$\Rightarrow p(x)=-p(1-x)+c$
at $x=0$ $p(0)=-p(1)+c \quad $
$ \Rightarrow 42=c$
now $p(x)=-p(1-x)+42$
$\Rightarrow p(x)+p(1-x)=42$
$I = \int\limits_0^1 p (x)dx = \int\limits_0^1 p (1 - x)dx$
$I = \int\limits_0^1 {\left( {42} \right)dx} \,\,\,\,\,\,\,\, \Rightarrow I = 21.$
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$R=\left\{(x, y): \max \left\{0, \log _{e} x\right\} \leq y \leq 2^{x}, \frac{1}{2} \leq x \leq 2\right\}$
is, $\alpha\left(\log _{e} 2\right)^{-1}+\beta\left(\log _{e} 2\right)+\gamma$, then the value of $(\alpha+\beta-2 \gamma)^{2}$ is equal to: