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converges material

Nicole18

New member
Dec 6, 2012
2
the problem says \(y=f(x)\) and find the value of \(\int_{-\infty}^{\infty}f(x)dx\) if it converges

\[f(x)=\begin{cases}2-e^{-0.2x}&\text{ for } x \ge 0\\0&\text{
otherwise} \end{cases}\]


.....help
 
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Jameson

Administrator
Staff member
Jan 26, 2012
4,041
the problem says y=f(x) and find the value of negative to positive infinityf(x)dx if it converges

f(x)={2-e^-.2x
{0
x is greater than or equal to 0
otherwise
.....help
Hi Nicole18,

Welcome to MHB. :)

Can you edit your post a bit? It's hard to fully understand what you mean. The first line could be $f(x)={2-e^{-.2}}x$ or $f(x)={2-e^{-.2x}}$. The second and third lines don't make any sense to me. What is greater than or equal to 0? Line 2? Line 1? Something else?

Lastly is this an integral or a sum?

Jameson
 

chisigma

Well-known member
Feb 13, 2012
1,704
the problem says y=f(x) and find the value of negative to positive infinityf(x)dx if it converges

f(x)={2-e^-.2x
{0
x is greater than or equal to 0
otherwise
.....help
Welcome on MHB Nicole!...

If You want to receive a precise answer You have to do a precise question. First : f(x) is like that...


$\displaystyle f(x) = \begin{cases} 2-e^{- .2\ x}\ \text{if } x \ge 0\\
0\ \text{otherwise} \end{cases}$ (1)

... or some else?... Second: do You ask if ...


$\displaystyle \int_{- \infty}^{+ \infty} f(x)\ dx$ (2)


... converges and if yes to what it converges?...


Kind regards


$\chi$ $\sigma$


P.S. If You click on 'replay with quote' You can have a good example of use of LaTex...
 

Nicole18

New member
Dec 6, 2012
2
yes the way chisigma wrote it is correct...i really have no idea how to even start this i need a precise answer. how do i tell if it converges i know converges means it is getting close to a number
 

chisigma

Well-known member
Feb 13, 2012
1,704
yes the way chisigma wrote it is correct...i really have no idea how to even start this i need a precise answer. how do i tell if it converges i know converges means it is getting close to a number
Very well!... we can write thye integral as $I= I_{1}+ I_{2}$ where...


$\displaystyle I_{1} = \int_{0}^{\infty} 2\ dx$ (1)


$\displaystyle I_{2}= - \int_{0}^{\infty} e^{- \frac{x}{5}}\ dx$ (2)

Starting from the second we have...

$\displaystyle I_{2} = \lim_{t \rightarrow \infty} - \int_{0}^{t} e^{- \frac{x}{5}}\ dx = \lim_{t \rightarrow \infty} 5\ |e^{- \frac{x}{5}}|_{0}^{t} = - 5$ (3)

... a now the first...

$\displaystyle I_{1} = \lim_{t \rightarrow \infty} \int_{0}^{t} 2\ dx = \lim_{t \rightarrow \infty} 2 |x|_{0}^{t} = + \infty$ (4)

The conclusion is: the integral doesn't converge...

Kind regards

$\chi$ $\sigma$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
the problem says \(y=f(x)\) and find the value of \(\int_{-\infty}^{\infty}f(x)dx\) if it converges

\[f(x)=\begin{cases}2-e^{-0.2x}&\text{ for } x \ge 0\\0&\text{
otherwise} \end{cases}\]


.....help

The integral diverges since the exponential term goes to \(0\) as \(x\) becomes large, so the right tail of the integral is like the integral of a non zero constant (and of course the left tail converges since the integrand is zero for \(x\lt 0\).

CB