Function Posted June 1, 2014 Posted June 1, 2014 Hello everyone Friday is part one of my maths exam.. One I can't seem to solve: [math]\int{e^{\sqrt{x}}dx}[/math] Could anyone help me? Thanks! F.
studiot Posted June 1, 2014 Posted June 1, 2014 Substitute t = root(x) Then integrate the resulting integral in t by parts.
Function Posted June 1, 2014 Author Posted June 1, 2014 (edited) Tried doing that, but expressing it with dt wasn't making it a bit simpler: [math]t=\sqrt{x}[/math] [math]\Leftrightarrow dt=\frac{dx}{2\sqrt{x}}[/math] [math]\Leftrightarrow dx=2\sqrt{x}dt[/math] [math]\int{\cdots}=\int{2e^{t}\sqrt{x}dt}=\int{2e^{t}tdt}[/math] (stupid.. could've just replaced the x in dx by t²) [math]\int{\cdots}=2\int{te^{t}dt}[/math] Partial integration: [math]\int{\cdots}=2\int{td\left(e^t\right)}[/math] [math]\int{\cdots}=2\cdot \left(t\cdot e^t-\int{e^t dt}\right)[/math] [math]\int{\cdots}=2\cdot\left(t\cdot e^t-e^t\right)+C[/math] [math]\int{\cdots}=2te^{t}-2e^{t}=2e^{t}(t-1)+C[/math] [math]\int{\cdots}=2e^{\sqrt{x}}\left(\sqrt{x}-1\right)+C[/math] With [math]C[/math] any constant the result may differ from in comparison to other possible results. Correct? Edited June 1, 2014 by Function
Function Posted June 1, 2014 Author Posted June 1, 2014 (edited) Glad to see it's correct. Can't say that I completely recognize the method the man uses in his video, but well.. Everyone has his own ways of solving problems, right.. Edited June 1, 2014 by Function
studiot Posted June 1, 2014 Posted June 1, 2014 It's the same method, just he used u for substitution, then found that u appears in the normal parts formula and regretted it. That was why I suggested t. Have a good exam.
Function Posted June 5, 2014 Author Posted June 5, 2014 Thanks! I'll let you know tomorrow how it went.
kenz Posted June 17, 2014 Posted June 17, 2014 Let t = sqrt(x) So dx = 2.t.dt Integrate 2 . e^t . t . dt u = 2t dv = e^t dt -> du = 2dt v = e^t -> integral(2e^t t dt) = 2 t e^t - integral(2 e^t dt) = 2 t e^t - 2 e^t = e^t (t-1) . 2 , t = sqrt_x as above
burgess Posted July 1, 2014 Posted July 1, 2014 Hi Function, Your procedure is absolutely correct, and one more addition for that, if you want to find out the 'C' value you can take limits for the integration and find out the C value.
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