Evaluate the following definite integrals using the Fundamental Theorem of Calculus.

$\displaystyle \int_{-6}^1 s\, \left|9 - s^2\right|\, ds$ =

$\displaystyle \int_{0}^{9 \pi^2} \frac{\sin(\sqrt{x})}{\sqrt{x}}\,dx$ =

$\displaystyle \int_{3}^{6}\frac{t-3}{t^2-6 t +10}\,dt$ =