Concavity of a function: y=2/3x^3-x^2+3x+5

Hollysmoke · 2006-06-27T15:39:18+00:00

I just wanted to make sure I did this right: y=2/3x^3-x^2+3x+5 y’=2x^2-2x+3 y’’=4x-2 =2(2x-1) 2x-1=0 2x=1 x=1/2 And then I substitute x into the original equation to find the IP, right?

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

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