# 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

$\displaystyle\int_{a}^{b}f(x)dx = (b-a)\int_{0}^{1}f\lfloor a+(b-a)x\rfloor dx$

Where, $f(x)$ is an integrable function in the range $[a,b]$ , $a$ and $b$ 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:

$\displaystyle\int_{a}^{b}f(x)dx = \int_{a+c}^{b+c}f(x-c) dx$; and

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

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, 4 months ago

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

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

Thirdly, \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.

- 4 years, 4 months ago

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

- 4 years, 4 months ago

Yep. You're welcome!

- 4 years, 4 months ago