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Math Help - Area under hyperbola

  1. #1
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    Area under hyperbola

    Hi!
    Today I took a test and saw an interesting thing while I was cheating
    It said that natural logarithm of x is the area under hyperbola 1/x starting from 1 to x.
    Is there an easy way to do this for other hyperbolas like 1/x^2, 2/x etc. ?
    Or is there a better way for doing this then using natural logarithms?
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  2. #2
    MHF Contributor ebaines's Avatar
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    Indeed the natural logarithm of A is defined as the area under the curve f(x) = 1/x from x=1 to x equals the value A:

    <br />
ln(A) = \int_1 ^A \frac {dx} x<br />

    Note that this is defined only for A>0.

    To find the area under the curve 1/x^2 you simply integrate:

    <br />
\int \frac 1 {x^2} dx = -\frac 1 x<br />

    So the area under the curve 1/x^2 from x = 1 to x = A is:

    <br />
1 - \frac 1 A<br />

    For the area under 2/x, it's simply two times the natural logarithm.

    You posted this question in a pre-calculus forum, so if this look like hieroglyphics to you, don't worry - you'll get to in school soon enough!
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  3. #3
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    Thanks!
    It's ok, I understand the basics of integrals and antiderivatives.
    The 1/x^2 bit confused me, but then I remembered that derivative of x^n = nx^(n-1).
    The sad thing is that I will never study this at school, because of the education programme, in the University we all need to already know this.
    This is a lot better way then finding in physics class the area under the curve of isothermic process graph then calculating small parts and summing them.
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