Given \( f(x) = \begin{cases} \frac{1}{2}(b^2 - a^2), & 0 \le x \le a \\[6pt] \frac{1}{2}b^2 - \frac{x^2}{6} - \frac{a^3}{3x}, & a<x \le b \\[6pt] \frac{1}{3} \cdot \frac{b^3 - a^3}{x}, & x>b \end{cases} \). Then:
$\displaystyle \int \frac{dx}{\sin(x-a)\cos(x-b)} =$
$\displaystyle \int_{\frac{1}{25}}^{1} x^{-2} e^{x^{-1/2}} dx =$
The area (in sq. units) bounded by the curves $y=\frac{8}{x}$, $y=2x$ and $x=4$ is