Another integral question

mfarr530 · #1 02-19-2008, 10:12 PM

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!

tommie frazier · #2 02-19-2008, 10:21 PM

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.

actas123 · #3 02-19-2008, 10:21 PM

Read this post its similar and should answer your question:
http://www.actuarialoutpost.com/actu...d.php?t=130542

tommie frazier · #4 02-19-2008, 10:21 PM

second one is another u sub, super easy.

mfarr530 · #5 02-19-2008, 10:31 PM

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".

atomic · #6 02-19-2008, 11:01 PM

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.

mfarr530 · #7 02-20-2008, 07:08 AM

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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