Q:

What is the function satisfying $\int_a^b f(x)\,dx =\int_a^b f'(x)\,dx$?

Let $$f:\mathbb R\rightarrow (0,\infty)$$ be a twice differentiable function  such  that  $f(0)=1$, and further, for all $a, b \in \mathbb{R}$ such that $a < b$,
$$\int_a^b f(x) \, dx =\int_a^b f'(x) \, dx$$
Then  which of the following is false?
$1.$ $f$ is bijective.
$2.$ The image of $f$ is compact.
$3.$ $f$ is unbounded.
$4.$ There is only  one such function.
Now since $e^x$  satisfies  the  given  equation  and the criterion, $1$ and $3$ are correct and $2$ is the false  statement. So, $4$  has  to  be  correct  and  if  so  $e^x$  is  the  only  such  function. Well, what I am asking is how do I prove the statement in $4$ $?$
Thanks.

A:

The fact that $f$ is differentiable entails that $f$ is continuous and the fact that $f$ is twice differentiable entails that $f'$ is continuous.  So we have two continuous functions $f$ and $g$ for which
$$
\forall a,b\in\mathbb R\  \int_a^b f(x)\,dx = \int_a^b g(x)\,dx.
$$
If $f$ is continuous then the fundamental theorem of calculus tells us that
$$
\lim_{b\,\to\, a} \frac 1 {b-a} \int_a^b f(x)\,dx = \left.\frac d {db}\int_a^b f(x)\,dx\right|_{b:=a} = f(a)
$$
and similarly for $g$, and since equality of the integrals means these limits for $f$ and $g$ must be equal, we conclude that $f(a)=g(a)$.  This is true of every real number $a$. Since $g(x)=f'(x)$, we have $f'(x) = f(x)$ for every $x\in\mathbb R$.
We know that one solution of this differential equation is $x\mapsto e^x$.  Suppose $f$ is any other solution.  Then we can apply the quotient rule:
$$
\frac d {dx} \frac{f(x)}{e^x} = \frac{e^x f'(x) - f(x) e^x}{e^{2x}}
$$
and this is everywhere $0$ since $f'(x)=f(x)$ and the denominator is nowhere $0$.  Consequently $\dfrac{f(x)}{e^x}$ is a constant.

