Need Some Reassurance

**TreeMoney** · June 18th 2006, 06:14 PM

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

**ThePerfectHacker** · June 18th 2006, 06:31 PM

Originally Posted by TreeMoney

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

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

**malaygoel** · June 18th 2006, 06:35 PM

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
$\int xe^{-2x}dx$ = $\frac{xe^{-2x}}{-2}$ $-\int \frac{e^{-2x}}{-2}\frac{dx}{dx}dx$

**CaptainBlack** · June 18th 2006, 08:58 PM

Originally Posted by ThePerfectHacker

This is what ya got,
$\int_{0}^{\infty}xe^{-2x}dx$
Which by definition is the limit,
$\lim_{N\to\infty}\int_0^Nxe^{-2x}dx$
Calculating the integral requires integration by parts.

No it does not, it requires a bit of observation and the fundamental
theorem of calculus.

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

so:

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

Hence, with a bit of limiting jiggery-pokery:

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

RonL

**malaygoel** · June 18th 2006, 10:10 PM

Originally Posted by CaptainBlack

No it does not, it requires a bit of observation and the fundamental
theorem of calculus.

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

so:

$\int xe^{-2x}dx=-\frac{1}{4}\int -4xe^{-2x^2}\ dx$

Observe the $\mbox{power of e}$ in the last line of the quoted work.
Keep Smiling
Malay

**CaptainBlack** · June 19th 2006, 12:05 AM

Originally Posted by malaygoel

Observe the $\mbox{power of e}$
Keep Smiling
Malay

Yes, I have observed it, now can you elaborate on what I should
be looking for?

RonL

**CaptainBlack** · June 19th 2006, 01:17 AM

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

$<br /> \int_{0}^{\infty}xe^{-2x}dx<br />$

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

RonL

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