another demo pls

$\LARGE \int_{-\infty}^{\infty} e^{-| x |} dx$

Question

another demo pls

$\LARGE \int_{-\infty}^{\infty} e^{-| x |} dx$

anonymous · Answer

@FoolForMath  pls help him!

anonymous · Answer

@blockcolder too :DD

zarkon · Answer

Find

$$\int\limits_{0}^{\infty}e^{-x}dx$$

zarkon · Answer

then multiply by 2

anonymous · Answer

uhmmm let -x = u so du = -dx

$\LARGE \int_{0}^{-t} e^u du$

that right?

zarkon · Answer

you are missing your limit and a negative sign

anonymous · Answer

oh yeah

$\LARGE -1 \lim_{x \rightarrow \infty} \int_{0}^{-t} e^u du$

that about right?

zarkon · Answer

I would just to this

$$\lim_{t\to\infty}\int\limits_{0}^{t}e^{-x}dx$$

$$=\lim_{t\to\infty}\left.-e^{-x}\right|_{0}^{t}=\cdots$$

anonymous · Answer

uhmm i think i know this.... $-e^{-\infty} = -1$ right?

zarkon · Answer

$$\lim_{t\to\infty}-e^{-t}+e^{0}=0+1=1$$

anonymous · Answer

oh...it's zero silly me....

zarkon · Answer

so the final answer is ...

anonymous · Answer

1 is the final answer right?

zarkon · Answer

no

zarkon · Answer

reread my second post

anonymous · Answer

oh oh yeah...that was just from 0 to infinity right?

zarkon · Answer

yes

anonymous · Answer

so the final answer is 2...this is confusing...divergent is when the answer is infinity right?

lalaly · Answer

$$\int\limits_{-a}^{a}(even function)=2\int\limits_{0}^{a}(same function)$$

zarkon · Answer

2 is correct

zarkon · Answer

divergent is when the integral does not coverge to some finite number

anonymous · Answer

oh oh....good to know thanks

zarkon · Answer

for example

$$\int_{0}^{\infty}\sin(x)dx$$

does not converge (and it is not infinity)

anonymous · Answer

isnt it infinity? i got $\infty - 1$