1. Need Some Reassurance

My professor gave the class a problem which is going to be on our exam and I want to make sure I get full credit for this problem. It's an improper integral, it is.....

I=Integrate(X*e^(-2x)) at (0,+infinity)

If someone could just tell me what integration technique needs to be used to solve this. I know it can be done using a formula, but i would rather show more work then that to ensure i get full credit for this problem. I tried integration by parts but it just seems to make the equation more complicated. Thanks in advance. John

2. Originally Posted by TreeMoney
I=Integrate(X*e^(-2x)) at (0,+infinity)
This is what ya got,
$\displaystyle \int_{0}^{\infty}xe^{-2x}dx$
Which by definition is the limit,
$\displaystyle \lim_{N\to\infty}\int_0^Nxe^{-2x}dx$
Calculating the integral requires integration by parts.
Given,
$\displaystyle \int xe^{-2x}dx$
Let,
$\displaystyle u=x$ then, $\displaystyle u'=1$
$\displaystyle v'=e^{-2x}$ then, $\displaystyle v=-1/2e^{-2x}$
Thus we have,
$\displaystyle -1/2xe^{-2x}+1/2\int e^{-2x}dx$
This is,
$\displaystyle -1/2xe^{-2x}-1/4e^{-2x}$
Thus,
$\displaystyle \left \lim_{N\to\infty}\int_0^Nxe^{-2x}dx=\lim_{N\to\infty} -1/2xe^{-2x}-1/4e^{-2x} \right|^N_0$
Thus,
$\displaystyle \lim_{N\to\infty} -1/2Ne^{-2N}-1/4e^{-2N}+1/2(0)e^{-2\cdot 0}+1/4e^{-2\cdot 0}$
Remember that an exponent always overtakes a polynomial thus,
$\displaystyle 0-0+0+1/4=1/4$

3. Originally Posted by TreeMoney
My professor gave the class a problem which is going to be on our exam and I want to make sure I get full credit for this problem. It's an improper integral, it is.....

I=Integrate(X*e^(-2x)) at (0,+infinity)

If someone could just tell me what integration technique needs to be used to solve this. I know it can be done using a formula, but i would rather show more work then that to ensure i get full credit for this problem. I tried integration by parts but it just seems to make the equation more complicated. Thanks in advance. John
I am not sure
$\displaystyle \int xe^{-2x}dx$=$\displaystyle \frac{xe^{-2x}}{-2}$$\displaystyle -\int \frac{e^{-2x}}{-2}\frac{dx}{dx}dx$

4. Originally Posted by ThePerfectHacker
This is what ya got,
$\displaystyle \int_{0}^{\infty}xe^{-2x}dx$
Which by definition is the limit,
$\displaystyle \lim_{N\to\infty}\int_0^Nxe^{-2x}dx$
Calculating the integral requires integration by parts.
No it does not, it requires a bit of observation and the fundamental
theorem of calculus.

$\displaystyle \frac{d}{dx}\ e^{-2x^2}=-4xe^{-2x^2}$,

so:

$\displaystyle \int xe^{-2x}dx=-\frac{1}{4}\int -4xe^{-2x^2}\ dx=-\frac{1}{4}e^{-2x^2}+C$

Hence, with a bit of limiting jiggery-pokery:

$\displaystyle \int_{0}^{\infty}xe^{-2x}dx=\frac{1}{4}$

RonL

5. Originally Posted by CaptainBlack
No it does not, it requires a bit of observation and the fundamental
theorem of calculus.

$\displaystyle \frac{d}{dx}\ e^{-2x^2}=-4xe^{-2x^2}$,

so:

$\displaystyle \int xe^{-2x}dx=-\frac{1}{4}\int -4xe^{-2x^2}\ dx$
Observe the $\displaystyle \mbox{power of e}$ in the last line of the quoted work.
Keep Smiling
Malay

6. Originally Posted by malaygoel
Observe the $\displaystyle \mbox{power of e}$
Keep Smiling
Malay
Yes, I have observed it, now can you elaborate on what I should
be looking for?

RonL

7. Originally Posted by CaptainBlack
Yes, I have observed it, now can you elaborate on what I should
be looking for?

RonL
OK I see it now the OP was asking for

$\displaystyle \int_{0}^{\infty}xe^{-2x}dx$

I had misread the power that $\displaystyle e$ is raised to

RonL