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Math Help - Integrals for exam

  1. #1
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    Integrals for exam

    Hey I have an exam in 2 days and I really need some help gettin these questions under control!

    integral (x^2*e^-x)dx
    I am not quite sure where to start, I was thinkin that I need to use the substitution where u=x^2
    but then I am just confused after that!

    integral (2x+4)/(x^3-2x^2)dx
    For this one I factorised and bottom line so I had
    integral (2x+4)/(x^2)(x-2) dx

    Then I sed that (2x+4)/(x^2)(x-2)= A/x+b/x+c/x-2

    Then I didnt really know wat to do after that!

    3. The functions S and T are defined by setting
    S(x):= (integral x^2 to 0) sqrt(1+t^2)dt

    I tried doin this one but when I substituted for 0 later on I got that it was undefined...

    If someone could show the workings or at least the start of these problems I would be much appreciative... then I would be able to use these as examples for the other practice exams!
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  2. #2
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    Quote Originally Posted by taryn View Post
    Hey I have an exam in 2 days and I really need some help gettin these questions under control!

    integral (x^2*e^-x)dx
    I am not quite sure where to start, I was thinkin that I need to use the substitution where u=x^2
    but then I am just confused after that!
    Integration by parts.
    \int x^2 e^{-x} dx
    Let, u=x^2 and v'=e^{-x}
    Thus,
    u'=2x and v=-e^{-x}
    Thus,
    uv-\int u'vdx
    Thus, (note the signs change )
    -x^2e^{-x}+2\int xe^{-x} dx
    Do the same integration by parts on this integral.
    u=x and v'=e^{-x}
    Thus,
    u'=1 and v=-e^{-x}
    Thus, (note the parantheses )
    -x^2e^{-x}+2\left( -xe^{-x}+\int e^{-x} dx\right)
    Thus,
    -x^2e^{-x}-2xe^{-x}-2e^{-x}+C
    Thus,
    -e^{-x}(x^2+2x+2)+C
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Integration by parts.
    \int x^2 e^{-x} dx
    Let, u=x^2 and v'=e^{-x}
    Thus,
    u'=2x and v=-e^{-x}
    Thus,
    uv-\int u'vdx
    Thus, (note the signs change )
    -x^2e^{-x}+2\int xe^{-x} dx
    [/tex]
    This is the only bit I am a little unsure of, Is there a reason y u can just take the uv part out of the integral?
    uv-\int u'vdx

    Thanks for ur help!
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  4. #4
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    Quote Originally Posted by taryn View Post
    This is the only bit I am a little unsure of, Is there a reason y u can just take the uv part out of the integral?
    uv-\int u'vdx

    Thanks for ur help!
    I do not understand.
    If you are asking whether you can do this,
    \int x^2 e^x dx=x^2\int e^x dx
    No because x^2 is not a constant function throughout the interval of integration.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    Integration by parts.
    \int x^2 e^{-x} dx
    Let, u=x^2 and v'=e^{-x}
    Thus,
    u'=2x and v=-e^{-x}
    Thus,
    uv-\int u'vdx
    Thus, (note the signs change )
    -x^2e^{-x}+2\int xe^{-x} dx
    Do the same integration by parts on this integral.
    u=x and v'=e^{-x}
    Thus,
    u'=1 and v=-e^{-x}
    Thus, (note the parantheses )
    -x^2e^{-x}+2\left( -xe^{-x}+\int e^{-x} dx\right)
    Thus,
    -x^2e^{-x}-2xe^{-x}-2e^{-x}+C
    Thus,
    -e^{-x}(x^2+2x+2)+C
    Quote Originally Posted by ThePerfectHacker View Post
    I do not understand.
    If you are asking whether you can do this,
    \int x^2 e^x dx=x^2\int e^x dx
    No because x^2 is not a constant function throughout the interval of integration.
    No I know thats not possible... I just mean how can u just take the uv out of the integral!
    Y are u able to do that! Obviously I know that x^2 is not a constant function throughout the interval... so does that mean u are sayin that uv is so thats y u can take it out?
    Is there a web site or somethin that I can read to help m e understand wat u have done?
    Last edited by taryn; November 20th 2006 at 04:44 PM. Reason: Looked like I was referencing to x^2
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  6. #6
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    Quote Originally Posted by taryn View Post
    No I know thats not possible... I just mean how can u just take the uv out of the integral!
    Y are u able to do that! Obviously I know that x^2 is not a constant function throughout the interval... so does that mean u are sayin that uv is so thats y u can take it out?
    Is there a web site or somethin that I can read to help m e understand wat u have done?
    Wikipedia explains it here.

    But be warned it sometimes like to overcomplicate things (I was told I do that too).

    Basically it is a special rule. You can take out that factor but then you need to change the integral. So you are not really taking the function out of the integral you are used making the integral simpler to work with.
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