Approximate \cos(x^2) using Composite Trapezoidal method Find an approximation to $$\int_2^3 \cos(x^2)dx$$ using the Composite Trapezoidal method with $r = 3$ and estimate the error. 
Attempt: using the composite trapezoidal formula I calculated the approximation was equal to $$\frac 16 [\cos(4) + \cos(9) + \cos(16) + \cos(9) $$ which is approximately -0.5723. However when calculating the error I am not sure how to find the maximum of the second derivative, $ -4x^2\cos(x^2) - 2\sin(x^2) $ and I am stuck here. Have I gone about this correctly so far, and how do I find the error?
 A: Assuming your "Composite Trapezoidal method" is my Trapezoidal rule, then you've made an error in applying the approximation.
With a uniform $$\Delta x = \frac{b-a}{n} = \frac{3-2}{3} = \frac{1}{3},$$
the trapezoidal rule gives
$$\begin{align*}\int_2^3 \cos\left(x^2\right)\,dx &\approx \frac{1}{6}\Big(\cos\left(2^2\right) + 2\cos\left((7/3)^2\right) + 2\cos\left((8/3)^2\right) + \cos\left(3^2\right)\Big) \\ &\approx 0.187472628\end{align*}$$

The error, as you've mentioned, is related to the second derivative
$$\frac{d^2}{dx^2} (\cos(x^2)) = \frac{d}{dx}(-2x\sin(x^2)) = -4x^2\cos(x^2) - 2\sin(x^2)$$
In particular, if we want to merely find an upper bound for the absolute value of the second derivative (which is what it sounds like you want), we can note
$$\left|-4x^2\cos(x^2) - 2\sin(x^2)\right| \leq 4x^2|\cos(x^2)| + 2|\sin(x^2)| \leq 4x^2 + 2 \leq 4(3^2)+2 = 38$$ (the actual maximum is $\approx 25.8$ at $x=3$)
Using our estimate, we then estimate the error as
$$\text{|Error|} \leq \frac{(b-a)^3}{12n^2}\cdot 38 = \frac{38}{108} \approx 0.351851852$$
(using the actual maximum of the second derivative updates this upper limit on the error to $0.239,$ and using the error formula on each subinterval could further refine the error estimate)
