Past exam of the numerical analysis course of the University of Oxford / 2015 / 1 / b

It suffices to assume $f\in C^2[a,b]$. The polynomial interpolation error for the endpoint linear interpolant gives, for every interior $x$, some ${\xi }_x\in (a,b)$ withLet $M={\max }_{[a,b]}|f^{\prime \prime }|$, which exists by continuity of the second derivative and the extreme value theorem. Since $(x-a)(b-x)\ge 0$,Integrating this pointwise inequality and using the triangle inequality for an integral yieldsThusThis derivation does not require choosing ${\xi }_x$ as a continuous function of $x$; the pointwise upper bound is sufficient.

The constant is sharp. Take the quadratic polynomial $f(x)=(x-a{)}^2$ and write $L=b-a$. Its exact integral is $L^3/3$, its trapezoidal rule value is $L^3/2$, and $\max |f^{\prime \prime }|=2$. ConsequentlyEquality holds, so no smaller universal constant than $1/12$ works under these assumptions. This is the sharp one-panel trapezoidal error bound.

The trapezoidal rule is exact for every affine function $f(x)=\alpha +\beta x$, equivalently every polynomial of polynomial degree at most one: the linear interpolant is the function itself. These functions have $f^{\prime \prime }=0$, so the preceding upper bound is also zero. There is no contradiction: zero error is precisely what the bound predicts.

The degree of precision of a quadrature rule is one, since the quadratic polynomial in part (i) is not integrated exactly. This does not exclude exactness for particular higher-degree or nonpolynomial functions. For example $f(x)=(x-m{)}^3$, $m=(a+b)/2$, has zero integral and opposite endpoint values, so the trapezoidal rule also gives zero. Its second derivative is generally nonzero, making the upper bound positive; an upper bound need not be attained. Signed quadrature error can cancel.