Answer :
Use the fundamental theorem of calculus.
[tex]f(t) = f(0) + \displaystyle \int_0^t f'(u) \, du[/tex]
In order to find velocity and position exactly, you need to know the initial velocity and position.
1) Integrate the acceleration function to get the velocity function. By the FTC,
[tex]v(t) = v(0) + \displaystyle \int_0^t a(u) \, du[/tex]
[tex]v(t) = v(0) + \displaystyle \int_0^t (3u^3+2u^2) \, du[/tex]
[tex]v(t) = \dfrac34 t^4 + \dfrac23 t^3 + v(0)[/tex]
At [tex]t=2\,\rm s[/tex], the velocity is
[tex]v(2) = \dfrac34 2^4 + \dfrac23 2^3 + v(0) = \boxed{\dfrac{52}3 + v(0)}[/tex]
2) Integrate the velocity function to the get the position function. By the FTC again,
[tex]x(t) = x(0) + \displaystyle \int_0^t v(u) \, du[/tex]
[tex]x(t) = x(0) + \displaystyle \int_0^t \left(\frac34 u^4 + \frac23 u^3 + v(0)\right) \, du[/tex]
[tex]x(t) = \displaystyle \frac3{20} t^5 + \frac16 t^4 + v(0) t + x(0)[/tex]
At [tex]t=4\,\rm s[/tex], the object's position is
[tex]x(4) = \dfrac3{20}4^5 + \dfrac16 4^4 + 4v(0) + x(0) = \dfrac{2944}{15} + 4v(0) + x(0)[/tex]
Fortunately, you don't need the initial position to find the displacement, since [tex]x(0)[/tex] cancels out in the end.
[tex]\Delta x = x(4) - x(0) = \boxed{\dfrac{2944}{15} + 4v(0)}[/tex]