Sorry Sam, but I disagree with your answer.
You are given a velocity equation and are asked for the position (equation), given a known starting point of s(0) = 3.
What is the relationship between position and velocity? The derivative of position is velocity, so since we were given velocity, we will need to integrate the velocity equation to get the position equation.
You wrote two different v(t) equations, the first subtracting e^4, the second multiplying e^4 with the sin equation. I'm going to assume you meant the first one and the second is a typo. If my assumption is wrong, this will be an Integration by Parts problem for the integral instead.
So we need to integrate v(t) = sin (2t) - e^4 + 1
We can break this up then as the integral of sin(2t) minus the integral of e^4 plus the integral of 1.
For the integral of sin (2t), we can treat this as a U-Substitution problem:
u = 2t
du = 2 dt
1/2 du = dt
So we are taking the integral of 1/2 sin (u) du = -1/2 cos (u) = -1/2 cos(2t) (+ C, but we will just add one +C at the end of the entire integral).
What about the integral of e^4? Don't be fooled by this, there is no variable here, this is just a constant. Therefore, the integral of e^4 = (e^4)t.
Finally the integral of 1 = t
Putting everything back together (with the proper signs of addition/subtraction in front of each part) we get:
The integral of sin(2t) - e^4 + 1 = -1/2cos(2t) - (e^4)t + t +C
This is where the initial condition of s(0) = 3 comes in, we use it to solve for C.
If s(0) = 3, then
3 = -1/2cos (2(0)) -(e^4)(0) + (0) +C
3 = -1/2 + C
C = 7/2 or 3.5
Making the final substitution for C, your final answer is:
s(t) = -1/2cos(2t) - (e^4)t + t + 7/2
P.S. Sorry, wasn't able to add the integral sign into my explanation.