1. **State the problem:** We want to find the constant $p$ and the error term for the quadrature formula: $$\int_{x_0}^{x_1} f(x) \, dx = \frac{h}{2}(f_0 + f_1) + p h^2 (f'_0 - f'_1)$$ where $x_1 = x_0 + h$, $f_0 = f(x_0)$, $f_1 = f(x_1)$, and $f'$ denotes the derivative with respect to $x$. Then, deduce the composite rule for integrating $$\int_a^b f(x) \, dx$$ with $a = x_0 < x_1 < \cdots < x_N = b$. 2. **Use Taylor expansions:** Expand $f(x)$ around $x_0$ to approximate $f_1$ and $f'_1$: $$f_1 = f(x_0 + h) = f_0 + h f'_0 + \frac{h^2}{2} f''_0 + \frac{h^3}{6} f'''_0 + O(h^4)$$ $$f'_1 = f'(x_0 + h) = f'_0 + h f''_0 + \frac{h^2}{2} f'''_0 + O(h^3)$$ 3. **Substitute expansions into the formula:** Calculate the right side: $$\frac{h}{2}(f_0 + f_1) + p h^2 (f'_0 - f'_1) = \frac{h}{2} \left(f_0 + f_0 + h f'_0 + \frac{h^2}{2} f''_0 + \frac{h^3}{6} f'''_0 \right) + p h^2 \left(f'_0 - \left(f'_0 + h f''_0 + \frac{h^2}{2} f'''_0 \right) \right) + O(h^5)$$ Simplify: $$= h f_0 + \frac{h^2}{2} f'_0 + \frac{h^3}{4} f''_0 + \frac{h^4}{12} f'''_0 + p h^2 (- h f''_0 - \frac{h^2}{2} f'''_0) + O(h^5)$$ $$= h f_0 + \frac{h^2}{2} f'_0 + \frac{h^3}{4} f''_0 + \frac{h^4}{12} f'''_0 - p h^3 f''_0 - \frac{p h^4}{2} f'''_0 + O(h^5)$$ 4. **Exact integral expansion:** The exact integral expanded about $x_0$ is: $$\int_{x_0}^{x_0 + h} f(x) \, dx = h f_0 + \frac{h^2}{2} f'_0 + \frac{h^3}{6} f''_0 + \frac{h^4}{24} f'''_0 + O(h^5)$$ 5. **Match terms to find $p$:** Equate the approximation to the exact integral up to order $h^4$: $$h f_0 + \frac{h^2}{2} f'_0 + \frac{h^3}{4} f''_0 + \frac{h^4}{12} f'''_0 - p h^3 f''_0 - \frac{p h^4}{2} f'''_0 = h f_0 + \frac{h^2}{2} f'_0 + \frac{h^3}{6} f''_0 + \frac{h^4}{24} f'''_0$$ Compare coefficients: - For $h^3 f''_0$: $$\frac{1}{4} - p = \frac{1}{6} \implies p = \frac{1}{4} - \frac{1}{6} = \frac{1}{12}$$ - For $h^4 f'''_0$: $$\frac{1}{12} - \frac{p}{2} = \frac{1}{24}$$ Substitute $p=\frac{1}{12}$: $$\frac{1}{12} - \frac{1}{24} = \frac{1}{24}$$ which holds true. 6. **Error term:** The leading error term is of order $h^4$, so the error is: $$E = O(h^4)$$ 7. **Composite rule:** Divide $[a,b]$ into $N$ subintervals of width $h = \frac{b-a}{N}$ with points $x_0, x_1, ..., x_N$. Apply the formula on each subinterval and sum: $$\int_a^b f(x) \, dx \approx \sum_{k=0}^{N-1} \left[ \frac{h}{2} (f_k + f_{k+1}) + \frac{h^2}{12} (f'_k - f'_{k+1}) \right]$$ where $f_k = f(x_k)$ and $f'_k = f'(x_k)$. This composite rule improves the trapezoidal rule by including derivative terms to increase accuracy. **Final answers:** $$p = \frac{1}{12}$$ Error term is of order $$O(h^4)$$. ---