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    Can anyone help me explain this?
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    (Original post by SWISH99)
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    Can anyone help me explain this?
    X^-2 is not continuous in the interval [-1,1] and so the integral isn't valid - you'd have to do the integral from -1 to 0 and add it to the integral from 0 to 1 for it to work - of cord rhr winter all doesn't exist as the area is infinite which explains why the answer is not -2.
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    X-2 has a singularity at 0, and cannot be integrated over [-1, 1] in the way shown in the grey box. Indeed, see that for a in (0, 1):

    int_{a}^{1} x^-2 dx = -(1)^-1 - (-a)^-1 = (1/a)-1
    int_{-1}^{-a} x^-2 dx = -(-a)^-1 - (-(-1)^-1) = (1/a)-1

    are both positive as expected. However, neither of those integrals converge at the limit where a goes to 0.
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    As others above this post have stated, f(x) = x^(-2), is not continuous in all of the interval [-1,1] due to the presence of a sigularity at x=0.
    Therefore, we can not use the Fundamental Theorem of Calculus.

    If we try to apply improper integration:

     \displaystyle \int^{1}_{-1} x^{-2} \, dx = \lim_{a \to 0^{+}} \int^{1}_{a} x^{-2} \, dx \, + \, \lim_{b \to 0^{-}} \int^{b}_{-1} x^{-2} \, dx = \Big( -1 +\frac{1}{a} \Big) + \Big( \frac{-1}{b} - 1 \Big),

    which does not converge as of course 1/a does not have a limit/converge as "a" goes to 0.
 
 
 
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