= Solution
Both rules use a uniform partition with $\Delta x=(b-a)/n$. An original implementation in pseudocode is:
``
function trapezoid(f, a, b, n):
dx = (b-a)/n
total = (f(a)+f(b))/2
for j in 1,...,n-1:
total += f(a+j*dx)
return dx*total
function midpoint(f, a, b, n):
dx = (b-a)/n
total = 0
for j in 0,...,n-1:
total += f(a+(j+1/2)*dx)
return dx*total
``
The <trapezoidal rule> gives each endpoint half weight; the <composite midpoint rule> uses one interior point per subinterval. Both converge for a continuous <integrand> on the closed interval.
\b[The second equality in the printed trapezoidal hint is missing a factor $1/2$.] The average of the left and right rectangle sums is required. For $f=1$, the uncorrected equality would return $2(b-a)$ instead of $b-a$.
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