Derive the approximation $\int_{a+h}^a f(x)\;dx =\frac{h}{2}[f(a) + f(a + h)] + \frac{h^2}{12} [f_0(a) − f_0(a + h)] + O(h^5)$

Question

Let $f$ be a smooth function. Derive the approximation $\int_{a+h}^a f(x)\;dx =\frac{h}{2}[f(a) + f(a + h)] + \frac{h^2}{12} [f_0(a) − f_0(a + h)] + O(h^5)$

Answer 1

We use the Hermite interpolating polynomial of order 3, and letting $z_0=z_1=a$ and $z_2=z_3=a+h$ \begin{align*} H_3(x) = &f[z_0]+f[z_0,z_1](x-z_0) + f[z_0,z_1,z_2](x-z_0)(x-z_1) \\ &+ f[z_0,z_1,z_2,z_3](x-z_0)(x-z_1)(x-z_2)\\=&f(a)+f'(a)(x-a) + \frac{\frac{f(a+h)-f(a)}{h}-f'(a)}{h}(x-a)^2 +\\ &\frac{f'(a+h)+f'(a)-2\frac{f(a+h)-f(a)}{h}}{h^2}(x-a)^2(x-(a+h)) \end{align*}

Thus $\int_{a+h}^a f(x)\;dx \approx H_3(x) = \frac{h}{2}[f(a) + f(a + h)] + \frac{h^2}{12} [f_(a) − f_(a + h)]$

We note the error term is $\int_{a+h}^a f(x)-H_3(x)\;dx = \int_{a+h}^a \frac{f^{(4)}(\epsilon(x))}{4!}(x-a)^2(x-(a+h))^2\;dx $

We note that since within $[a,a+h], |x-a| \text{ and } |x-(a+h)|$ is less than $h$, the error is (by comparison test) not more than

$K\int_{a+h}^a h^4\;dx = Kh^5$, for some constants $h,K$, which is $O(h^5)$, as required.