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Say we have the definite integral: $$\int_a^b{f(x)\, \mathrm{d}x} = \alpha$$ Given, $a, b,$ and $\alpha \in \mathbb{R}$, is it possible to get the functions $f(x)$ in general case?

Thank you

  • For just one triple $(a,b, \alpha)$? Or do you have this information for arbitrary $a,b$? – M10687 Jun 21 '16 at 22:07
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    No.${}{}{}{}{}{}$ – David Mitra Jun 21 '16 at 22:07
  • Let $\forall n \in \mathbb{N}^*: \ f_n:x\mapsto\begin{cases}n \ (x<\frac{1}{n})\ 0 \ (x \geq \frac{1}{n})\end{cases}$. Calculate $\int_{0}^1 f_n(x) dx$. How many different functions are there for this $a, b, \alpha$ combination? – gpr1 Jun 21 '16 at 22:12
  • I think it's clear that the OP is looking for a way to describe the functions that satisfy the given inequality. It's the only way to interpret the question, really, given that he used the term "functions" and not "function". – Git Gud Jun 21 '16 at 22:18

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No, there will be many functions that have the same definite integral over a given range. In fact if you take any $f(x$ with $$ \int_a^b f(x) \,dx = \alpha $$ and form $$ g(x) = f(x) + \frac{2x}{b+a}-1 $$ you will find that $$ \int_a^b f(x) \,dx = \alpha $$

Mark Fischler
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