Integrate 2/(3x-k) between k and 3k

[apple2357]

When i looked at this problem graphically, i realised the answer was always the same regardless of the value of k.
So for k=1:



and if k=2



Etc for all values of k,the answer is 0.92

I can show this algebraically but i can't get my head around what is happening graphically, there appears to be a stretch ( or is it translation?) to the graph as k changes.
Is the fact that the problem is independent of k obvious?

 

[Dr.Peterson]

It isn't obvious to me!

But if you rewrite the function as [MATH]f(x) = \frac{\frac{2}{k}}{3\frac{x}{k} - 1} = \frac{1}{k}\frac{2}{3\frac{x}{k} - 1}[/MATH], you can see that it can be obtained from [MATH]f(x) = \frac{2}{3x - 1}[/MATH] by stretching horizontally by k and compressing vertically by 1/k, which explains your observation.

 

[pka]

Nor is it obvious to me either. But the title says integrate.
So \[\int_k^{3k} {\frac{2}{{3x - k}}dx = \frac{2}{3}\log \left| {3x - k} \right|_k^{3k} = \frac{2}{3}\left( {\log (8k} \right) - \log (2k) = \frac{2}{3}\log (4)} \]

 

[apple2357]

It isn't obvious to me!

But if you rewrite the function as [MATH]f(x) = \frac{\frac{2}{k}}{3\frac{x}{k} - 1} = \frac{1}{k}\frac{2}{3\frac{x}{k} - 1}[/MATH], you can see that it can be obtained from [MATH]f(x) = \frac{2}{3x - 1}[/MATH] by stretching horizontally by k and compressing vertically by 1/k, which explains your observation.

Is this the same reasoning as the discussion i opened up on a similar idea?

https://www.freemathhelp.com/forum/...-x-dx-why-is-a-a-a-b-a-ab.115301/#post-450280

 

[Dr.Peterson]

That's related, certainly; but not quite the same reasoning, as it involved some additional issues and wasn't centrally a matter of rewriting the formula.

The related idea is that f(x) = 1/x can be either stretched vertically, k*1/x, or stretched horizontally, 1/(x/k) with identical results, k/x.