Definitely not a semicircle

Calculus Level 3

1 ( 1 x ) 1 x x x 1-(1-x)^{1-x}x^x The curve looks like a semicircle with a radius 0.5.

Which is larger, 0 1 1 ( 1 x ) 1 x x x d x \displaystyle\int_0^1 1-(1-x)^{1-x}x^x dx or π 8 \dfrac{\pi}8 ?

0 1 1 ( 1 x ) 1 x x x d x \displaystyle\int_0^1 1-(1-x)^{1-x}x^x dx Can't determine. π 8 \dfrac{\pi}8

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1 solution

Vincent Moroney
Jun 18, 2018

π 8 = 0 1 1 2 ( 1 ( 2 x 1 ) 2 ) d x \frac{\pi}{8} = \int_0^1\frac{1}{2}\sqrt{(1-(2x-1)^2)}\,dx For all x [ 0 , 1 2 ) ( 1 2 , 1 ] we have 1 2 ( 1 ( 2 x 1 ) 2 ) > 1 ( 1 x ) 1 x x x . \text{For all } x \in \Big[0,\frac{1}{2}\Big) \cup \Big(\frac{1}{2}, 1\Big] \text{ we have } \frac{1}{2}\sqrt{(1-(2x-1)^2)} > 1-(1-x)^{1-x}x^x. So it must be true that 0 1 1 ( 1 x ) 1 x x x < π 8 . \text{So it must be true that } \int_0^11-(1-x)^{1-x}x^x < \frac{\pi}{8}. I observed this via graphing.

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