75. Put $H(x)=\exp(-\int_0^xG(t)dt)$. Mutiplying the inequality with $H(x)$ and noticing that $H>0$, we get
\[
\frac{d}{dx}(u(x)H(x))\leq0\Longrightarrow u(x)H(x)-u(0)H(0)=\int_0^1\frac{d}{dt}(u(t)H(t))dt\leq0,
\]
by the fundamental theorem of calculus. Thus $u(x)\leq H^{-1}(x)u(0)H(0)=0$ as needed.
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