Improper intergral help

[max]

I need help understanding the integral with lower limit negative infinity and upper limit 0 for xe^(-4x)dx

I solved using integration by parts to get. the limit as b goes to negative infinity of [-1/4(xe^(-4x))-1/16(e^(-4x)) evaluated with x =0 minus the same quantity with x = b ]
Please forgive me. I don't know if my wording is clear.
When I do that, I get -1/16 -( infinity/infinity - 1/infinity) I used L'Hopital rule for the indeterminate form to get a final answer of -1/16. My book shows -infinity as the answer. I don't understand.

Also I don't see how they simplified the limit to limit as b approaches negative infinity of [-1/16 + b/4 +1/16(e^(-4b))] With that, I see the negative infinity. But, shouldn't b/4 be multiplied by e^(-4b) also? which will give the indeterminate form?

 

[pappus]

I need help understanding the integral with lower limit negative infinity and upper limit 0 for xe^(-4x)dx

I solved using integration by parts to get. the limit as b goes to negative infinity of [-1/4(xe^(-4x))-1/16(e^(-4x)) evaluated with x =0 minus the same quantity with x = b ]
Please forgive me. I don't know if my wording is clear.
When I do that, I get -1/16 -( infinity/infinity - 1/infinity) I used L'Hopital rule for the indeterminate form to get a final answer of -1/16. My book shows -infinity as the answer. I don't understand.

Also I don't see how they simplified the limit to limit as b approaches negative infinity of [-1/16 + (b/4 +1/16)(e^(-4b))] With that, I see the negative infinity. But, shouldn't b/4 be multiplied by e^(-4b) also? which will give the indeterminate form?

In my opinion there is missing a pair of parantheses in the answer of your book.

1. Your result is OK. Transcribe the term to:

\(\displaystyle -\frac14(xe^{-4x})-\frac1{16}(e^{-4x}) = -e^{-4x}\left(\frac x4+\frac1{16}\right)\)

2.
\(\displaystyle \int_b^0\left( xe^{-4x} \right) dx = \left[-e^{-4x}\left(\frac x4+\frac1{16}\right)\right]_b^0 = -\frac1{16}+e^{-4b}\left(\frac b4+\frac1{16}\right)\)

3.
\(\displaystyle \lim_{b \to -\infty}\left( -\frac1{16}+\underbrace{e^{-4b}}_{approches \ +\infty}\left(\underbrace{\frac b4}_{approaches \ -\infty}+\frac1{16}\right) \right) = -\infty\)

 

[max]

In my opinion there is missing a pair of parantheses in the answer of your book.

1. Your result is OK. Transcribe the term to:

\(\displaystyle -\frac14(xe^{-4x})-\frac1{16}(e^{-4x}) = -e^{-4x}\left(\frac x4+\frac1{16}\right)\)

2.
\(\displaystyle \int_b^0\left( xe^{-4x} \right) dx = \left[-e^{-4x}\left(\frac x4+\frac1{16}\right)\right]_b^0 = -\frac1{16}+e^{-4b}\left(\frac b4+\frac1{16}\right)\)

3.
\(\displaystyle \lim_{b \to -\infty}\left( -\frac1{16}+\underbrace{e^{-4b}}_{approches \ +\infty}\left(\underbrace{\frac b4}_{approaches \ -\infty}+\frac1{16}\right) \right) = -\infty\)

Thanks for the reply!

If you rewrite e^(-4x) to 1/e^(4x), wouldn't use end up having negative infinity over negative infinity in the problem?

 

[pappus]

Thanks for the reply!

If you rewrite e^(-4x) to 1/e^(4x), wouldn't use end up having negative infinity over negative infinity in the problem?

No, because \(\displaystyle e^{-4x} = \frac1{e^{4x}}>0\ ,\ x\in\mathbb{R}\) , even though x is negative.