Show that $\exists c\in (0,1)$ such that $e^{f'(c)}=f(c)+1.$

WLOG, we suppose $f(0)>0,f(1)>0$. Define $$F(x):=e^{f'(x)}-f(x)-1, \forall x\in(0,1).$$ On the one hand, $\int_0^1f(x)dx=0$ implies $f$ can attain its Minimum value at some point $x_0\in(0,1)$, such that $f(x_0)<0,$ and $f'(x_0)=0$(Fermat's lemma). So $$F(x_0)=-f(x_0)>0.$$

On the other hand, $f(0)>0,f(x_0)<0$ implies that: the zeros set $$\{x\mid x\in[0, x_0], f(x) =0\}\ne \emptyset.$$ (by intermediate value theorem)

Thanks @ Paramanand Singh for pointing out that: the zeros set $$\{x\mid x\in[0, x_0], f(x) =0\}$$ has minimum and maximum element no matter it is finite set and infinite set.(Proof can be found Intermediate value theorem and supremum)

Let $$x_1=\max\{x\mid x\in[0, x_0], f(x) =0\},$$ and obviously $x_1\in(0, x_0)$. So $f(x_1)=0$ and $f(x)<0$ for $x\in(x_1,x_0)$ . Consider the derivative $f'(x_1)$, we know that $$f'(x_1)=\lim_{x\to x_1^+}\frac{f(x)-f(x_1)}{x-x_1} =\lim_{x\to x_1^+}\frac{f(x)}{x-x_1}\leq 0.$$

If $f'(x_1)=0$, take $c=x_1$, we can get $e^{f'(c)}=f(c)+1$.

If $f'(x_1)<0$, then $F(x_1)=e^{f'(x_1)}-1<0$, by intermediate value theorem, we can conclude that $\exists\ c\in(x_1,x_0)\subset(0,1)$, such that $$F(c)=0\iff e^{f'(c)}=f(c)+1.$$

If such $c$ doesn't exist, either always $e^{f'}>f+1$ or always $e^{f'}<f+1$. Whenever $f=0$, the former means $f'>0$ always and the latter means $f'<0$ always. However $f$ must change sign at least twice, and this is impossible.