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     \frac{x}{(3-4x-4x^2)^{\frac{1}{2}}}

    The limits of integration on this are from -0.5 to 0.

    I completed the square and used the Trig Substitution 2x+1 = 2SinQ


    At the end after integrationI was left with  \frac{-CosQ}{2} - \frac{Q}{4}

    Now I dont particularly like changing everything back into x just to evaluate the integral so I changed the limits to suit my substitution.

    Basically when x = -0.5 , Q = 0
    when x = 0 Q = pi/6

    However my final answer is not right according to the book. It is right however if I substitute all the way back and use the original x values. So my question is, what is wrong with my new limits of integration?
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    (Original post by Zilch)
     \frac{x}{(3-4x-4x^2)^{\frac{1}{2}}}

    The limits of integration on this are from -0.5 to 0.

    I completed the square and used the Trig Substitution 2x+1 = 2SinQ


    At the end after integrationI was left with  \frac{-CosQ}{2} - \frac{Q}{4}

    Now I dont particularly like changing everything back into x just to evaluate the integral so I changed the limits to suit my substitution.

    Basically when x = -0.5 , Q = 0
    when x = 0 Q = pi/6

    However my final answer is not right according to the book. It is right however if I substitute all the way back and use the original x values. So my question is, what is wrong with my new limits of integration?
    I haven't checked your integration but you should get exactly the same answer if you've substituted the limits correctly. Is your answer vastly different, or just a negative sign out, or something else?
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    (Original post by davros)
    I haven't checked your integration but you should get exactly the same answer if you've substituted the limits correctly. Is your answer vastly different, or just a negative sign out, or something else?
    I did it myself at first and thought it was an algebra error but then I did it online on wolfgram and noticed the same difference.

    The answer I'm getting is roughly 0.032 and the actual answer is -0.06
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    (Original post by Zilch)
    I did it myself at first and thought it was an algebra error but then I did it online on wolfgram and noticed the same difference.

    The answer I'm getting is roughly 0.032 and the actual answer is -0.06
    Well I just put your Q limits into your Q function and got -0.06391 approximately so the error is in your arithemtic. I ended up evaluating:

    (-0.5cos(pi/6) - (pi/24)) - (-0.5 - 0)

    Note that your original function is negative throughout the range of integration, so you know that the answer has to be negative!
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    (Original post by davros)
    Well I just put your Q limits into your Q function and got -0.06391 approximately so the error is in your arithemtic. I ended up evaluating:

    (-0.5cos(pi/6) - (pi/24)) - (-0.5 - 0)

    Note that your original function is negative throughout the range of integration, so you know that the answer has to be negative!
    Right, thanks. Made a slight mistake in the end where I integrated x/4 instead of 1/4. That made it x^2/8 in the online calc.

    Thanks for the help!
 
 
 
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