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Math Help - help with basic integral problem?

  1. #1
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    help with basic integral problem?

    \int_{-5}^5 \frac{2}{x^3} dx

    the answer i got is 0.

    the answer in the back of the book says: Does not exist

    what did i do incorrectly?

    This is what I did: \int_{-5}^5 \frac{2}{x^3} dx
    = \frac{-1}{x^2}|_{-5}^5
    = \frac{-1}{(5)^2}-(\frac{-1}{(-5)^2})
    = \frac{-1}{25}+\frac{1}{25}
    =0
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  2. #2
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    Quote Originally Posted by yoman360 View Post
    \int_{-5}^5 \frac{2}{x^3} dx

    the answer i got is 0.

    the answer in the back of the book says: Does not exist

    what did i do incorrectly?
    Note* this post will be updated soon so i can show you my work
    The update will be unnecessary since you probably neglected to recognise it as an improper integral (the integrand is undefined at x = 0 which lies in the interval of integration). The first step is therefore to write:

    \lim_{\alpha \rightarrow 0} \int_{-5}^{\alpha}\frac{2}{x^3} \, dx + \lim_{\beta \rightarrow 0} \int_{\beta}^{5}\frac{2}{x^3} \, dx .

    (A pre-emptive) by the way, statements like \infty - \infty make no sense and certainly do not equal zero ....
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    The update will be unnecessary since you probably neglected to recognise it as an improper integral (the integrand is undefined at x = 0 which lies in the interval of integration). The first step is therefore to write:

    \lim_{\alpha \rightarrow 0} \int_{-5}^{\alpha}\frac{2}{x^3} \, dx + \lim_{\beta \rightarrow 0} \int_{\beta}^{5}\frac{2}{x^3} \, dx .

    (A pre-emptive) by the way, statements like \infty - \infty make no sense and certainly do not equal zero ....
    ah i see since f is not continuous on [a,b] I can't use the fundamental theorem of calculus:
    \int_a^b f(x)dx=F(b)-F(a)

    where F is any antiderivative of f, that is, a function such that F'=f

    Thank you so much for pointing that out I just forgot to check if its continuous on [-5,5] since its discontinuous at x=0 the integral does not exist
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