Definite Integral Party

Hello, fellow members of the human race!

I came across the following digression(while studying organic chemistry....).

And I would love it if anyone could prove it for me because I seem to have no clue whatsoever. Here goes

abf(x)dx=(ba)01fa+(ba)xdx\displaystyle\int_{a}^{b}f(x)dx = (b-a)\int_{0}^{1}f\lfloor a+(b-a)x\rfloor dx

Where, f(x)f(x) is an integrable function in the range [a,b][a,b] , aa and bb are real numbers and \lfloor\cdot\rfloor denotes the Greatest Integer Function or the Floor function.

By the way, the hints suggested were the following properties:

abf(x)dx=a+cb+cf(xc)dx\displaystyle\int_{a}^{b}f(x)dx = \int_{a+c}^{b+c}f(x-c) dx; and

abf(x)dx=(1k)kakbf(xk)dx\displaystyle\int_{a}^{b}f(x)dx = (\frac{1}{k})\int_{ka}^{kb}f(\frac{x}{k}) dx

I had absolutely no clue as to how to the use the hints.........

You might..........

Also everyone give a shout out to Deeparaj for getting a seat in CMI. Let's congratulate him on his amazing achievement!!!!!

Note by Anirudh Chandramouli
4 years, 11 months ago

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First off,  Thank you!  \textbf{ Thank you! }

Secondly, similar things happened to me while trying to read organic seriously :P

Thirdly, (ba)01f(a+(ba)x)dx=0baf(a+u)duu=x(ba)=abf(v)dvv=u+aabf(x)dx( in general ) \begin{aligned} (b-a) \int_0^1 f \left (\left \lfloor a + (b-a)x \right \rfloor \right) \, dx &= \int_0^{b-a} f \left (\left \lfloor a+u \right \rfloor \right) \, du \quad u=x(b-a) \\&= \int_a^b f \left (\left \lfloor v \right \rfloor \right) \, dv \quad v=u+a \\& \neq \int_a^b f(x) \, dx \quad (\text{ in general } )\end{aligned}

So, most probably, you misread square brackets as the floor function.

A Former Brilliant Member - 4 years, 11 months ago

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Thanks for the solution. I was completely dumbfounded by the perceived GIF and hence chose to resort to foreign help to solve the question. You are probably right about my mistake in reading the square brackets as a GIF. The property is trivial when the GIF is absent thanks anyways

Anirudh Chandramouli - 4 years, 11 months ago

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Yep. You're welcome!

A Former Brilliant Member - 4 years, 11 months ago

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