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.

Substitute t = root(x)

 

Then integrate the resulting integral in t by parts.

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, 201411 yr by Function

http://www.youtube.com/watch?v=P-o6iFzJLpw

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, 201411 yr by Function

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.

 

Thanks! I'll let you know tomorrow how it went.

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

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.