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Math Help - improper integral

  1. #1
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    improper integral

    need a little help starting this problem:
    evalute:
    \int_{1}^{\infty} x e^{-x}dx
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  2. #2
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    Quote Originally Posted by viet View Post
    need a little help starting this problem:
    evalute:
    \int_{1}^{\infty} x e^{-x}dx
    Let u=x \mbox{ and }v'=e^{-x}

    Then u' = 1 \mbox{ and }v=-e^{-x}

    Thus,

    -xe^{-x}\big|_1^{\infty} + \int_1^{\infty} e^{-x} dx

    Note -xe^{-x}\big|_1^{\infty} = \lim_{N\to \infty} - Ne^{-N} + e^{-1} = e^{-1}

    And, \int_1^{\infty} e^{-x} dx = -e^{-x}\big|_1^{\infty} = \lim_{N\to \infty} -e^{-N} + e^{-1} = e^{-1}
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  3. #3
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    Hello, viet!

    I got a different answer . . .


    Evalute: . \int_{1}^{\infty} x e^{-x}dx
    Integrate by parts . . .

    . . \begin{array}{ccccccc}u & = & x & \quad & dv & = & e^{-x}dx \\<br />
du & = & dv & \quad & v & = & -e^{-x} \end{array}

    We have: . -xe^{-x} + \int e^{-x}dx \;=\;-xe^{-x} - e^{-x}\;=\;-\frac{x+1}{e^x}


    Evaluate from 1 to \infty

    . . \underbrace{\lim_{x\to\infty}\left(-\frac{x+1}{e^x}\right)} - \underbrace{\left(-\frac{1 + 1}{e}\right)}

    . . . . = \;\;\;0\qquad\quad-\;\;\left(-\frac{2}{e}\right)

    . . . . . . = \quad\frac{2}{e}

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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    Let u=x \mbox{ and }v'=e^{-x}

    Then u' = 1 \mbox{ and }v=-e^{-x}

    Thus,

    -xe^{-x}\big|_1^{\infty} + \int_1^{\infty} e^{-x} dx

    Note -xe^{-x}\big|_1^{\infty} = \lim_{N\to \infty} - Ne^{-N} + e^{-1} = e^{-1}

    And, \int_1^{\infty} e^{-x} dx = -e^{-x}\big|_1^{\infty} = \lim_{N\to \infty} -e^{-N} + e^{-1} = e^{-1}
    Quote Originally Posted by Soroban View Post
    Hello, viet!

    I got a different answer . . .


    Integrate by parts . . .

    . . \begin{array}{ccccccc}u & = & x & \quad & dv & = & e^{-x}dx \\<br />
du & = & dv & \quad & v & = & -e^{-x} \end{array}

    We have: . -xe^{-x} + \int e^{-x}dx \;=\;-xe^{-x} - e^{-x}\;=\;-\frac{x+1}{e^x}


    Evaluate from 1 to \infty

    . . \underbrace{\lim_{x\to\infty}\left(-\frac{x+1}{e^x}\right)} - \underbrace{\left(-\frac{1 + 1}{e}\right)}

    . . . . = \;\;\;0\qquad\quad-\;\;\left(-\frac{2}{e}\right)

    . . . . . . = \quad\frac{2}{e}

    You both have the same answers. TPH simply did not add the two pieces together at the end.

    -Dan
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  5. #5
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    Use integration by parts
    Attached Thumbnails Attached Thumbnails improper integral-qq.gif  
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  6. #6
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    its easier than i expected thanks guys
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