= Solution
{c}
The midpoint error annihilates affine <functions>, so here $n=1$. Apply the functional to $(x-t)_+$:
$$
K(t)=\frac12(h-t)^2-h(h/2-t)_+.
$$
Thus
$$
\boxed{K(t)=
\begin{cases}t^2/2,&0\leq t\leq h/2,\\
(h-t)^2/2,&h/2\leq t\leq h.\end{cases}}
$$
It is nonnegative, and symmetry gives $\int_0^hK=2\int_0^{h/2}t^2/2\,dt=h^3/24$. Therefore
$$
\boxed{|X[u]|\leq\frac{h^3}{24}\|u''\|_\infty.}
$$
For a quadratic with constant nonzero second <derivative> equality holds, so this <midpoint quadrature rule> error constant is sharp.
\Image[/past-exam-of-the-approximation-theory-and-finite-element-methods-course-of-the-university-of-oxford/2008-midpoint-peano.png]
{title=The positive midpoint Peano kernel has area h cubed over 24}
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