HELP WITH PHYSICS..............!

A car is initially at rest on a straight road. The histogram below shows the car's acceleration along that road as a function of time.
1. Calculate the distance traveled during the first 5 s.
2. Calculate the distance traveled from t=10 s to t=14 s.


HERE IS THE HISTOGRAM:

http://img337.imageshack.us/img337/7789/screenshot20120907at307.png
3.Calculate the car's average speed from t = 6 s to t=9 s.

use 2 formula's:
Vf = Vi + a×t
and
d = (Vi + Vf) × t/2
where
Vf = final velocity
Vi = initial velocity
a = acceleration
t = time
d = distance

from t = 0 s to t = 1 s
Vi = 0 m/s
a = 0 m/s²
t = 1 - 0 = 1 s
Vf = 0 + 0×1 = 0 m/s
d = (0 + 0) × 1/2 = 0 m

from t = 1 s to t = 2 s
Vi = 0 m/s
a = 2 m/s²
t = 2 - 1 = 1 s
Vf = 0 + 2×1 = 2 m/s
d = (0 + 2) × 1/2 = 1 m

from t = 2 s to t = 4 s
Vi = 2 m/s
a = 3 m/s²
t = 4 - 2 = 2 s
Vf = 2 + 3×2 = 8 m/s
d = (2 + 8) × 2/2 = 10 m

from t = 4 s to 5 s
t = 5 - 4 = 1 s
Vi = 8 m/s
a = -1 m/s²
Vf = 8 + (-1)×1 = 7 m/s
d = (8 + 7) × 1/2 = 7.5 m

from t = 5 s to t = 6 s
t = 6 - 5 = 1 s
Vi = 7 m/s
a = -2 m/s²
Vf = 7 + (-2)×1 = 5 m/s
d = (7 + 5) × 1/2 = 6 m

from t = 6 s to t = 7 s
t = 7 - 6 = 1 s
Vi = 5 m/s
a = 1 m/s²
Vf = 5 + 1×1 = 6 m/s
d = (5 + 6) × 1/2 = 5.5 m

from t = 7 s to t = 8 s
t = 8 - 7 = 1 s
Vi = 6 m/s
a = 3 m/s²
Vf = 6 + 3×1 = 9 m/s
d = (6 + 9) × 1/2 = 7.5 m

from t = 8 s to t = 9 s
t = 9 - 8 = 1 s
Vi = 9 m/s
a = -1 m/s²
Vf = 9 + (-1)×1 = 8 m/s
d = (9 + 8) × 1/2 = 8.5 m

from t = 9 s to t = 10 s
t = 10 - 9 = 1 s
Vi = 8 m/s
a = -1 m/s²
Vf = 8 + (-1)×1 = 7 m/s
d = (8 + 7) × 1/2 = 7.5 m

from t = 10 s to t = 14 s
t = 14 - 10 = 4 s
Vi = 7 m/s
a = 0 m/s²
Vf = 7 + 0×4 = 7 m/s
d = (7 + 7) × 4/2 = 28 m

Q1)
d = 0 + 1 + 10 + 7.5 = 18.5 m

Q2)
d = 28 m

Q3)
d = 5.5 + 7.5 + 8.5 = 21.5 m
t = 9 - 6 = 3 s
v = 21.5 / 3 = 7.16 m/s