One of the points of intersection of the curves $$y=1+3 x-2 x^2$$ and $$y=\frac{1}{x}$$ is $$\left(\frac{1}{2}, 2\right)$$. Let the area of the region enclosed by these curves be $$\frac{1}{24}(l \sqrt{5}+\mathrm{m})-\mathrm{n} \log _{\mathrm{e}}(1+\sqrt{5})$$, where $$l, \mathrm{~m}, \mathrm{n} \in \mathbf{N}$$. Then $$l+\mathrm{m}+\mathrm{n}$$ is equal to

The area of the region enclosed by the parabolas $$y=4 x-x^2$$ and $$3 y=(x-4)^2$$ is equal to :

The area of the region $$\left\{(x, y): y^2 \leq 4 x, x<4, \frac{x y(x-1)(x-2)}{(x-3)(x-4)}>0, x \neq 3\right\}$$ is