Trapezoidal integration rule error analysis.

Question

Let $f(x)\in C^2[a,b]$ and $p\in P_1$ its Lagrange interpolation polynom for nodes $a,b$: $$p(a) = f(a), ~p(b) = f(b).$$ Then the interpolation error is $$f(x) - p(x) = \frac{1}{2}(x-a)(x-b)f''(\xi(x)), ~\xi(x)\in[a,b]$$ It follows from calculation of $p(x)$ that $$\int_a^bf(x)dx = \frac{b-a}{2}(f(a)+f(b)) + \int_a^b\frac{1}{2}(x-a)(x-b)f''(\xi(x))dx$$ Now, in my homework I am to prove that $$\int_a^bf(x)dx = \frac{b-a}{2}(f(a)+f(b)) + \int_a^b\frac{1}{2}(x-a)(x-b)f''(x)dx$$ Is this a misprint? And if not, how can one possible derive that representation? I also tried the divided difference error representation with no success.

Answer 1

Calculate backwards, with partial integration: \begin{align} \int_a^b(x-a)(x-b)f''(x)dx &=[(x-a)(x-b)f'(x)]_a^b-\int_a^b(2x-a-b)f'(x)dx\\ &=0-0-[(2x-a-b)f(x)]_a^b+\int_a^b2f(x)dx\\ &=-(b-a)(f(b)+f(a))+2\int_a^bf(x)dx \end{align} so that indeed the claimed identity holds.