# Thread:

1. ## 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!

2. $-x=u$

$dx=-du$

$x=0\Rightarrow u=0$

$x=1\Rightarrow u=-1$

3. 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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