Let $$I_{n}(x)=\int_{0}^{x} \frac{1}{\left(t^{2}+5\right)^{n}} d t, n=1,2,3, \ldots .$$ Then :

The area enclosed by the curves $$y=\log _{e}\left(x+\mathrm{e}^{2}\right), x=\log _{e}\left(\frac{2}{y}\right)$$ and $$x=\log _{\mathrm{e}} 2$$, above the line $$y=1$$ is:

Let $$f(x)=2+|x|-|x-1|+|x+1|, x \in \mathbf{R}$$.

Consider

$$(\mathrm{S} 1): f^{\prime}\left(-\frac{3}{2}\right)+f^{\prime}\left(-\frac{1}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)+f^{\prime}\left(\frac{3}{2}\right)=2$$

$$(\mathrm{S} 2): \int\limits_{-2}^{2} f(x) \mathrm{d} x=12$$

Then,

The area of the region enclosed by $$y \leq 4 x^{2}, x^{2} \leq 9 y$$ and $$y \leq 4$$, is equal to :