Prove that $\exists$ a unique differentiable function $f: R \to R$ s.t. $f'(x) = e^{x^2} \forall x \in R$ and $f(0) = 0$
I'm having trouble showing the uniqueness part. This is what I have so far.
Write
$F(x) = \int_{a}^{x} e^{t^2} dt$
Choose $a = 0$
$F(x) = \int_{0}^{x} e^{t^2} dt$
We are told $F(0) = 0$, so
$F(0) = 0 = \int_{0}^{0} e^{t^2} dt = 0$
So the integral exists. Now we must show it is unique. I am really not sure where to go from here. I think the $e^{x^2}$ is throwing me off because there is no anti-derivative. So I'm not sure how I would show it is unique here. Any help is appreciated!
Best Answer
Assume $G$ is another such function. Then $H:=G-F$ is differentiable with $H'\equiv 0$ and $H(0)=0$. Then $H(x)=0$ for all $x$. Indeed, by the Mean Value Theorem, there exists $\xi $ between $0$ and $x$ such that $\frac{H(x)-H(0}{x-0}=H'(\xi)=0$.