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My question is just for fun, but I want also to verify if I understand something in variation calculus...

I want to know if it is possible to calculate this :

$$ \int_{0}^{1} x \mathrm{d}2x $$

A geometric argument is enough to conclude the area is an half, but here I have a two in my integral and I hope I can't find a result of 1... like this

$$ \int_{0}^{1} x \mathrm{d}2x = 2\int_{0}^{1} x\mathrm{d}x = 1 $$

So it looks like a false result ! Can you explain me why I'm wrong ?

ParaH2
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4 Answers4

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Well it is not a contradiction, actually the real value is 1, because in your case you can apply $$ \int f(x) dg = \int f(x)\cdot g'(x) dx. $$This gives exactly $$ \int_{0}^1 x d(2x) = \int_{0}^1 x\cdot 2 dx = 2\int_{0}^1 x dx =1 $$ OBS: Since you are working on a Riemann-Stieltjes integral you are NOT computing areas,also the change of parameters needs to be justified to be a valid approach

Alonso Delfín
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  • Well here there is two answer to my question and they are different so what I have to think ? I precise I'm not doing maths in my life but chemistry, but sometimes because I like maths I have some specific question ^^ – ParaH2 Apr 17 '15 at 18:55
  • Look that in both answers the result is 1. Then you were wrong thinking that it was 1/2. However, @PransunBiswas answer does not justify the Change of Variables as noted in the coments. In my answer I'am just using a classical result of Riemann-Stieltjes integration. – Alonso Delfín Apr 17 '15 at 18:58
  • @Shadock Since you are not doing maths and I am doing maths may I suggest you that this solution posted by Alonso Delfín is the correct one and one getting up votes is not correct (provided Prasun Biswas is not using Change of Variables for Riemann-Stiltjes Integral, which I am pretty sure he is not)... – Urban PENDU Apr 17 '15 at 18:58
  • @UrbanPENDU thank you, and for the first time I create a dessension here lol now I have an other solution 1/4 x) – ParaH2 Apr 17 '15 at 19:02
  • 1/4 is wrong he forgot to change limits even in change of variables – avz2611 Apr 17 '15 at 19:04
  • The only solution is 1. The one with 1/4 is getting seriously down voted, It makes no sense at all :). However as noted by @UrbanPENDU you can see that sometimes an answer can give the correct result by luck but have an unjustified approach. – Alonso Delfín Apr 17 '15 at 19:05
  • yes the approach was not justified , but i guess we can use change of variables (standard form ) whenever $g(x)=kx$ where k is some constant , as it still evaluates area under the curve of a compressed function – avz2611 Apr 17 '15 at 19:07
  • @avz2611Right, however the OP, may think changing variables on Riemann-Stiltjes Integral works as well as on Riemann Integrals, which is not true. – Alonso Delfín Apr 17 '15 at 19:11
  • I promise I don't ask the same question if I put a power on the "d" like $\int x d^2 x$ or better with fractionnal calculus like this $\int x d^{3/8} x$ :P – ParaH2 Apr 17 '15 at 19:12
  • @AlonsoDelfín Oh, change of variables in R-S integral works just fine...when done correctly, of course. – Timbuc Apr 17 '15 at 19:17
  • @Timbuc Yes indeed, but you have to check more hypothesis than with the usual integral, what I meant is that you can not take it as lightly as one may think, for example if $g$ is not an increasing continuous function. – Alonso Delfín Apr 17 '15 at 19:21
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using the integration by parts of Riemann–Stieltjes integral we get $$\int_{a}^{b}f(x)dg(x)=-f(a)g(a)+f(b)g(b)-\int_{a}^{b}g(x)df(x)$$ now substitute $a=0$ , $b=1$ and $f(x)=x$ and $g(x)=2x$

avz2611
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May I propose you another, more geometric, approach?:

$$\int_0^1 x\,d(2x)=\frac12\int_0^1 2x\,d(2x)$$

Think of the above integral as usually done in high school: that gives the area between the identity function $\;f(2x)=2x\;$ and the $\;2x$-axis .

In this "scaled up" axis system, we have the area of a straight angle triangle with vertices at the origin $\;(0,0)\;$, at $\;\left(0,2\right)\;$ and at $\;(2,2)\;$ , and thus its area is $\;\frac{2\cdot2}2=2\;$ . This, times $\;\frac12\;$ , gives us the correct solution of $\;1\;$ .

Timbuc
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$$\int\limits_0^1x\,\mathrm d(2x)\stackrel{u=2x}=\frac{1}{2}\int\limits_0^2u\,\mathrm du$$

How are the limits decided? This way:

$$u=2x\implies \begin{cases}x=0\implies u=0\\ x=1\implies u=2\end{cases}$$

Can you continue?

Prasun Biswas
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  • Yes I can, thank you ! I forget to change the limits ! – ParaH2 Apr 17 '15 at 18:48
  • @Prasun Biswas Are you using Change of Variables for Riemann-Stiltjes Integral??? – Urban PENDU Apr 17 '15 at 18:52
  • he is not using it – avz2611 Apr 17 '15 at 18:53
  • If he is not using it then the solution is not well justified, he just got lucky to get the answer – Urban PENDU Apr 17 '15 at 18:53
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    Seriously why are you guys voting up this not well-justified solution. The solution given by Alonso Delfín is a correct solution.... – Urban PENDU Apr 17 '15 at 18:55
  • @UrbanPENDU although change of variables is justified here as it is of form $kx$ which means it still evaluates area under the curve just a one that is compressed , although if function inside was maybe $x^2$ or something , i guess it is not justified – avz2611 Apr 17 '15 at 19:04
  • yes you know that and I also know that but the authors solution doesn't show that whether he is aware of such a thing or not. Change of Variables needs to be mentioned at least. – Urban PENDU Apr 17 '15 at 19:08
  • i agree with that – avz2611 Apr 17 '15 at 19:09