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  1. #1
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    evaluate

    Given that $\displaystyle y=2xe^{2x-1}$ and $\displaystyle \frac{dy}{dx}=(4x+2)(e^{2x-1})$, hence evaluate $\displaystyle \int^1_0[2xe^{2x-1}]dx$
    Last edited by Punch; Oct 8th 2010 at 04:59 PM.
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  2. #2
    MHF Contributor Amer's Avatar
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    by parts dv = exp(2x-1) u=x

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

    $\displaystyle \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, $\displaystyle \frac{1}{e}=3.67$ which is wrong
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  4. #4
    MHF Contributor harish21's Avatar
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    Use integration by parts:

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

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

    Now,

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

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

    $\displaystyle \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 $\displaystyle y=2xe^{2x-1}$ and $\displaystyle \frac{dy}{dx}=(4x+2)(e^{2x-1})$, hence evaluate $\displaystyle \int^1_0[2xe^{2x-1}]dx$
    Don't use integration by parts, you're asked to use the information given.

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

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


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

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

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

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

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

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

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


    Now evaluate $\displaystyle \displaystyle{\int_0^1{2x\,e^{2x - 1}\,dx}}$.
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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

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

    $\displaystyle \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 $\displaystyle v = \frac{exp(2x-1)}{2}$

    and applying integration by parts would give:

    $\displaystyle uv - \int v. du$

    $\displaystyle \displaystyle \int 2x e^{2x-1} \; dx =\frac{2x e^{2x-1}}{2} - 2\int \frac{e^{2x-1}}{2} dx = \frac{2xe^{2x-1}}{2} - \frac{2e^{2x-1}}{4}+C$
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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.

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

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


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

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

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

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

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

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

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


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

    Also, I did understand this part!!

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

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

    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 :

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

    So,

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


    Regarding your second question:

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

    and,

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

    So,

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

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

    $\displaystyle \displaystyle{2x\,e^{2x - 1} - e^{2x-1} = \int{4x\,e^{2x - 1}\,dx} $
    Last edited by harish21; Oct 8th 2010 at 07:56 PM.
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  9. #9
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    Great!! i love MFH
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