For (x), you are given that $$\mu_x= \begin{cases} 0.01, &\text{if}& x<50\\ 0.02, &\text{if}& x>60\end{cases}$$
and $\mu_x$ is continuous ad linear on [50,60]. Calculate $_{10}p_{55}\\$.
Here is part of my solution:
Note that $_{10}p_{55}= e^{-\int_{55}^{65} \mu_y dy}$
Calculating the integral part:
$\int_{55}^{65} \mu_y dy = \int_{55}^{60} \mu_y dy + \int_{60}^{65} \mu_y dy = \int_{55}^{60} \mu_y dy + \int_{60}^{65} 0.02 dy $
I have no idea how to solve the first integral from 55 to 60 as its force of mortality is not given. Maybe by using the assumption that $\mu_x$ is continuous and linear on [50,60], I might be able to solve it?
If $50\le x\le 60$, $$\mu_x= 0.01 + \tfrac{0.02-0.01}{60-50}(x-50)=0.001x-0.04.$$