Show that $F$ and $G$ differ by a constant Suppose $F$ and $G$ are differentiable functions defined on $[a,b]$ such that $F'(x)=G'(x)$ for all $x\in[a,b]$. Using the fundamental theorem of calculus, show that $F$ and $G$ differ by a constant. That is, show that there exists a $C\in\mathbb R$ such that $F(x)-G(x)=C$.
I'm assuming this is quite simple, but I can't seem to figure it out.

By the Fundamental Theorem of Calculus,
$$
\begin{align}
[F(x)-G(x)]-[F(a)-G(a)]
&=\int_a^x(F'(t)-G'(t))\mathrm{dt}\\
&=\int_a^x0\,\mathrm{dt}\\[6pt]
&=0
\end{align}
$$
Therefore, for all $x\in[a,b]$
$$
F(x)-G(x)=F(a)-G(a)
$$
