## anonymous 4 years ago prove the following

1. anonymous

$\int\limits_{x}^{2x}(1/t)dt$ is constant on interval (0, infinity)

2. 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)

3. 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).

4. Phantom

ln2 is not 0

5. anonymous

oh, you're right. idk, someone else I knew had that first proof...