prove the following

Question

prove the following

anonymous · Answer

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

phantom · Answer

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 · Answer

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).

phantom · Answer

ln2 is not 0

anonymous · Answer

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