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Thread: Integral

  1. #1
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    Integral

    How do I find the indefinite integral to $\displaystyle \int e^{\sqrt x}dx$?
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  2. #2
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    Consider the following

    $\displaystyle
    \int e^{f( x)}dx = \frac{e^{f( x)}}{f'( x)}+c
    $
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  3. #3
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    I must be an idiot, because I dismissed that idea immediately without even trying!

    Maybe it was the star beside the question that screwed up my mind.
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  4. #4
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    u = √x ---> uČ = x ---> 2 u du = dx
    substitute then use integration by parts ..
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  5. #5
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    Radicals do have a strange way of throwing people off track
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    Differentiate $\displaystyle 2e^{\sqrt x } \sqrt x - 2e^{\sqrt x } $.
    What do you get?
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  7. #7
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    Quote Originally Posted by chengbin View Post
    How do I find the indefinite integral to $\displaystyle \int e^{\sqrt x}dx$?
    $\displaystyle u = \sqrt{t}$

    $\displaystyle du = \frac{1}{2\sqrt{t}} \, dt = \frac{1}{2u} \, dt$

    $\displaystyle dt = 2u \, du$


    $\displaystyle 2\int ue^u \, du$

    integrate by parts ...

    $\displaystyle 2e^u(u - 1) + C$

    $\displaystyle 2e^{\sqrt{t}}(\sqrt{t} - 1) + C$
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  8. #8
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    Quote Originally Posted by pickslides View Post
    Radicals do have a strange way of throwing people off track
    Is this for me ?
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  9. #9
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    Quote Originally Posted by TWiX View Post
    Is this for me ?

    I'd say no, seeing as your method was correct.
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  10. #10
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    Quote Originally Posted by pickslides View Post
    Consider the following

    $\displaystyle
    \int e^{f( x)}dx = \frac{e^{f( x)}}{f'( x)}+c
    $
    that's a wrong formula.
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  11. #11
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    Quote Originally Posted by Krizalid View Post
    that's a wrong formula.
    yep it is, was confused with $\displaystyle y= e^{f(x)} \Rightarrow \frac{dy}{dx}= f'(x)e^{f(x)}$

    Gentle reminder to myself not to post on MHF the morning after a big night on the sauce.
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