Finding final position using acceleration vs time graph

I am given initial position (x,y)=(2m,7m) and initial velocity (9m/s,-5m/s). What is the final position vector?
The acceleration vs time graph shows the acceleration (for x) is 2 m/s2 for 2 seconds, then -1m/s2 for the next 4 seconds, then back to 2 m/s2. (total time is 8 seconds).

My question is how to "extract" the acceleration from this graph in order to incorporate it into the proper kinematic equation?

**hopefully this link shows the graph
http://www.webassign.net/zinphysls2/2-p-014.gif

x: initial d=2m , v=9m/s
d= 2+9*2+1/2*2*2^2= 24m
v= 9+2*2= 13m/s
d= 24+13*4+1/2(-1)*4^2= 68m
v= 13-1*4= 9m/s
d= 68+9*2+1/2*2*2^2= 90[m]
v= 9+2*2= 13[m/s]

y: initial d=7m , v=-5m/s
d= 7+(-5)*2= -3m
v= -5+0= -5m/s
d= -3+(-5)*3+1/2*1*3^2= -13.5m
v= -5+1*3= -2m/s
d= -13.5+(-2)*3+1/2(-2)*3^2= -28.5[m]
v= -2+(-2)*3= -8[m/s]

d(x,y)= 90m , -28.5m
v(x,y)= 13m/s , -8m/s