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anonymous
 4 years ago
prove the following
anonymous
 4 years ago
prove the following

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{x}^{2x}(1/t)dt\] is constant on interval (0, infinity)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hope 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
 4 years ago
Best ResponseYou've already chosen the best response.0So 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):\[F(x) = \int\limits_{x}^{2x}(1/t)dt\]\[F'(x)= [\ln t]_{x}^{2x} = \ln 2x  \ln x = \ln2x/x = \ln 2 = 0\] 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).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh, you're right. idk, someone else I knew had that first proof...
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