Midpoint Rule Simpsons Rule Trap Rule Error Bounds 
I need help on the error part. I know the formula for finding the error but I am having trouble determining what $k$ should be. Is $k$ just any number?
 A: Let's do the analysis for TRAP. The formula that you wrote that you know probably reads something like
$$\frac{(b-a)^3}{12n^2}|f''(\xi)|,$$
where $\xi$ is a number between $a$ and $b$. 
In our case, $b=2$, and $a=0$. Also, our function $f(x)$ is $\cos(x^3)$. Then $f'(x)=- 3x^2\sin(x^3)$ and therefore $f''(x)=-9x^4\cos(x^3)-6x\sin(x^3)$. We want to find an upper bound $B$ on $|f''(x)|$ in our interval. It is not clear how hard we should work to get a cheap upper bound. A crude one is that since $\sin$ and $\cos$ are never bigger than $1$, we have $|f''(x)|\le 9(2^4)+12=156$.
So our error if we use $n$ intervals is $\le \frac{8}{12n^2}(156)=\frac{104}{n^2}$.
To make sure this is less than $10^{-4}$, we solve the inequality $\frac{104}{n^2}\lt 10^{-4}$, or equivalently $n^2\gt (104)(10000)$. The smallest integer that works is $n=1020$.
Remark: Note that our calculation does not show that $n=1020$ is needed, only that for sure it will do the job. 
For one thing, the maximum of the second derivative over our interval is under $50$.
Far more relevant is that when we use the maximum for our estimate of $|f''(\xi)|)$ we are for most functions being very pessimistic. The error will usually be far smaller than the pessimistic estimate we made, and the required $n$ therefore far smaller. The estimates are not used in practical work. Roughly speaking, one doubles the number of points used until the answer becomes stable enough. 
A: For the midpoint and trapezoidal rules $K$ is the maximum magnitude of the second derivative over the interval.
$$K=\max_{x\in[a,b]}|f''(x)|.$$
For Simpsons rule is of the fourth derivative
$$K_S=\max_{x\in[a,b]}|f^{iv}(x)|.$$
I am guessing that the functions have to be twice respectively four times continuously differentiable as your one is.
For part (b) they have given you an upper bound for $K$.
