# Thread: help with integrating e^-x over the interval [0,1]

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}
\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 $f(x)=(-1)e^{-x}+C=-e^{-x}+C$.

Alternatively, we could let


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

to obtain

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