Creating a Low-Maintenance Garden

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Can someone see if my explanation is correct?
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Just a minor remark: You do not need to check the second derivative at the local extremum as you are anyway computing its value. Comparing the values of the local extrema and the boundary values will give you the global min and max.
 
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Orodruin said:
Just a minor remark: You do not need to check the second derivative at the local extremum as you are anyway computing its value. Comparing the values of the local extrema and the boundary values will give you the global min and max.
Thank you; I accidentally didn't finish writing the question properly, but it's been fixed.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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