Continuous Probability distribution problem

entire problem in the figure below:

http://sphotos-e.ak.fbcdn.net/hphotos-ak-ash4/1005492_641089589237598_1910688690_n.jpg

please help :(

Note first that integral(x=0 to 2)(f(x) dx) = 1, so the function is already normalized.

a) For this we need to calculate integral(x=0 to 1.2)(f(x) dx) =

integral(x=0 to 1)(x dx) + integral(x=1 to 1.2)((2 - x) dx) =

(x^2 / 2)(from x=0 to 1) + (2x - (x^2 /2))(from x=1 to x=1.2) =

(0.5) + [(2.4 - 0.72) - (2 - (0.5))] = 0.68, or 68%.

b) For this we need to calculate integral(x=0.5 to 1)(f(x) dx) =

integral(x=0.5 to 1)(x dx) = (x^2 /2)(from x=0.5 to 1) =

(0.5) - (0.125) = 0.375, or 37.5%.

I am taking a stab from my general knowledge of these problems:

p(<120 hours) = integral(0,1) x dx + integral(1,1.2) 2-x dx
Note: hours are in terms of hundreds so the boundaries are 1 and 1.2 to represent 100 and 120 respectively.

p(<120 hours)= 0.68 when you calculate


P(50