maybe try u-substituting u = 1/x. Link to comment Share on other sites More sharing options...

Ceasium Posted October 21, 2013 Author Share Posted October 21, 2013 (edited) Thanks for your reply. It got me somewhere, yet I am stuck again on the unsolvable part. I've attached my working out here, and on the second page I try to work out the integral, but then come across a problem. Can you see where my computation went wrong? Or should I try a whole different approach? -edit: now attached the files math hw 5 calc 5 (1).pdf Math hw 5 calc 1 (2).pdf Edited October 21, 2013 by Ceasium Link to comment Share on other sites More sharing options...

daniton Posted October 21, 2013 Share Posted October 21, 2013 (edited) I don't know how you do it but it is wrong. Use u= 1/x the substitute this and dx the do the integrations it is pretty easy. Edited October 21, 2013 by daniton Link to comment Share on other sites More sharing options...

Cap'n Refsmmat Posted October 21, 2013 Share Posted October 21, 2013 It is not true that if u=1/x, then [math]\int \frac{e^{1/x}}{x^2} \, dx = \int u^2 e^u \, du[/math] You need to figure out what du is in terms of dx. Link to comment Share on other sites More sharing options...

overtone Posted October 21, 2013 Share Posted October 21, 2013 Is visual recognition of a chain rule derivative an allowed suggestion? Seems like the plethora of techniques available has led to confusion - KISS. You just want something that gives you that derivative - doesn't matter how you find it. Trial and error is a perfectly good approach. Link to comment Share on other sites More sharing options...

Ceasium Posted October 21, 2013 Author Share Posted October 21, 2013 I don't know how you do it but it is wrong. Use u= 1/x the substitute this and dx the do the integrations it is pretty easy. It is not true that if u=1/x, then [math]\int \frac{e^{1/x}}{x^2} \, dx = \int u^2 e^u \, du[/math] You need to figure out what du is in terms of dx. I used your approach, and it worked, see the attached files. But if I now try to evaluate the integral on the given points, all I get is nonsense. If I evaluate the integral on the given numbers, I get a non- true answer. (see attached files in my previous post). Can someone see what I am doing wrong whilst evaluating the limits Link to comment Share on other sites More sharing options...

Cap'n Refsmmat Posted October 21, 2013 Share Posted October 21, 2013 Where did you use it? In the files attached to post #3, you did your u-substitution incorrectly, which forced you to use integration by parts. You should not have to integrate by parts. Remember: if u=1/x, then du=1/x^{2}dx. You need to account for this when you replace dx with du. Link to comment Share on other sites More sharing options...

Ceasium Posted October 23, 2013 Author Share Posted October 23, 2013 Thanks, that was the step that I needed to solve it correctly Link to comment Share on other sites More sharing options...

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