Tetration Without Integration

Note that $\displaystyle \int_0^1 x \, dx = \frac{1}{2}$ .

We will show that $x ^ { x^ x } \geq x$ ,
At $x=1$ and $x=0$ , the two values are equal. Consider $0 < x < 1$ .
Recall for that all $0 < a , b < 1,$ $a^b < 1^b = 1$ and $a^b > a^1 = a$ hence $a < a^b < 1$ .
Let $y = x^x < 1$ . We get that $x^y > x$ , and thus $x^ { x^x } > x$ .

Hence, we can conclude that

$\displaystyle \int_0^1 x ^{x^x} \, dx > \int_0^1 x = \frac{1}{2}.$

In a x-y plane if we join the origin to the point (1,1), join 1,1 to 1,0. and join 1,0 to 0,0 again we get a triangle whose area is 1/2 . If you see x^x^x ,for every value of x between 0 and 1 has a greater value than x except at x=1 . So the curve x^x^x between x=0,1 always lies above line x=y which is the top side of the triangle considered so its integral which is the area is greater than 1/2