Area of a triangle using definite integral from a to x

I am doing a question that's asking me to evaluate the value of an integral from -1 to 0 and the area under the graph happens to be a triangle; when I do it however I get 3 instead of 3/2 (since the area of a triangle is bh/2) but I don't get it, isn't my method supposed to work?? here is the link in case you want to see it

http://www.collegeboard.com/prod_downloads/ap/students/calculus/calculus_ab_frq_02.pdf
section 2 question 4) a) evaluate g(-1)
what I did:
g(-1)= - integral from -1 to 0 of f(t) dt
-((f(-1)-f(0)) = -(0-3) = 3

unofficial scoring guidelines ( http://home.roadrunner.com/~askmrcalculus/ABBC124.html) say otherwise

HELP!

line of the form y=mx+c
m =3-(-3)/0-(-2) = -6/-2 =3
c = 3 (intercept when t =0)
g (t) = 3t +3

area= integral from -1 to 0 of f(t) dt
area= integral from -1 to 0 of [3t +3] dt
area = [3t^2/2 +3] between limits of -1 and 0
area= [3 x (0)^2 + 3 x 0]-[(3 x (-1)^2/2) +3(-1)]
area = [0 -0]- [(3 x 1/2) -3]
area = -3/2 +3
area = -3/2 +6/2
area = 3/2

g(x) = ∫[0,x] f(t) dt
g'(x) = f(x)
g"(x) = f '(x)

Rather than writing out a piecewise function for f(t), i'll just use basic shapes.

a) g(-1) = A_Δ_above + A_Δ_below = (1/2)(3)(1) + (1/2)(-3)(1) = 0
g'(-1) = f(-1) = 0
g"(-1) = f '(-1) = Slope of f(t) at x = -1 → (3 - 0)/(0 - (-1)) = 3