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Thread: 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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    $\displaystyle -x=u$

    $\displaystyle dx=-du$

    $\displaystyle x=0\Rightarrow u=0$

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

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

    $\displaystyle \begin{aligned}
    \frac{d}{dx}(-1)e^{-x}&=\frac{d}{dx}(-1)e^{(-1)x}\\
    &=(-1)\frac{d}{dx}e^{(-1)x}\\
    &=(-1)(-1)e^{(-1)x}\\
    &=1\cdot e^{(-1)x}\\
    &=e^{(-1)x}\\
    &=e^{-x}.
    \end{aligned}$

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

    Alternatively, we could let

    $\displaystyle
    \begin{aligned}
    u&=-x\\
    du&=-dx
    \end{aligned}$

    to obtain

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