Understanding Upper Bound & Sup in Theorem Proving

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This calc book that I am reading uses words like "upper bound" and "sup" a lot when proving theorems. I have never heared these terms before so it makes it hard for me to understand the proofs.

I think it has to deal with max's values of a graph: For example given a set S of all elements c in a ≤ c ≤ b would the the upper bound in the following graph in [a,b] be c? and also c = Sup S...
is this correct?
 

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Upper bound is any value that is greater than or equal to all the values in a set.

sup is short for supremum, the least upper bound of a set of numbers. The range of a function might be such a set. The supremum is not necessarily a member of the set: for example, the supremum of the range of the sigmoid function \frac{1}{1+e^{-x}} is 1, but that is not a value of the function. 1 is an upper bound to this function, but so are 2, 100 and 100!2
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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