Prove the obvious!

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mathman
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Let ##f(x)## be an integrable function defined on ##[0,1]## with the following property: ##a=inf(f(x))\lt f(x) \lt b=sup(f(x))##. Prove ##a\lt \int_0^1f(x)dx \lt b##. It is obviously true, but how does one prove it?
 

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Let ##f(x)## be an integrable function defined on ##[0,1]## with the following property: ##a=inf(f(x))\lt f(x) \lt b=sup(f(x))##. Prove ##a\lt \int_0^1f(x)dx \lt b##. It is obviously true, but how does one prove it?
Mean value theorem for integration: https://en.wikipedia.org/wiki/Mean_value_theorem#Mean_value_theorems_for_integration
It requires a continuous function ##f(x)##, so one probably has to have a look on the proof. But if ##f## is continuous, we have ##\int_0^1f(x)dx = f(\xi)\cdot (1-0)## with a mean value ##a<f(\xi)<b##.

The general version has two functions: ##g## continuous, and ##f## integrable, and says ##\int_0^1f(x)g(x)dx=g(\xi)\int_0^1f(x)dx##. Maybe one can find an appropriate ##g## and apply this general version.
 
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Alternatively, write the integral as limit of a sequence of Riemann sums. To get the strict inequality, you will have to play with some ##\epsilon##'s though.
 

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