= Solution
{c}
Again let $G=\int_a^bg(t)\,dt\geq0$ and $I=\int_a^bf(t)g(t)\,dt$. Since $f$ is a nondecreasing <monotonic function>, $f(a)\leq f(t)\leq f(b)$ on $[a,b]$. The condition $g\geq0$ and the <monotonicity of the Riemann integral> give
$$
f(a)G\leq I\leq f(b)G.
$$
Set $H(x)=\int_a^xg(t)\,dt$. By part (a), $H$ is <continuous>. Therefore the <function>
$$
K(x)=f(a)H(x)+f(b)\bigl(G-H(x)\bigr)
$$
is <continuous> on $[a,b]$. Its endpoint values are $K(a)=f(b)G$ and $K(b)=f(a)G$, which enclose $I$. The <intermediate value theorem> provides $d\in[a,b]$ with $K(d)=I$, so
$$
\boxed{\int_a^bf(t)g(t)\,dt=f(a)\int_a^dg(t)\,dt+f(b)\int_d^bg(t)\,dt.}
$$
If the two endpoint values coincide, the preceding bounds force $I$ to equal that common value and either endpoint works. Thus the proof also covers $G=0$, $f(a)=f(b)$ and $a=b$, without dividing by any possibly zero quantity.
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