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.

As you said, based on the fundamental theorem of calculus: $\int_a^xF'(x) = F(x) + C_1$ and $\int_a^xG'(x) = G(x) + C_2$, where $C_1$ and $C_2$ are constants from $\mathbb R$. As $F'(x) = G'(x)$, the above expressions are equal too, hence: $F(x) + C_1 = G(x) + C_2 \Leftrightarrow F(x) - G(x) = C_2 - C_1 = C_3$, where C_3 is some constant from $\mathbb R$. Q.E.D.
