Note that $e^y + y$ is increasing and because of that for $x\in[0,e]$ the equation $e^y+y=x+1$ has unique solution with respect to $y$, thus identifying a function $y(x)\,,\, x\in [0,e]$. It means for any $f\in \mathcal{F}$ it holds $f(x)\leq y(x),x\in[0,e]$. Hence, $I(f)$ takes it's maximal value when $f=y(x)$. It remains to calculate $\int_0^e y(x)\,dx$. Now, $y(x)$ is an increasing function for $x\in[0,e]$ taking values in $[0,1]$ and let $x(y)$ be its inverse when $x\in[0,1]$. We have: $$\int_0^1 (e^y+y)\,dy = \int_0^1 (x(y)+1)\,dy \implies$$$$e-1+\frac{1}{2} = \int_0^1 x(y)\,dy +1 \implies e-1+\frac{1}{2} = e\cdot 1 -\int_0^e y(x)\,dx +1$$Hence $\int_0^e y(x)\,dx = \frac{3}{2}$.

dgrozev wrote: Note that $e^y + y$ is increasing and because of that for $x\in[0,e]$ the equation $e^y+y=x+1$ has unique solution with respect to $y$, thus identifying a function $y(x)\,,\, x\in [0,e]$. It means for any $f\in \mathcal{F}$ it holds $f(x)\leq y(x),x\in[0,e]$. Hence, $I(f)$ takes it's maximal value when $f=y(x)$. It remains to calculate $\int_0^e y(x)\,dx$. Now, $y(x)$ is an increasing function for $x\in[0,e]$ taking values in $[0,1]$ and let $x(y)$ be its inverse when $x\in[0,1]$. We have: $$\int_0^1 (e^y+y)\,dy = \int_0^1 (x(y)+1)\,dy \implies$$$$e-1+\frac{1}{2} = \int_0^1 x(y)\,dy +1 \implies e-1+\frac{1}{2} = e\cdot 1 -\int_0^e y(x)\,dx +1$$Hence $\int_0^e y(x)\,dx = \frac{3}{2}$. could you explain your solution, I couldnt understand it at all, I mean not even a bit