Solution:

Since the total probability is 1

Let $\int_{-\infty}^\infty P(x) dx =1 $

= $\int_{0}^3 (\frac{x}{6} +k)dx =1 $

= $[\frac{x^2}{6\times 2} +kx]_0^3 =1$

$[\frac{x^2}{12} +kx]_0^3 =1$

=$[\frac{3^2}{12}+3k]-[\frac{0}{12}+k(0)]=1$

$\frac {9}{12} +3k=1$

$3k=1-\frac {9}{12}$

$k = \frac {1}{12} $

$p(X=x) = \begin{cases} \frac{x}{6}+k, \ if \ & 0 \leq x\leq3 \\ 0, & elsewhere \end{cases}$

$P(1\le x\le 2)$= $\int_1^2 (\frac {x}{6} +\frac {1}{12})dx $

[$\frac {x^2}{6\times 2}+\frac{x}{12}]_1^2 $

upper limit -lower limit

= $\frac {4}{12} +\frac {2}{12}]-[\frac {1}{12}+\frac {1}{12}]$

= $\frac {1}{12} [4+2-1-1]$

= $\frac {1}{12}(4)$

$P(1\le x\le 2)=\frac{1}{3}$