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    For the following link for question 36 I don't understand what you can get rid of a term in an integral due to an odd integrand- how do u know it is odd. Thank you very much
    http://madasmaths.com/archive/maths_...plications.pdf
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    (Original post by runny4)
    For the following link for question 36 I don't understand what you can get rid of a term in an integral due to an odd integrand- how do u know it is odd. Thank you very much
    http://madasmaths.com/archive/maths_...plications.pdf
    Link does not work for me.
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    (Original post by runny4)
    For the following link for question 36 I don't understand what you can get rid of a term in an integral due to an odd integrand- how do u know it is odd. Thank you very much
    http://madasmaths.com/archive/maths_...plications.pdf
    Link is disrupted, I can't solve if you can't resolve...
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    (Original post by runny4)
    For the following link for question 36 I don't understand what you can get rid of a term in an integral due to an odd integrand- how do u know it is odd. Thank you very much
    http://madasmaths.com/archive/maths_...plications.pdf
    (Original post by 16Characters....)
    Link does not work for me.
    (Original post by Pilate VII)
    Link is disrupted, I can't solve if you can't resolve...
    as this is one of mine but I am a bit busy ...
    see if this link works

    http://madasmaths.com/archive/maths_...plications.pdf
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    (Original post by runny4)
    For the following link for question 36 I don't understand what you can get rid of a term in an integral due to an odd integrand- how do u know it is odd. Thank you very much
    http://madasmaths.com/archive/maths_...plications.pdf
    For just that part, I'll call it f(u)

    f(-u) = (-u)(r^2-(-u)^2)^\frac{1}{2}=-u(r^2-u^2)^\frac{1}{2}=-f(u)

    Hence an odd function.
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    Sub u = -t to see how that term is zero.

    \displaystyle \begin{aligned} I & =  \int_{-r}^{r} \frac{u}{\sqrt{r^2-u^2}}\;{du}  = \int_{r}^{-r} \frac{t}{\sqrt{r^2-(-t)^2}}\;{dt} \\& = -\int_{-r}^{r} \frac{t}{\sqrt{r^2-t^2}}\;{dt} = -\int_{-r}^{r} \frac{u}{\sqrt{r^2-u^2}}\;{du} \\& = -I. \end{aligned}
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    Let k=r^2-u^2 ,(dk/-2u)=du,substitute in to the integral and,then easy to integrate and final answer will be r(r-1) or r^2-r
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    (Original post by ghostwalker)
    For just that part, I'll call it f(u)

    f(-u) = (-u)(r^2-(-u)^2)^\frac{1}{2}=-u(r^2-u^2)^\frac{1}{2}=-f(u)

    Hence an odd function.
    thanks
 
 
 
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