Let $f:(1, \infty) \rightarrow \mathbf{R}$ be a function defined as $f(x)=\frac{x-1}{x+1}$. Let $f^{i+1}(x)=f\left(f^i(x)\right), i=1,2, \ldots, 25$, where $f^1(x)=f(x)$. If $g(x)+f^{26}(x)=0, x \in(1, \infty)$, then the area of the region bounded by the curves $y=g(x), 2 y=2 x-3, y=0$ and $x=4$ is :

The area of the region $\left\{(x, y): x^2-8 x \leq y \leq-x\right\}$ is :

The area of the region $\left\{(x, y): 0 \leq y \leq 6-x, y^2 \geq 4 x-3, x \geq 0\right\}$ is :

Let $e$ be the base of natural logarithm and let $f:\{1,2,3,4\} \rightarrow\left\{1, e, e^2, e^3\right\}$ and $\mathrm{g}:\left\{1, e, e^2, e^3\right\} \rightarrow\left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right\}$ be two bijective functions such that $f$ is strictly decreasing and $g$ is strictly increasing. If $\phi(x)=\left[f^{-1}\left\{g^{-1}\left(\frac{1}{2}\right)\right\}\right]^x$, then the area of the region $\mathrm{R}=\left\{(x, y): x^2 \leq y \leq \phi(x), 0 \leq x \leq 1\right\}$ is :