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Math Help - improper integration

  1. #1
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    improper integration

    I am trying to do this
    integral sign going from 0 to 4 (0 on the bottom, 4 on the top) 1/(4-x) dx
    i got stuck.
    Thanks a lot!!!
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  2. #2
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    \int_0^4\frac1{4-x}~dx=\lim_{a\to4}\int_0^a\frac1{4-x}~dx
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    \int_0^4\frac1{4-x}~dx=\lim_{a\to4}\int_0^a\frac1{4-x}~dx
    Yeah i was able to get that far, but not so sure how to solve it.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by davecs77 View Post
    Yeah i was able to get that far, but not so sure how to solve it.
    If you got that far, what's the problem?


    \lim_{a \to 4} \int_0^a \frac {1}{4 - x}~dx

    = \lim_{a \to 4} \left[ - \ln (4 - x) \right]_{0}^{a}

    = \lim_{a \to 4} \left( - \ln (4 - a) + \ln 4 \right)

    Now take the limit
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