anonymous
  • anonymous
prove the following
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[\int\limits_{x}^{2x}(1/t)dt\] is constant on interval (0, infinity)
phantom
  • phantom
hope u dont give a closed interval bro ...j/k..its anitiderivative in ln t. and ln (2x) - ln( x) = ln 2 ==a constant (+ve or -ve ??? its -ve)
anonymous
  • anonymous
So I realized the answer, *ahem* proof to this about an hour after I asked. If anyone saw this and wanted to know what to do, here are two proofs. Proof 1 (using 1st Fundamental Thm of Calc):\[F(x) = \int\limits_{x}^{2x}(1/t)dt\]\[F'(x)= [\ln |t|]_{x}^{2x} = \ln |2x| - \ln |x| = \ln|2x/x| = \ln 2 = 0\] Proof 2 (using 2nd Fundamental Thm of Calc):\[F(x) = \int\limits_{x}^{2x}(1/t)dt = \int\limits_{x}^{a}(1/t)dt + \int\limits_{a}^{2x}(1/t)dt = -\int\limits_{a}^{x}(1/t)dt + \int\limits_{a}^{2x}(1/t)dt\]\[F'(x) = -(1/x) + (1/2x)(2) = -(1/x) + (1/x) = 0\] Derivatives represent slope, so here F'(x) has zero or no slope. Therefore, it is constant (over all the domain of all positive real numbers).

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phantom
  • phantom
ln2 is not 0
anonymous
  • anonymous
oh, you're right. idk, someone else I knew had that first proof...

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