This is the question given in the book and is related to the error associated with the trapezoidal rule: Consider the integral $I(h)=\int_{a}^{a+h} f(x) dx$. Establish an expression for the error term for each of the following rules:
a. $I(h) ~ hf(a+h)$
b. $I(h)$ ~ $hf(a+h)-1/2h^2f'(a)$
c. $I(h)$~$hf(a)$
d. $I(h)$~hf(a)+$1/2h^2f'(a)$.
For each, determine, the corresponding general rule and error terms for the integral: $\int_{a}^{b} f(x) dx$, where the partition is uniform; that is, $x_i=a+ih$ and $h=(b-a)/n$ for $0\leq i \leq n$.
Here's the book answer for parts a and b:
Is the answer for part b incorrect? Shouldn't it be $I(h)=\int_{a}^{a+h} f(x) dx$-($hf(a+h)-1/2h^2f'(a)$) =$(1/6-1/2)h^3f''(c_i) $=$-1/3h^3f''(c_i)$, not $-1/6h^3f''(c_i)$, which the book has? Also isn't the answer for part c the same as the answer for part a? While the answer for part b should be the same answer as part b: $-1/3h^3f''(c_i)$?
If you assume that for an $n$-point formula, the error is $Ch^{n+1}f^{(n)}(\xi)$, which isn't always true then we can plug in an $n^{\text{th}}$ degree polynomial for which the quadrature formula predicts a value of $0$ and so get the error directly. $$\int_a^{a+h}(x-a-h)dx=\left.\frac12(x-a-h)^2\right|_a^{a+h}=-\frac12h^2=hf(a+h)-\frac12h^2f^{\prime}(\xi)$$ $$\begin{align}\int_a^{a+h}\left[(x-a)^2-h^2\right]dx&=\left.\left[\frac13(x-a)^3-h^2x\right]\right|_a^{a+h}=\frac13h^3-h^3=-\frac23h^3\\ &=hf(a+h)-\frac12h^2f^{\prime}(a)-\frac13f^{\prime\prime}(\xi)\end{align}$$ $$\int_a^{a+h}(x-a)dx=\left.\frac12(x-a)^2\right|_a^{a+h}=\frac12h^2=hf(a)+\frac12h^2f^{\prime}(\xi)$$ $$\int_a^{a+h}(x-a)^2dx=\left.\frac13(x-a)^3\right|_a^{a+h}=\frac13h^3=hf(a)+\frac12h^2f^{\prime}(a)+\frac16f^{\prime\prime}(\xi)$$ So you weren't quite careful enough with parts $c)$ and $d)$, getting a sign and a factor of $2$ off. Actually we can show that the error formulas are correct here but it gets a little ugly, especially in part $b)$. We can start with $$q(x)=\frac{\left[(x-a)^2-(y-a)^2\right]f(a+h)+\left[h^2-(x-a)^2\right]f(y)+(a+h-y)(y-x)(x-a-h)f^{\prime}(a)}{\left[h^2-(y-a)^2\right]}$$ For which $q(y)=f(y)$, $q(a+h)=f(a+h)$, and $q^{\prime}(a)=f^{\prime}(a)$, and assuming $a<y<a+h$ we can apply Rolle's theorem twice to find that there is some $a<u<a+h$ such that $f^{\prime\prime}(u)-q^{\prime\prime}(u)=f^{\prime\prime}(u)+\frac{-2f(a+h)+2f(y)+2(a-h-y)f^{\prime}(a)}{h^2-(y-a)^2}=0$, then solve for $f(y)$ and integrate from $a$ to $a+h$ and we get our error formula. It only works because the denominator in the last expression has the same same for $a<y<a+h$... but that's probably more in depth than required today.