Calculus please please help me :)

1) Find the position function s(t) given acceleration a(t)=3t if V(2)=0 and s(2)=1
2) If F''(x)= -4sin2x, f(0)=0 and f'(0)=2, find the value of f(pi/4)

1) V(t) = integral a(t) dt
V(t) = integral 3t dt
V(t) = 3t^2/2 + C
Put V(2) = 0
0 = 3 * 2^2/2 + C
0 = 6 + C
C = -6
Therefore, V(t) = 3t^2/2 - 6

s(t) = integral V(t) dt
s(t) = integral 3t^2/2 - 6 dt
s(t) = 3/2 * t^3/3 - 6t + K
s(t) = 1/2 * t^3 - 6t + K

Put s(2) = 1
1 = 1/2 * 2^3 - 6*2 + K
1 = -8 + K
K = 9
s(t) = 1/2 * t^3 - 6t + 9 (answer)

2) f '' (x) = -4sin2x
f '(x) = integral -4 sin2x dx
f '(x) = -4 * 1/2 * (-cos 2x) + C
f '(x) = 2 cos2x + C
Put f '(0) = 2
2 = 2 cos0 + C
2 = 2 + C
C = 0
Therefore f '(x) = 2 cos 2x
f(x) = integral f '(x) dx = integral 2 cos2x dx
f(x) = 2 * 1/2 * sin 2x + K
f(x) = sin 2x + K
Put f(0) = 0
0 = sin0 + K
0 = 0 + K
K = 0
Therefore, f(x) = sin 2x
f(pi/4) = sin 2(pi/4) = sin(pi/2) = 1 (answer)