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Math Help - Evaluating an Integral

  1. #1
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    Evaluating an Integral

    Please help me evaluate

    e^(x)^1/2/ (x^1/2)

    I started out with u substitution. First I let radical x be u, but I couldn't work it out properly because I didn't know what to cross out. Please give me some insight on this!
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  2. #2
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    Quote Originally Posted by kikiya View Post
    Please help me evaluate

    e^(x)^1/2/ (x^1/2)

    I started out with u substitution. First I let radical x be u, but I couldn't work it out properly because I didn't know what to cross out. Please five me some insight on this!
    Your substitution is correct.

    \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx

    u=\sqrt{x} \implies du=\frac{dx}{2\sqrt{x}}

    If you sub this in you should get

    \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=2\int e^{u}du
    Last edited by TheEmptySet; February 25th 2011 at 11:40 AM. Reason: derivative error
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  3. #3
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    You're substitution is correct, what did you find \displaystyle \frac{du}{dx} to be?

    Edit: TES beat me to the punch.
    Last edited by pickslides; February 25th 2011 at 11:57 AM.
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    Your substitution is correct.

    \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx

    u=\sqrt{x} \implies du=\frac{dx}{\sqrt{x}}

    If you sub this in you should get

    \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=\int e^{u}du

    I get

    \displaystyle u=\sqrt{x} \implies du=\frac{dx}{2\sqrt{x}}

    \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=2\int e^{u}du
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  5. #5
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    Quote Originally Posted by pickslides View Post
    I get

    \displaystyle u=\sqrt{x} \implies du=\frac{dx}{2\sqrt{x}}

    \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=2\int e^{u}du
    and you are correct!
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  6. #6
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    I'm sorry, but where did you get the 2 that's outside of the integral? Please explain your process lol
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  7. #7
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    O.K given \displaystyle u = \sqrt{x}what do you get for \displaystyle \frac{du}{dx} ?
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  8. #8
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    Sorry I'm late but is it 1/ 2 (x^1/2) ?
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  9. #9
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    Exactly. Unfortunately, there isn't a 1/2 inside the integral, so you can't substitute du = \frac{1}{2\sqrt{x}}dx yet. Fortunately, you can place a 1/2 inside the integral as long as you place a 2 outside the integral (to balance the 1/2). Now, the problem looks like:

    2\int \frac{e^{\sqrt{x}}}{2\sqrt{x}} \, dx

    Substituting u = \sqrt{x} and du = \frac{1}{2\sqrt{x}}dx gives:

    2\int e^u \, du
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  10. #10
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    Quote Originally Posted by kikiya View Post
    Please help me evaluate

    e^(x)^1/2/ (x^1/2)
    Strictly speaking, what you wrote here was
    \frac{\left(e^x)^{1/2}}{x^{1/2}}= \frac{e^{\frac{x}{2}}}{x^{1/2}}
    but apparently you meant
    e^{x^{1/2}}}{x^{1/2}}
    which would be e^(x^(1/2))/x^(1/2)

    I started out with u substitution. First I let radical x be u, but I couldn't work it out properly because I didn't know what to cross out. Please give me some insight on this!
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  11. #11
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    Oh, I see the error! Thank you!
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  12. #12
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    Thank you for walking me through it! I was confiused about the balancing out, but now I get it! Thank you!
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