= Solution
The order-$k$ <divided difference> defining $M_i$ is a finite linear combination of values at its knots. Hence integration can be interchanged with it directly. For a knot value $u\in[a,b]$,
$$
\int_a^b(u-t)_+^{k-1}\,dt=\frac{(u-a)^k}{k}.
$$
It follows that
$$
\begin{aligned}
\int_a^bM_i(t)\,dt
&=k[t_i,\ldots,t_{i+k}]\left(u\mapsto\int_a^b(u-t)_+^{k-1}\,dt\right)\\
&=[t_i,\ldots,t_{i+k}](u-a)^k=1.
\end{aligned}
$$
The final equality holds because the order-$k$ <divided difference> of a degree-$k$ monic <polynomial> is its <leading coefficient>, one. Thus
$$
\boxed{\int_a^bM_i(t)\,dt=1.}
$$
This proves the <unit-integral normalization of a B-spline>. Its <support> is contained in $[t_i,t_{i+k}]$: below all its knots, the sampled truncated power agrees with a degree-$k-1$ <polynomial>, annihilated by the order-$k$ divided difference; above all its knots, every sample is zero. No mass is omitted by integration over $[a,b]$.
Since $N_i=(t_{i+k}-t_i)M_i/k$, the corresponding partition-normalized <B-spline> has integral $(t_{i+k}-t_i)/k$. Its normalization is different from that of $M_i$.
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