
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
Starting over...
You get out what you put in.
edit again
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; 12042008 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.
0004
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...
any ideas?
Last edited by Reko; 12042008 at 05:02 PM.
Neeerds!
This:
Video
Last edited by KarstenDD; 12042008 at 05:20 PM.
Roll Tide.
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....
Good luck
You have two definite integrals, Int[1,1, f(x) dx] = F(1)  F(1)
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 rewritten 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
Starting over...
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've gone through calc III...and retained just about none of it. BTW, calc III was one year ago.
**** calculus!
i took college alegbra 1 twice.
Last edited by Brad08; 12052008 at 01:53 PM.
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