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Math Help - Could somebody please help me integrate this please.

  1. #1
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    Could somebody please help me integrate this please.

    f(x)=40*e^(-0.5x)
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  2. #2
    Member McScruffy's Avatar
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    Quote Originally Posted by chevy900ss View Post
    f(x)=40*e^(-0.5x)
    It should be -80e^{\frac{-1}{2}x}+C. When you u-substitute for -1/2x.
    Last edited by McScruffy; July 26th 2009 at 03:11 PM.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by chevy900ss View Post
    f(x)=40*e^(-0.5x)
    Let u=-0.5x, then du=-0.5dx\Rightarrow\int{40}e^{-0.5}dx=\int\frac{40}{-0.5}e^udu=-80\int{e}^udu=-80e^{-0.5}+C
    Last edited by VonNemo19; July 26th 2009 at 02:50 PM.
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  4. #4
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    To determine how much of a drug is absorbed into the body, researchers measure the difference between the dosage H and the amount of the drug excreted from the body. The total amount excreted is found by integrating the excretion rate r(t) from 0 to infinity. Therefore, the amount of the drug absorbed by the body is
    H Integral (r(t)dt) from 0 to infinity_
    If the initial dose is H = 200 milligrams (mg), and the excretion rate is r(t)=40e^(-0.5t)mg per hour, find the amount of the drug absorbed by the body.


    Is the answer 120 mg?
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  5. #5
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    Quote Originally Posted by chevy900ss View Post
    To determine how much of a drug is absorbed into the body, researchers measure the difference between the dosage H and the amount of the drug excreted from the body. The total amount excreted is found by integrating the excretion rate r(t) from 0 to infinity. Therefore, the amount of the drug absorbed by the body is
    H Integral (r(t)dt) from 0 to infinity_
    If the initial dose is H = 200 milligrams (mg), and the excretion rate is r(t)=40e^(-0.5t)mg per hour, find the amount of the drug absorbed by the body.


    Is the answer 120 mg?
    Yes.

    CB
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  6. #6
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    Thanks for all the help.
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