Are local extremum possible at endpoints of a closed bounded interval?

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I thought local extremum did not exist at the endpoints of a closed bounded interval, however my textbook claims this.

Wikipedia:

"A continuous (real-valued) function on a compact set always takes maximum and minimum values on that set. An important example is a function whose domain is a closed (and bounded) interval of real numbers (see the graph above). The neighborhood requirement precludes a local maximum or minimum at an endpoint of an interval."
 
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If you consider a 'neighborhood' to be a 'neighborhood in the domain' your textbook is right. If you consider it to be a 'neighborhood in the reals' then Wikipedia is right. There are a lot of terms that are defined somewhat differently in different references. I think you'd better live by your textbooks definition. Until you change textbooks.
 
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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