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Math Help - integral with e to the sq. x

  1. #1
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    integral with e to the sq. x

    Hi,

    To solve the int. 1 to 4 e to the sq. x dx you would first do a u substitution, right?

    u=e to the x to the 1/2

    du=e to the x(1-2x to the -1/2)

    Is this correct?

    Thank you
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    Quote Originally Posted by chocolatelover View Post
    Hi,

    To solve the int. 1 to 4 e to the sq. x dx you would first do a u substitution, right?

    u=e to the x to the 1/2

    du=e to the x(1-2x to the -1/2)

    Is this correct?

    Thank you
    note that \int e^{\sqrt {x}}~dx = \int \frac {2 \sqrt {x}}{2 \sqrt {x}}e^{\sqrt {x}}~dx

    now use the substitution, u = \sqrt {x}
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  3. #3
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    How did you know that? Is it a rule?

    Would du be e to sq. x(1/2x to the -1/2)dx??????????

    Thank you
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  4. #4
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    \int e^{\sqrt x}\,dx

    u=\sqrt x\implies du=\frac1{2\sqrt x}\,dx, the integral becomes to

    \int e^{\sqrt x}\,dx=2\int ue^u\,du

    Next step, it's integration by parts.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post


    How did you know that? Is it a rule?

    Would du be e to sq. x(1/2x to the -1/2)dx??????????

    Thank you
    an explanation is given by TPH in the second post here

    EDIT: Ah, I see Krizalid already answered
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  6. #6
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    Would u=e to sq. x and dv=tdt????
    du=e to the sq. x(1/2x to -1/2)
    v=t2/2

    Thank you
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  7. #7
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    I suggest you get practice on LaTeX, not hard.

    This way, it'll be better to check your answers.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Would u=e to sq. x and dv=tdt????
    du=e to the sq. x(1/2x to -1/2)
    v=t2/2

    Thank you
    what are you talking about? the integral you are doing by parts is 2 \int u e^u~du. where did \sqrt {x} come from? pay attention
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  9. #9
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    I got

    u=e to the u
    dv=u
    du=e to the u
    v=u2/2

    then (e to the u)u2/2-2int.1 to 4(u2/2)e to the u

    Now I can just plug in 1 and 4 right?
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  10. #10
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    Better you set y=u & dv=e^u\,du
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Better you set y=u & dv=e^u\,du
    yeah, the u's will get mixed up. use t for the first substitution.

    so you want to find 2 \int t e^t~dt

    now go ahead



    EDIT: and if you won't use LaTex, please write in math shorthand as i showed you in one of your earlier posts. reading through your work is annoying when you use sentences to describe math symbols
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  12. #12
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    I got ue to the u-2int. 1 to 4 e to the u u2/2

    Then I would do it again, right?

    Thank you
    Last edited by chocolatelover; September 3rd 2007 at 07:04 PM.
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  13. #13
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    Is the final answer ueu-16e4-6e
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  14. #14
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    Quote Originally Posted by chocolatelover View Post
    Is the final answer ueu-16e4-6e
    the final answer is 2 e^{\sqrt {x}}(\sqrt {x} - 1)
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  15. #15
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    Quote Originally Posted by chocolatelover View Post
    int. 1 to 4 e to the sq. x dx you would first do a u substitution, right?
    Jhevon, it's a definite one
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