How to determine the velocity of an object by the graph of its force? The multiple choice question reads:

A 6-kg object is at rest on a frictionless horizontal plane.

If the object is moved from rest by the force shown, what is the velocity of the object at time t = 4?
(A) 2 m/s
(B) 6 m/s
(C) 12 m/s
(D) 16 m/s
(E) 32 m/s

My assumption to solve this is to integrate to determine the area under the curve, then divide by the mass of the object, which would result in a value of 4.67 m/s. The alleged value is 6 m/s, or choice B. Is my approach correct or am I missing a crucial component?
 A: Yep, as this is the net force as a function of time, the area would be the impulse (change in momentum). You can find the object’s change if velocity. It’s not clear what “time images” mean as pointed out. The approach is fine but question is not clear.
A: I got the same answer as you: $4/6+3\cdot8/6=4.667 $
I think I'd be questioning that question. 
A: $$F = ma \implies a = {F \over m}$$
As $F$ is changing with time $t$, $a$ is changing with time $t$, too:
$$a(t) = {F(t) \over m}$$
The speed $v$ is an integral of the acceleration:
$$v(4) = \int_0^4 a(t)\, \mathrm dt = \int_0^4 {F(t) \over m}\, \mathrm dt = \frac 1 m\int_0^4 {F(t)}\, \mathrm dt\tag1\label 1$$
$\int_0^4 {F(t)}\, \mathrm dt$ is the area under your graph, i. e. the area of the  firs triangle and one rectangle, so
$$\int_0^4 {F(t)}\, \mathrm dt = {1 \cdot 8 \over 2}+ 3\cdot 8 = 28$$
Substituting it and the mass into $\eqref 1$ you will obtain
$$v(4) = \frac 1 6\cdot 28 = 4\cfrac {_{2}} {^3}$$
A: At the instant $t=4$, force $F = 8N$
As
$$F = \frac{dp}{dt}$$
$$p = \int{Fdt} = \int{8dt} = 8t$$
Now, $p = mv$
$$v = {p \over m} = \frac{8t}{6}$$
At t = 4 s,
$$v = \frac{8\cdot4}{6} = 5.333\; \mathrm {ms^{-1}} $$
