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

**LexiRae** · April 5th 2009, 06:01 AM

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

thanks!

**red_dog** · April 5th 2009, 06:18 AM

$-x=u$

$dx=-du$

$x=0\Rightarrow u=0$

$x=1\Rightarrow u=-1$

**Scott H** · April 5th 2009, 06:22 AM

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

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

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