MHB Find Value of \(\int_{-\infty}^{\infty} f(x) dx\) for \(y=f(x)\)

[Nicole18]

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

 

[Jameson]

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]

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]

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]

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]

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