a. In order for [tex]f[/tex] to be a proper density function, its integral over its domain must evaluate to 1:
[tex]\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=k\int_0^1(1-x)\,\mathrm dx=\frac k2=1\implies k=2[/tex]
[tex]f[/tex] also must be non-negative over its support, which is the case here.
I assume the instruction regarding the distribution function doesn't apply to parts (b) and (c).
(b) Integrate the density over the interval [0.1, 0.2]:
[tex]P(0.1<x<0.2)=\displaystyle\int_{0.1}^{0.1}2(1-x)\,\mathrm dx=0.17[/tex]
(c) Integrate the density over the interval [0.5, 1]:
[tex]P(X>0.5)=\displaystyle\int_{0.5}^12(1-x)\,\mathrm dx=0.25[/tex]
The distribution function is obtained by integrating the density:
[tex]F(x)=\displaystyle\int_{-\infty}^xf(t)\,\mathrm dt=\begin{cases}0&\text{for }x<0\\2x-x^2&\text{for }0\le x<1\\1&\text{for }x\ge1\end{cases}[/tex]
(d) Using the distribution function, we have
[tex]P(X<0.3)=F(0.3)=0.51[/tex]
(e) Using [tex]F[/tex] again, we get
[tex]P(0.4<X<0.6)=P(X<0.6)-P(X<0.4)=F(0.6)-F(0.4)=0.84-0.64=0.2[/tex]
(f) The mean is
[tex]E[X]=\displaystyle\int_{-\infty}^\infty x\,f(x)\,\mathrm dx=\int_0^12x(1-x)\,\mathrm dx=\frac13[/tex]
The variance is
[tex]V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2[/tex]
where
[tex]E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f(x)\,\mathrm dx=\int_0^12x^2(1-x)\,\mathrm dx=\frac16[/tex]
so that the variance is
[tex]V[X]=\dfrac16-\left(\dfrac13\right)^2=\dfrac1{18}[/tex]
