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Thread: improper integral

  1. #1
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    improper integral

    need a little help starting this problem:
    evalute:
    $\displaystyle \int_{1}^{\infty} x e^{-x}dx$
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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:
    $\displaystyle \int_{1}^{\infty} x e^{-x}dx$
    Let $\displaystyle u=x \mbox{ and }v'=e^{-x}$

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

    Thus,

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

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

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

    I got a different answer . . .


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

    . . $\displaystyle \begin{array}{ccccccc}u & = & x & \quad & dv & = & e^{-x}dx \\
    du & = & dv & \quad & v & = & -e^{-x} \end{array}$

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


    Evaluate from $\displaystyle 1$ to $\displaystyle \infty$

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

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

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

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

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

    Thus,

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

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

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

    I got a different answer . . .


    Integrate by parts . . .

    . . $\displaystyle \begin{array}{ccccccc}u & = & x & \quad & dv & = & e^{-x}dx \\
    du & = & dv & \quad & v & = & -e^{-x} \end{array}$

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


    Evaluate from $\displaystyle 1$ to $\displaystyle \infty$

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

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

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

    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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