Reducing an integral equation to a differential one

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In my course about differential equations I have the following problem:

Find all the functions $f:\mathbb{R} \longrightarrow \mathbb{R}^+$ such that the area below the graphic of the function in an interval $[a,b]$ (b>a) is proportional to the length of the interval itself. (with $f \in \mathcal{C}^1(\mathbb{R}))$

I have enunciate it as follows:

$\int_a^b{f(x)}dx = K (b-a)$

but I cant solve that with my knowledge in differential equiations. I have read in serveral places that integral equations and differential ones can be related between them but I can't think of any way to transform it to a differential equation

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One form of the Fundamental Theorem of Calculus says: $$\frac{d}{db} \int_a^b f(x)\,dx = f(b)\tag1$$ where $f$ is assumed continuous, and $b$ is taken as a variable with respect to which the derivative is taken.

Differentiating both sides of $$\int_a^b{f(x)}dx = K (b-a)\tag2$$ with respect to $b$ and using (1), you will arrive at the conclusion.