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Math Help - help with integrating e^-x over the interval [0,1]

  1. #1
    Junior Member LexiRae's Avatar
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    help with integrating e^-x over the interval [0,1]

    integrate e^-xdx over the interval [0,1]
    let u=?
    du= ?
    when x= 0, u=?
    when x = 1, u = ?

    thanks!
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  2. #2
    MHF Contributor red_dog's Avatar
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    -x=u

    dx=-du

    x=0\Rightarrow u=0

    x=1\Rightarrow u=-1
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  3. #3
    Senior Member
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    First, you may note that \frac{d}{dx}e^x=e^x, and therefore that \frac{d}{dx}e^{kx}=ke^{kx} by the Chain Rule.

    When k=-1 and f'(x)=e^{-x}, we want the derivative f'(x) to end up with the positive coefficient 1 instead of -1. To do this, we add a -1 as a coefficient to the antiderivative so that we will end up with 1:

    \begin{aligned}<br />
\frac{d}{dx}(-1)e^{-x}&=\frac{d}{dx}(-1)e^{(-1)x}\\<br />
&=(-1)\frac{d}{dx}e^{(-1)x}\\<br />
&=(-1)(-1)e^{(-1)x}\\<br />
&=1\cdot e^{(-1)x}\\<br />
&=e^{(-1)x}\\<br />
&=e^{-x}.<br />
\end{aligned}

    The general antiderivative is therefore f(x)=(-1)e^{-x}+C=-e^{-x}+C.

    Alternatively, we could let

    <br />
\begin{aligned}<br />
u&=-x\\<br />
du&=-dx<br />
\end{aligned}

    to obtain

    -\int_0^{-1} e^u\,du =\int_{-1}^0 e^u\,du.
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