The use of min and max in statistics

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Homework Statement



These are taken from my class's lecture slides. I'm trying to understand the entire thought process and reasoning, so I'm just wondering what min and max mean in the g(x) function examples.

Here is the first example: http://gyazo.com/7f9d707df0c8512bfa91a1f200dc3429.png

Here is the other one: http://gyazo.com/91ebcc5fcdb4cbb405449860db0c1404.png

They are both based on the same problem.

Thank you for any help!

Homework Equations





The Attempt at a Solution

 
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penguinnnnnx5 said:

Homework Statement



These are taken from my class's lecture slides. I'm trying to understand the entire thought process and reasoning, so I'm just wondering what min and max mean in the g(x) function examples.

Here is the first example: http://gyazo.com/7f9d707df0c8512bfa91a1f200dc3429.png

Here is the other one: http://gyazo.com/91ebcc5fcdb4cbb405449860db0c1404.png

They are both based on the same problem.

Thank you for any help!

Homework Equations





The Attempt at a Solution


The value of max{a,b} is the larger of the two values of a and b. min{a,b} is the smaller. It's what you might have guessed.
 
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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