this is an extra credit question on my take home test..
i think i attached it to the post here but it would read...
the integral from -1 to 1 of [ f(x) - f(-x) ] / 2
he wants us to explain why it is = 0 for any f(x)
I'm pretty sure it has to do with the numerator being 0 because you are subtracting the opposite of f(x) from f(x) ... ? I think
You get out what you put in.
it does cancel out as the exponents that get added will allow the signs to flip and let you cancel it out. use an example with X and N (for exponent) if necessary to show that.
Last edited by Reko; 12-04-2008 at 12:35 PM.
also, dont forget to show C in your work to get full credit!
substitute t for -x as a dummy variable of the f(-x) integral.
Oh ok. If you do the substitution of t=-x and -dt=dx, then you end up with identical integrals except with different dummy variables, and you can just cancel out both integrals without having to integrate anything.
are you sure that works? because then you have F(x) - F(t), where t = -x, as opposed to plain x.
I am def. rusty on this stuff, especially the u sub methods. I'll double check my older notes tonight if i remember.
Here is what I was thinking:
Can you leave the -x as -x and then integrate the parts separately (breaking it up into sparate integrals of f(x)/2 - f(-x)/2 first), which if you use X as x you would have (x^[n+1])/(2*[n+1]) - ([-x]^(n+1)/(2*[n+1])?
If n+1 is positive, then you would have a positive - positive = 0
If n+1 is odd, then since you are multiplying negative, you would have a negative - negative = 0 as well.
I'm just not sure if you can keep that -x in the integral without having to pull it out a -1...
Last edited by Reko; 12-04-2008 at 05:02 PM.
Last edited by KarstenDD; 12-04-2008 at 05:20 PM.
Well, its extra credit right? I can't figure anything out on it other than that as an idea, or maaaaaaybe finding a way to get
involved, but I'm not sure that applies here. Never know though....
The other, Int[-1,1, f(-x) dx], after doing a few manipulation rules...
Let u = -x; du = -dx; x = -1 => u = 1; x = 1 => u = -1
So it can be re-written as Int[1,-1, -f(u) du] = Int[-1,1, f(u) du]
So the 2nd integral evaluates to F(1) - F(-1) as well.
So you end up subtracting (F(1) - F(-1)) - (F(1) - F(-1)) = 0. The 1/2 is irrelevant as far as I can see.
Currently Recomposing... - Weight: 190 lbs | BF: ~15%
eps was right
You get out what you put in.
Wow guys, this was awsome seeing a lifting community with brains to go with their strength. We've got it all here.
|A:26 | H:6'5'' | W:198| GW: 215|
Stats: Start | Current | Goal
Bench: 135lb |170lb| 225lb
Squat: 135lb |185lb| 225lb
Dead: 180 |185lb| 315lb
OH Press: 85lb |100lb| 150lb
(all 3 x 5 reps)
It has been 13 years since I had that class and couldn't even begin to tell what any of this means - translation = calculus is worthless once you leave college
lol I'm gonna have to relearn most of it actually.
i took college alegbra 1 twice.
Last edited by Brad08; 12-05-2008 at 01:53 PM.