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Math Help - Need Some Reassurance

  1. #1
    Junior Member
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    Need Some Reassurance

    My professor gave the class a problem which is going to be on our exam and I want to make sure I get full credit for this problem. It's an improper integral, it is.....


    I=Integrate(X*e^(-2x)) at (0,+infinity)


    If someone could just tell me what integration technique needs to be used to solve this. I know it can be done using a formula, but i would rather show more work then that to ensure i get full credit for this problem. I tried integration by parts but it just seems to make the equation more complicated. Thanks in advance. John
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  2. #2
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    Quote Originally Posted by TreeMoney
    I=Integrate(X*e^(-2x)) at (0,+infinity)
    This is what ya got,
    \int_{0}^{\infty}xe^{-2x}dx
    Which by definition is the limit,
    \lim_{N\to\infty}\int_0^Nxe^{-2x}dx
    Calculating the integral requires integration by parts.
    Given,
    \int xe^{-2x}dx
    Let,
    u=x then, u'=1
    v'=e^{-2x} then, v=-1/2e^{-2x}
    Thus we have,
    -1/2xe^{-2x}+1/2\int e^{-2x}dx
    This is,
    -1/2xe^{-2x}-1/4e^{-2x}
    Thus,
    \left \lim_{N\to\infty}\int_0^Nxe^{-2x}dx=\lim_{N\to\infty} -1/2xe^{-2x}-1/4e^{-2x} \right|^N_0
    Thus,
    \lim_{N\to\infty} -1/2Ne^{-2N}-1/4e^{-2N}+1/2(0)e^{-2\cdot 0}+1/4e^{-2\cdot 0}
    Remember that an exponent always overtakes a polynomial thus,
    0-0+0+1/4=1/4
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  3. #3
    Super Member malaygoel's Avatar
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    Quote Originally Posted by TreeMoney
    My professor gave the class a problem which is going to be on our exam and I want to make sure I get full credit for this problem. It's an improper integral, it is.....


    I=Integrate(X*e^(-2x)) at (0,+infinity)


    If someone could just tell me what integration technique needs to be used to solve this. I know it can be done using a formula, but i would rather show more work then that to ensure i get full credit for this problem. I tried integration by parts but it just seems to make the equation more complicated. Thanks in advance. John
    I am not sure
    \int xe^{-2x}dx= \frac{xe^{-2x}}{-2}  -\int \frac{e^{-2x}}{-2}\frac{dx}{dx}dx
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    This is what ya got,
    \int_{0}^{\infty}xe^{-2x}dx
    Which by definition is the limit,
    \lim_{N\to\infty}\int_0^Nxe^{-2x}dx
    Calculating the integral requires integration by parts.
    No it does not, it requires a bit of observation and the fundamental
    theorem of calculus.

    \frac{d}{dx}\ e^{-2x^2}=-4xe^{-2x^2},

    so:

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

    Hence, with a bit of limiting jiggery-pokery:

    \int_{0}^{\infty}xe^{-2x}dx=\frac{1}{4}

    RonL
    Last edited by CaptainBlack; June 18th 2006 at 09:30 PM.
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  5. #5
    Super Member malaygoel's Avatar
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    Quote Originally Posted by CaptainBlack
    No it does not, it requires a bit of observation and the fundamental
    theorem of calculus.

    \frac{d}{dx}\ e^{-2x^2}=-4xe^{-2x^2},

    so:

    \int xe^{-2x}dx=-\frac{1}{4}\int -4xe^{-2x^2}\ dx
    Observe the \mbox{power of e} in the last line of the quoted work.
    Keep Smiling
    Malay
    Last edited by malaygoel; June 19th 2006 at 12:55 AM.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by malaygoel
    Observe the \mbox{power of e}
    Keep Smiling
    Malay
    Yes, I have observed it, now can you elaborate on what I should
    be looking for?


    RonL
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack
    Yes, I have observed it, now can you elaborate on what I should
    be looking for?


    RonL
    OK I see it now the OP was asking for

    <br />
\int_{0}^{\infty}xe^{-2x}dx<br />

    I had misread the power that e is raised to

    RonL
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