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Math Help - solving definite integrals, and solving systems using the Gaussian Elimination method

  1. #1
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    solving definite integrals, and solving systems using the Gaussian Elimination method

    not sure if this is the correct section to post but seemed most suitable to me...please move if needed.

    here's the problem integral problem


    the way i've been doing it, and the way i want to is to find the antiterivative and then sub in the limits, however, have some trouble doing so.
    i'l use '~' as the integral symbol
    and 'sqrt' as square root symbol

    first i let u = sinx where du=cosxdx
    therefore

    =-sinx 0~1/2 sqrt(u)*du if x=0, u=0 if x=1/2, u=0.009

    = -sinx 0~0.009 u^1/2*du

    =-sinx [2/3U^2/3]

    =[-sinx(2/3sinx^3/2)]

    and the answers are completely wrong. i have check it on my calculator and it should be 0.031 or thereabouts.

    secondly, solving the systems using the Gaussian elimination method, i haven't been able to come across any results that satisfy and of the equations.

    here is the equations giving me trouble


    and here is what i worked out




    if someone could smash up what they thinks right, thatd be awesome as i'm just blind to what i've done wrong (most people are to their own mistakes!), or how i should go about it. these are the methods i need to use to solve them to so please dont smash up easier ways, i'd be using them if i could!
    thanks,
    dave
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  2. #2
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    Hello, 1600dave!

    Here's #59 . . .


    Code:
    2x + 4y - 6z  =  2
          y + 2z  =  4
     x      + 4z  = -6

    Divide the first equation by 2 and we have:
    Code:
              1   2  -3   |   1
              0   1   2   |   4
              1   0   4   |  -6
    
    
              1   2  -3   |   1
              0   1   2   |   4
      R3-R1   0  -2   7   |  -7
    
    
    
    R1-2ĚR2   1   0  -7   |  -7
              0   1   2   |   4
    R3+2ĚR2   0   0  11   |   1
    
    
              1   0  -7   |  -7
              0   1   2   |   4
    R3/11     0   0   1   | 1/11
    
    
    R1+7ĚR3   1   0   0   | -70/11
    R2-2ĚR3   0   1   0   |  42/11
              0   0   1   |   1/11
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by 1600dave View Post
    not sure if this is the correct section to post but seemed most suitable to me...please move if needed.

    here's the problem integral problem


    the way i've been doing it, and the way i want to is to find the antiterivative and then sub in the limits, however, have some trouble doing so.
    i'l use '~' as the integral symbol
    and 'sqrt' as square root symbol

    first i let u = sinx where du=cosxdx
    therefore

    =-sinx 0~1/2 sqrt(u)*du if x=0, u=0 if x=1/2, u=0.009

    = -sinx 0~0.009 u^1/2*du

    =-sinx [2/3U^2/3]

    =[-sinx(2/3sinx^3/2)]

    and the answers are completely wrong. i have check it on my calculator and it should be 0.031 or thereabouts.
    The solution to your integral problem is quite simple. When you calculated the integral by hand you are implicitly assuming the argument of the sine function is in radians. Apparently when you did the problem on your calculator you were in "degree" mode. (I replicated your answer of 0.031 using degree mode.) Except that the negative sign shouldn't be there, the integration you provided is correct.

    -Dan
    Last edited by topsquark; October 11th 2006 at 05:35 AM. Reason: Negative sign
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  4. #4
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    Thankyou, only just realised!!
    turns out the corrct answer is 0.221

    cheers, and thanks everyone,
    dave
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