improper integrals, infinite limit

[needhelpwithcalculus]

I'm solving the integral from 0 to infinity xe^(-5x) dx
I got:

-1/5x(e^-5x) (x+1/5) as my integral
now i have to solve the limit as T goes to infinity.
so:

-1/5x(e^-5T) (T+1/5) - [-1/5e^(-5*0) (0+1/5)
= -1/5(e^-5T)(T+1/5) + 1/25
the answer is 1/25, so as T-> infinite, the first part has to be 0 right?
so why does -1/5e^(-5T) (T+1/5) = 0 instead of D.N.E?

 

[tkhunny]

Please be a littlle more careful with your language and understanding

-1/5x(e^-5x) (x+1/5) as my integral

No, your integral is what you had before you did this.

now i have to solve the limit as T goes to infinity.

There is no "T" in the given expression.

-1/5x(e^-5x) (x+1/5) as my integral

There is nothing about the integral argument that could be construed as negative. How did your result manage to be negative?

Please try again and be much more careful.

 

[galactus]

I would use an L, but I think the T comes from \(\displaystyle \lim_{T\to \infty}\int_{0}^{T}xe^{-5x}dx\)

Then, upon integrating \(\displaystyle \frac{-T}{5e^{5T}}-\frac{1}{25e^{5T}}+\frac{1}{25}\)

Now, take the limit \(\displaystyle \lim_{T\to \infty}\left(\frac{-T}{e^{5T}}-\frac{1}{25e^{5T}}+\frac{1}{25}\right)\)

Is that what you're up to?.

 

[Hanajanda]

REGARDING

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[needhelpwithcalculus]

I would use an L, but I think the T comes from \(\displaystyle \lim_{T\to \infty}\int_{0}^{T}xe^{-5x}dx\)

Then, upon integrating \(\displaystyle \frac{-T}{5e^{5T}}-\frac{1}{25e^{5T}}+\frac{1}{25}\)

Now, take the limit \(\displaystyle \lim_{T\to \infty}\left(\frac{-T}{e^{5T}}-\frac{1}{25e^{5T}}+\frac{1}{25}\right)\)

Is that what you're up to?.

yess and I get the integration part but whenever I find the equation as T goes to infinity, I get confused. when you plug in Infinity for T wouldn't it result to DNE instead of 0?

 

[galactus]

No, the e's in the denominator grow exceedingly large and those terms result to 0, so that all is left is the 1/25.

You take the limit, you don't literally plug in infinity.