I suppose the density is supposed to be
[tex]f(x)=\begin{cases}\frac k{x^4}&\text{for }x\ge1\\0&\text{otherwise}\end{cases}[/tex]
(since [tex]kx^4[/tex] diverges as [tex]x\to\infty[/tex], and 0 is conveniently left out of the domain)
For this to be a proper density function, it must be positive (it is) and its integral over the support [tex][1,\infty)[/tex] must evaluate to 1. We have
[tex]\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=\int_1^\infty\frac k{x^4}\,\mathrm dx=-\frac k{3x^3}\bigg|_1^\infty=\frac k3=1[/tex]
which means [tex]k=3[/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<1\\1-\frac1{x^3}&\text{for }x\ge1\end{cases}[/tex]
The probability that the component exceeds 2 years in its lifetime is
[tex]P(X>2)=1-P(X\le 2)=1-F(2)=1-\left(1-\dfrac18\right)=\dfrac18[/tex]
