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  1. #1
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    evaluate

    Given that y=2xe^{2x-1} and \frac{dy}{dx}=(4x+2)(e^{2x-1}), hence evaluate \int^1_0[2xe^{2x-1}]dx
    Last edited by Punch; October 8th 2010 at 05:59 PM.
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  2. #2
    MHF Contributor Amer's Avatar
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    by parts dv = exp(2x-1) u=x

    \displaystyle \int 2x e^{2x-1} \; dx =\frac{2x e^{2x-1}}{2} - 2\int e^{2x-1} dx  = \frac{2xe^{2x-1}}{2} - \frac{2e^{2x-1}}{2}+C

    \displaystyle \int_{0}^{1} 2xe^{2x-1} dx = e^{2-1} - e^{2-1} - ( 0 - e^{0 -1 }) = \frac{1}{e}
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  3. #3
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    answer is 1.54, \frac{1}{e}=3.67 which is wrong
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  4. #4
    MHF Contributor harish21's Avatar
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    Use integration by parts:

    Let u = 2x and dv=e^{2x-1}

    so, du = 2dx and  v = \frac{e^{2x-1}}{2}

    Now,

     \int 2x e^{2x-1} dx = 2x.\frac{1}{2} e^{2x-1} - \int \frac{1}{2}e^{2x-1}2dx

     = x(e^{2x-1}) - \frac{e^{2x-1}}{2}

     \int_{0}^{1} 2xe^{2x-1} dx = \left[ e - \frac{e}{2} \right] - \left[0 - \frac{1}{2e}\right] = \frac{e}{2} + \frac{1}{2e} = 1.543
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  5. #5
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    Quote Originally Posted by Punch View Post
    Given that y=2xe^{2x-1} and \frac{dy}{dx}=(4x+2)(e^{2x-1}), hence evaluate \int^1_0[2xe^{2x-1}]dx
    Don't use integration by parts, you're asked to use the information given.

    \displaystyle{\frac{dy}{dx} = (4x + 2)e^{2x - 1}}

    \displaystyle{\frac{dy}{dx} = 4x\,e^{2x - 1} + 2e^{2x - 1}}


    You should know \displaystyle{y = \int{\frac{dy}{dx}\,dx}}, so

    \displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x - 1} + 2e^{2x - 1}\,dx}}

    \displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x - 1}\,dx} + \int{2e^{2x - 1}\,dx}}

    \displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x- 1}\,dx} + e^{2x - 1}}

    \displaystyle{2x\,e^{2x - 1} - e^{2x - 1} = \int{4x\,e^{2x - 1}\,dx}}

    \displaystyle{2x\,e^{2x - 1} - e^{2x - 1} = 2\int{2x\,e^{2x - 1}\,dx}}

    \displaystyle{x\,e^{2x - 1} - \frac{e^{2x - 1}}{2} = \int{2x\,e^{2x - 1}\,dx}}.


    Now evaluate \displaystyle{\int_0^1{2x\,e^{2x - 1}\,dx}}.
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  6. #6
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Amer View Post
    by parts dv = exp(2x-1) u=x

    \displaystyle \int 2x e^{2x-1} \; dx =\frac{2x e^{2x-1}}{2} - 2\int e^{2x-1} dx  = \frac{2xe^{2x-1}}{2} - \frac{2e^{2x-1}}{2}+C

    \displaystyle \int_{0}^{1} 2xe^{2x-1} dx = e^{2-1} - e^{2-1} - ( 0 - e^{0 -1 }) = \frac{1}{e}
    The answer came out to be wrong here because Amer supposed dv = exp(2x-1),, which gives v = \frac{exp(2x-1)}{2}

    and applying integration by parts would give:

     uv - \int v. du

    \displaystyle \int 2x e^{2x-1} \; dx =\frac{2x e^{2x-1}}{2} - 2\int \frac{e^{2x-1}}{2} dx  = \frac{2xe^{2x-1}}{2} - \frac{2e^{2x-1}}{4}+C
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  7. #7
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    Quote Originally Posted by Prove It View Post
    Don't use integration by parts, you're asked to use the information given.

    \displaystyle{\frac{dy}{dx} = (4x + 2)e^{2x - 1}}

    \displaystyle{\frac{dy}{dx} = 4x\,e^{2x - 1} + 2e^{2x - 1}}


    You should know \displaystyle{y = \int{\frac{dy}{dx}\,dx}}, so

    \displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x - 1} + 2e^{2x - 1}\,dx}}

    \displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x - 1}\,dx} + \int{2e^{2x - 1}\,dx}}

    \displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x- 1}\,dx} + e^{2x - 1}}

    \displaystyle{2x\,e^{2x - 1} - e^{2x - 1} = \int{4x\,e^{2x - 1}\,dx}}

    \displaystyle{2x\,e^{2x - 1} - e^{2x - 1} = 2\int{2x\,e^{2x - 1}\,dx}}

    \displaystyle{x\,e^{2x - 1} - \frac{e^{2x - 1}}{2} = \int{2x\,e^{2x - 1}\,dx}}.


    Now evaluate \displaystyle{\int_0^1{2x\,e^{2x - 1}\,dx}}.
    after evaluating, ans is 5.436 which is still wrong

    Also, I did understand this part!!

    from \displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x- 1}\,dx} + e^{2x - 1}}

    to \displaystyle{2x\,e^{2x - 1} - e^{2x - 1} = \int{4x\,e^{2x - 1}\,dx}}

    Lastly, indeed i am asked to use the information given!! pls do not use integration by part thanks!
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  8. #8
    MHF Contributor harish21's Avatar
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    Prove It has already shown that :

      \int{2x\,e^{2x - 1}\,dx}} = \displaystyle{x\,e^{2x - 1} - \frac{e^{2x - 1}}{2}

    So,

     \displaystyle{\int_0^1{2x\,e^{2x - 1}\,dx}} = \left[ x e^{2x-1} - \frac{e^{2x-1}}{2} \right]_0^1 gives you 1.543


    Regarding your second question:

    \displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x - 1}\,dx} + \int{2e^{2x - 1}\,dx}}

    and,

     \int 2e^{2x - 1} dx= 2 \int e^{2x - 1}dx = 2 \times \frac{e^{2x-1}}{2} = e^{2x-1}

    So,

    \displaystyle{2x\,e^{2x - 1} = \int{4x\,e^{2x - 1}\,dx} +  e^{2x-1}

    then,  e^{2x-1} is subtracted form both the sides to get:

    \displaystyle{2x\,e^{2x - 1} - e^{2x-1} = \int{4x\,e^{2x - 1}\,dx}
    Last edited by harish21; October 8th 2010 at 08:56 PM.
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  9. #9
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    Great!! i love MFH
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