The Five Biggest Contradictions in Fitness
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The Five Biggest Contradictions in Fitness

Itís no secret that when people contradict themselves, it has the effect of making the flaws in their actions or statements seem glaringly obvious. But what about when WE ourselves get caught contradicting ourselves by someone else?

By: Nick Tumminello Added: January 6th, 2014
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  1. #1
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    calculus question (integrals)

    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
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    You get out what you put in.

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  3. #2
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    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; 12-04-2008 at 12:35 PM.

  4. #3
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    also, dont forget to show C in your work to get full credit!

  5. #4
    Senior Member Doobs's Avatar
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    substitute t for -x as a dummy variable of the f(-x) integral.

  6. #5
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    0004

  7. #6
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    Quote Originally Posted by Doobs View Post
    substitute t for -x as a dummy variable of the f(-x) integral.
    X and -X would be better...

  8. #7
    Senior Member Doobs's Avatar
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    Quote Originally Posted by Reko View Post
    X and -X would be better...
    I don't get what you're saying

  9. #8
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    Quote Originally Posted by Doobs View Post
    I don't get what you're saying
    ITs easier to keep it all in terms of X dx rather than have a Xdx and a t dt in the same problem.

  10. #9
    Senior Member Doobs's Avatar
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    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.

  11. #10
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    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; 12-04-2008 at 05:02 PM.

  12. #11
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    Neeerds!

    This:
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    Last edited by KarstenDD; 12-04-2008 at 05:20 PM.
    Roll Tide.

  13. #12
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    Quote Originally Posted by Reko View Post
    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?
    I think this is what i was trying for.. so if this IS possible. I thnk its it
    Starting over...

    You get out what you put in.

  14. #13
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    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

  15. #14
    teh skinny fat :-( eps's Avatar
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    Quote Originally Posted by beaverfootball View Post
    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 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 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%

  16. #15
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    Quote Originally Posted by eps View Post
    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 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.

    By golly!

    I assume that would work for anything that is X though? Since it calls for f(x)?

  17. #16
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    eps was right
    Starting over...

    You get out what you put in.

  18. #17
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    Wow guys, this was awsome seeing a lifting community with brains to go with their strength. We've got it all here.
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  19. #18
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    Quote Originally Posted by Reko View Post
    WHAT SORT OF WIZARDRY IS THIS???

  20. #19
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    Quote Originally Posted by Brad08 View Post
    WHAT SORT OF WIZARDRY IS THIS???
    Definite integrals.

  21. #20
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    Quote Originally Posted by Reko View Post
    Definite integrals.
    math is ****ing ghey.

  22. #21
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    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

  23. #22
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    lol I'm gonna have to relearn most of it actually.

  24. #23
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    I've gone through calc III...and retained just about none of it. BTW, calc III was one year ago.

    **** calculus!

  25. #24
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    Quote Originally Posted by borracho View Post
    I've gone through calc III...and retained just about none of it. BTW, calc III was one year ago.

    **** calculus!
    Awesome. I thought I was bad for taking two semesters in college and forgetting, from 13 years ago.

  26. #25
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    i took college alegbra 1 twice.
    Last edited by Brad08; 12-05-2008 at 01:53 PM.

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