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#1
02-19-2008, 10:12 PM
 mfarr530 Member Join Date: Sep 2007 Location: Tampa, FL Studying for MFE Favorite beer: is in a green bottle Posts: 129
Another integral question

My study guide often skips steps and leaves me stuck. If anyone can fill in the blanks on this integral, I would greatly appreciate it.

a) 324*integral from 0 to infinity[x(x+3)^-5 dx]

c) 81*integral from 0 to infinity[(x+3)^-4 dx]

What step am I missing that removes the x and divides 324 by 4? I think it may be adding 3 and subtracting 3 to get another (x+3), but I can't get there. Thanks!
#2
02-19-2008, 10:21 PM
 tommie frazier Member Join Date: Aug 2003 Favorite beer: The kind with 2 e's Posts: 19,175

u = x+3
du = dx

x = u-3

first one is integral from a to b (figure those out or switch back at the end) of (u-3)/u^5 du. simplify and integrate. i think.
#3
02-19-2008, 10:21 PM
 actas123 Join Date: Jan 2008 Posts: 25

http://www.actuarialoutpost.com/actu...d.php?t=130542
#4
02-19-2008, 10:21 PM
 tommie frazier Member Join Date: Aug 2003 Favorite beer: The kind with 2 e's Posts: 19,175

second one is another u sub, super easy.
#5
02-19-2008, 10:31 PM
 mfarr530 Member Join Date: Sep 2007 Location: Tampa, FL Studying for MFE Favorite beer: is in a green bottle Posts: 129

Quote:
 Originally Posted by tommie frazier second one is another u sub, super easy.
Sorry, I guess I didn't explain it extremely well.

a) is the first step in the problem.

c) is the second step in the problem

I was looking for a step b) to show how to get from a) to c). I can get from c) to the answer, which is "1".
#6
02-19-2008, 11:01 PM
 atomic Member CAS Join Date: Jul 2006 Posts: 4,088

Quote:
 Originally Posted by mfarr530 Sorry, I guess I didn't explain it extremely well. a) is the first step in the problem. c) is the second step in the problem I was looking for a step b) to show how to get from a) to c). I can get from c) to the answer, which is "1".
The methodology that was used to go from the first integral to the second is integration by parts, with the choice

$u = x$
$du = dx$
$dv = (x+3)^{-5} \,dx$
$v = \frac{(x+3)^{-4}}{-4}$

Therefore,

$\int_0^\infty u \, dv = \Bigl[uv\Bigr]_{x=0}^\infty \,- \int_0^\infty v \, du$

implies

$324 \int_0^\infty x (x+3)^{-5} \, dx = 324 \Bigl[x(x+3)^{-5}\Bigr]_{x=0}^\infty \, + 81 \int_0^\infty (x+3)^{-4} \, dx$.

Since

$\lim_{x\rightarrow\infty}\frac{x}{(x+3)^5} = 0$,

we obtain the claimed result.

Obviously, this is not the most straightforward method of solution. It is far easier to employ the linear substitution previously mentioned.
#7
02-20-2008, 07:08 AM
 mfarr530 Member Join Date: Sep 2007 Location: Tampa, FL Studying for MFE Favorite beer: is in a green bottle Posts: 129

Quote:
 Originally Posted by atomic The methodology that was used to go from the first integral to the second is integration by parts, with the choice $u = x$ $du = dx$ $dv = (x+3)^{-5} \,dx$ $v = \frac{(x+3)^{-4}}{-4}$ Therefore, $\int_0^\infty u \, dv = \Bigl[uv\Bigr]_{x=0}^\infty \,- \int_0^\infty v \, du$ implies $324 \int_0^\infty x (x+3)^{-5} \, dx = 324 \Bigl[x(x+3)^{-5}\Bigr]_{x=0}^\infty \, + 81 \int_0^\infty (x+3)^{-4} \, dx$. Since $\lim_{x\rightarrow\infty}\frac{x}{(x+3)^5} = 0$, we obtain the claimed result. Obviously, this is not the most straightforward method of solution. It is far easier to employ the linear substitution previously mentioned.
No, makes sense. Thanks!

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