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    I get how to do the process but I just can't seem to understand why some have finite answers and some are undefined. Can anyone help?

    E.g. why does the integral of 1 to infinity of x^-2 have an answer but the integral of 1 to infinity of x^-1/2 not?
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    (Original post by Lunu)
    I get how to do the process but I just can't seem to understand why some have finite answers and some are undefined. Can anyone help?

    E.g. why does the integral of 1 to infinity of x^-2 have an answer but the integral of 1 to infinity of x^-1/2 not?
    \displaystyle \int \dfrac{1}{x^2} \ dx = -\dfrac{1}{x} + C

    So, ignoring the +C, you evaluate between "infinity" and 1 by using the limit as x tends to infinity.

    \displaystyle \int \dfrac{1}{\sqrt{x}} \ dx = 2\sqrt{x} + C


    doesn't converge when evaluated between 1 and "infinity" (root x doesn't have a limit as x gets massive).

    Does that make sense?
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    (Original post by Lunu)
    I get how to do the process but I just can't seem to understand why some have finite answers and some are undefined. Can anyone help?

    E.g. why does the integral of 1 to infinity of x^-2 have an answer but the integral of 1 to infinity of x^-1/2 not?
    It is a matter of convergence - they all have answers, (per se) its just that as the upper limit tends to infinity some integrals converge, others do not.
    It also depends on what restrictions you place on the domain as to whether the integral converges.
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    I'm still really confused. Sorry I'm awful at FP1. :/
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    The answer to the first one is -1/n +1
    The answer to the second is 2sqrtx -2

    The first one has an answer of 1 and the second one doesn't have an answer. Why is that?
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    I get on the first one that dividing by infinity make the answer really small so tends towards zero but I don't get the second one
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    (Original post by Lunu)
    I get on the first one that dividing by infinity make the answer really small so tends towards zero but I don't get the second one
    Consider the square root of a massive number. Say: 10000000000000000000000000000000 000000.
    What can you tell me about its square root.
    If I were to make this number bigger, what would happen to its square root - then what if it were infinite?
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    (Original post by Lunu)
    I'm still really confused. Sorry I'm awful at FP1. :/

    \displaystyle\int_{1}^{\infty} \dfrac{1}{x^{2}} dx = \left[-\dfrac{1}{x}\right]_{1}^{\infty}

    Now obviously as x becomes very large, \frac{1}{x} becomes very small. You could think about it as

    \displaystyle\lim_{a \to \infty} \left[-\dfrac{1}{x}\right]_{1}^{a}= \displaystyle\lim_{a \to \infty} \left(-\dfrac{1}{a} - - \dfrac{1}{1}\right)

    and that

    \displaystyle\lim_{a \to \infty} \dfrac{1}{a} \to 0

    Thus it effectively becomes \left(0 - - \dfrac{1}{1}\right) = 1

    Now think about the other integral

    \displaystyle\int_{1}^{\infty} \dfrac{1}{\sqrt{x}} dx = \left[2\sqrt{x} \right]_{1}^{\infty}

    Again, think about it as

    \displaystyle\lim_{a \to \infty} \left[2\sqrt{x} \right]_{1}^{a}

    = \displaystyle\lim_{a \to \infty} \left(2 \sqrt{a} - 2 \sqrt{1} \right)

    But

    \displaystyle\lim_{a \to \infty} 2 \sqrt{a} doesn't converge, hence the integral can't be evaluated
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    (Original post by joostan)
    Consider the square root of a massive number. Say: 10000000000000000000000000000000 000000.
    What can you tell me about its square root.
    If I were to make this number bigger, what would happen to its square root - then what if it were infinite?
    The square root is very small? It converges to a fixed number?
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    (Original post by Lunu)
    The square root is very small? It converges to a fixed number?
    No it doesn't. Why would it? If the square root function was bounded you can just go 1 above the bound and square that numer to get a contradiction.
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    (Original post by james22)
    No it doesn't. Why would it? If the square root function was bounded you can just go 1 above the bound and square that numer to get a contradiction.
    what? :confused:
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    (Original post by Lunu)
    what? :confused:
    The point is as x becomes very large \sqrt{x} doesn't converge to a limit.
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    so what happens to it?
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    (Original post by Lunu)
    I get how to do the process but I just can't seem to understand why some have finite answers and some are undefined. Can anyone help?

    E.g. why does the integral of 1 to infinity of x^-2 have an answer but the integral of 1 to infinity of x^-1/2 not?
    As far as further maths is concerned, you just have to evaluate the integral as a function of x and then see what happens as x gets larger and larger.

    You have the same situation with infinite sums - for example adding up all the numbers of the form 1/n (1/1, 1/2, 1/3 etc) gives you a number that just gets bigger and bigger and doesn't tend to a limit, but if you add up the reciprocals of the square numbers (1/1, 1/4, 1/9 etc) then you do get a finite value.

    An integral is just like a great big wibbly-wobbly sum!
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    (Original post by davros)
    An integral is just like a great big wibbly-wobbly sum!
    Wibbly-wobbly timey-wimey sum?
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    (Original post by davros)
    As far as further maths is concerned, you just have to evaluate the integral as a function of x and then see what happens as x gets larger and larger.

    You have the same situation with infinite sums - for example adding up all the numbers of the form 1/n (1/1, 1/2, 1/3 etc) gives you a number that just gets bigger and bigger and doesn't tend to a limit, but if you add up the reciprocals of the square numbers (1/1, 1/4, 1/9 etc) then you do get a finite value.

    An integral is just like a great big wibbly-wobbly sum!
    So it's similar to the series stuff in core 2?
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    (Original post by Lunu)
    what? :confused:
    Basically, as x gets very large, sqrt(x) also get very large so it will not converge.
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    (Original post by Lunu)
    so what happens to it?
    It continues to get infinitely large - think about the graph y = \sqrt{x} it just increases and tends to infinity as x approaches infinity
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    (Original post by Lunu)
    So it's similar to the series stuff in core 2?
    sort of.

    If you've done geometric series then you know that the ratio of terms determines whether or not the infinite sum makes sense.

    If you've done the infinite binomial series then you know that it only converges for certain values.

    Integrals are basically a generalization of a big sum - imagine working out the area under a curve by approximating it with loads and loads of rectangles, then increasing the number of rectangles to infinity while their width gets smaller and smaller. Sometimes this sum will converge, sometimes it won't.

    Wibbly-wobbly, timey-wimey stuff
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    (Original post by james22)
    Basically, as x gets very large, sqrt(x) also get very large so it will not converge.
    I think I understand now
 
 
 
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