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    Why is the integration of infinity 0?
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    (Original post by FrostedStarzzz)
    Why is the integration of infinity 0?
    wat?

    you're claiming this?

    \displaystyle \int \ \infty =0 ?
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    (Original post by FrostedStarzzz)
    Why is the integration of infinity 0?
    What's your question? What I see doesn't make any sense.
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    (Original post by RDKGames)
    What's your question? What I see doesn't make any sense.
    http://pmt.physicsandmathstutor.com/...ation%20MS.pdf


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    The infinity is on the bottom of the fraction. Anything divided by infinity is zero.
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    infinity isn't an actual value itself it describes how big something is

    so what it's describing is what happens when x get super super big

    well if x is on the bottom of a fraction then we can essentially say that the fraction is 0 because x is so small it's pretty much negligible so we say it's 0
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    \displaystyle I= \int_{2}^{\infty}x^{-3}.dx = \lim_{a \rightarrow \infty} \int_{2}^a x^{-3} .dx = \lim_{a \rightarrow \infty}\left[ -\frac{1}{2x^2} \right]_2^a

    So we get \displaystyle I = \lim_{a \rightarrow \infty} \left( -\frac{1}{2a^2} \right) +\frac{1}{8}

    Now what is \displaystyle \lim_{a \rightarrow \infty} \left( -\frac{1}{2a^2} \right)??

    In other words, what value does  -\frac{1}{2a^2} approach as a tends to infinity?


    Side note: it might be tempting for many students encountering infinity in this context to just sub it in, but infinity is not a number so we cannot just sub it in. We need to examine what happens to our function as our variable TENDS TO infinity, which is what I've shown here.
 
 
 
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