A random variable $X$ has the following probability distribution

$$ \begin{array}{|l|c|c|c|c|c|} \hline \mathrm{X}: & 0 & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}): & \mathrm{k} & 2 \mathrm{k} & 4 \mathrm{k} & 2 \mathrm{k} & \mathrm{k} \\ \hline \end{array} $$

then the value of $\mathrm{P}(1 \leqslant \mathrm{X}<4 \mid \mathrm{X} \leqslant 2)=$

The area of the region bounded by $\frac{x^2}{9}+\frac{y^2}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$ is

If $\mathrm{f}(x)=\left\{\begin{array}{ll}\operatorname{m} x+1, & x \leqslant \frac{\pi}{2} \\ \sin x+\mathrm{n}, & x>\frac{\pi}{2}\end{array}\right.$, is continuous at $x=\frac{\pi}{2},(\mathrm{~m}, \mathrm{n} \in \mathbb{Z})$ then

$$ \int_{-2}^2\left|x^2-x-2\right| \mathrm{d} x= $$