Here is the problem:
Let $H_5(x)$ be the Hermite polynomial that interpolates $f(x) = \text{sin}x$ at the points $-1, 0, 1$. If $\int_{-1}^{1} H_5(x)dx$ is used to approximate $\int_{-1}^{1} \text{sin}xdx$, derive the best upper bound you can for $|\int_{-1}^{1} f(x)-H_5(x)dx|$.
I did this problem by finding $H_5(x)$ using divided differences. Then I integrated , and found $\int_{-1}^{1} H_5(x)dx = .00647$. Now,
$|\int_{-1}^{1} f(x)-H_5(x)dx| = |\int_{-1}^{1} \text{sin}xdx - \int_{-1}^{1} H_5(x)dx| = |\int_{-1}^{1}H_5(x)dx| \leq .00647$.
Is there something wrong with this (or is there an easier way to do it, without computing the polynomial with divided differences?)
The best upper bound for $\left|\int_{-1}^1 f(x)-H_5(x)\,dx\right|$ is zero, because both $\sin$ and the interpolating polynomial are odd functions on a symmetric interval. It's not a perfect estimate, but the errors in the positive half of the interval are exactly balanced by the errors in the negative half of the interval.
The polynomial $H_5(x)$ is of the form $x+ax^3+bx^5$. That form matches $f(0)$ and $f'(0)$ automatically. For the other two coefficients, we solve a system to match $f(1)$ and $f'(1)$: $1+a+b=\sin 1$ and $1+3a+5b=\cos 1$, so $a=\frac52\sin 1-\frac12\cos 1-2$ and $b=\frac12\cos 1-\frac32\sin 1+1$. Then $$H_5(-1)=-1-(\frac52\sin 1-\frac12\cos 1-2)-(\frac12\cos 1-\frac32\sin 1+1)=-\sin 1=f(-1)$$ $$H_5'(-1)=-1-3(\frac52\sin 1-\frac12\cos 1-2)-5(\frac12\cos 1-\frac32\sin 1+1)=-\cos 1=f'(-1)$$ We have matched $f(-1),f(0),f(1),f'(-1),f'(0),f'(1)$ with a polynomial of degree $5$, so $H_5(x)=x+(\frac52\sin 1-\frac12\cos 1-2)x^3+(\frac12\cos 1-\frac32\sin 1+1)x^5$ is the interpolating polynomial we seek.
Now that we have it, it's obviously odd. As such, $\int_{-1}^1 H_5(x)=0$. We also have that $\int_{-1}^1 \sin x=0$, so the integral of the difference is exactly zero.
Alternatively, what happens if we bring the absolute values inside? For that, we can't rely on odd symmetry, and instead need the theoretical bounds on the accuracy of the approximation. For any $x\in [-1,1]$, $f(x)-H_5(x)=\frac{f^{(6)}(c)}{6!}\left[(x-1)^2\cdot x^2\cdot(x+1)^2\right]$ for some $c\in [-1,1]$. For our $f=\sin$, it's sixth derivative is $-\sin$, and thus $|f^{(6)}(c)|\le \sin 1$. The polynomial we multiply that derivative by is nonnegative, so it's a perfect square and $|f(x)-H_5(x)|\le \frac{\sin 1}{720}(x^3-x)^2$. Integrating, $$\int_{-1}^1 |f(x)-H_5(x)|\,dx \le \int_{-1}^1 \frac{\sin 1}{6!}(x^3-x)^2\,dx=\frac{\sin 1}{6!}\int_{-1}^1 x^6-2x^4+x^2\,dx$$ $$\int_{-1}^1 |f(x)-H_5(x)|\,dx \le \frac{\sin 1}{6!}\cdot\left(\frac27-\frac45+\frac23\right)=\frac{\sin 1}{720}\cdot\frac{30-84+70}{105}=\frac{\sin 1}{4725}$$ Numerically, $\frac{\sin 1}{4725}\approx 1.781\cdot 10^{-4}$. That's the level of error we're working with.
I went to the trouble of calculating $\int_{-1}^1 |\sin x-H_5(x)|\,dx$ numerically. It's about $1.599\cdot 10^{-5}$, less than a tenth of the theoretical bound. While that sixth derivative could in theory be evaluated anywhere in the interval, in practice it's always near the center where $\sin$ is small. The error is largest for $x\approx \pm 0.65$, with values approximately $\frac{0.09}{720}(x^3-x)^2$.
(From the comments)
This is... Why on earth would you ever deliberately introduce more roundoff error by calculating to a very low precision? Roundoff error is something we can always control by just putting more effort into the calculation and increasing precision. Sure, there are ways to estimate its impact, or order calculations to reduce that impact - but it's not an intrinsic limit of the approximation method.
When we estimate the error of a method, we're talking about the truncation error - how good that estimate would be if we had theoretically perfect calculations. In this case, that truncation error is the difference $f(x)-H_5(x)$, with $H_5$ calculated exactly. Integrate that (with absolute values), and we get $\approx 1.599\cdot 10^{-5}$. Your estimate $6.47\cdot 10^{-3}$ is all roundoff error, and is entirely unrelated to the truncation error we're interested in.