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Math Help - stuck on integral

  1. #1
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    stuck on integral

    my teacher has given us this integration

    \int_0^x \frac{2. \lambda exp(- \lambda.x}{(1+exp(- \lambda.x)^2} dx

    I dont get how this equals

    \int_\frac{1}{2}^{\frac{1}{(1+exp(- \lambda.x)}} 2 du

    the domain of integration can change? can anyone explain this me?
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  2. #2
    MHF Contributor ebaines's Avatar
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    Your equations seem to have a few errors. I think what you meant is this:

    <br />
\int _ 0 ^ X \frac {2 \lambda e ^{- \lambda x}} {(1 + e ^{- \lambda x} ) ^2 } dx<br />

    Is that right? You are changing the variable from x to u using:

    <br />
u = \frac 1 {1 + e^ {-\lambda x}}<br />

    From this you get:
    <br />
x = \frac {-1} {\lambda} ln( \frac {1-u} u), and  dx = \frac {1} {\lambda u (1-u)} du

    When you do this you have to change the limits of integration to conform to this new variable. So where initially you had a lower limit of x = 0, you now have  u = \frac 1 {1 + e^0} = \frac 1 2 . Likewise, the upper end becomes u = \frac 1 {1 + e ^ {- \lambda X}}. And the new integral is:

    <br />
2\int _ {\frac 1 2 } ^ \frac 1 {1 + e ^ {- \lambda X}} du<br />
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  3. #3
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    Quote Originally Posted by notgoodatmath View Post
    my teacher has given us this integration

    \int_0^x \frac{2. \lambda exp(- \lambda.x}{(1+exp(- \lambda.x)^2} dx

    I dont get how this equals

    \int_\frac{1}{2}^{\frac{1}{(1+exp(- \lambda.x)}} 2 du

    the domain of integration can change? can anyone explain this me?

    How the integration variable appears as upper limit in your integral??

    Tonio
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  4. #4
    MHF Contributor ebaines's Avatar
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    I think it's a typo. The OP used a lower case x both as his variable and to define the upper limit. In my response I used upper case X (not lower case x) in the upper llimit, so as to not confuse the two.
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