# 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. $\displaystyle -x=u$

$\displaystyle dx=-du$

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

$\displaystyle x=1\Rightarrow u=-1$

3. 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

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