Let $f:(0,1)\rightarrow\mathbb{R}$ be a continuous function with $$\int_0^1f(x)dx=\frac{2^{2018}-1}{2018}$$ Show that there $\exists a\in(0,1)$ so that $f(a)=(a+1)^{2017}$.
It is clear that $$ \int_0^1f(x)dx=\frac{2^{2018}-1}{2018}=\int_0^1(x+1)^{2017}dx$$ but does that imply $f(x)=(x+1)^{2017}$?
EDIT: I think that it is so only if $f(x)-(x+1)^{2017}\ge0$. Still, how could the existence of such an $a$ be proven?
On
Assume there is no such $a$. Then $f(a)<(a+1)^{2017}$ for any $a \in (0,1)$, or $f(a)>(a+1)^{2017}$. By continuity you have same inequality for $a=0,1$, with the possibility of equality.
By integrating you get that your integral is either strictly bigger or smaller(depending on the two cases above) than $\frac{2^{2018}-1}{2018}$, which is a contradiction.
You have strict inequality for the integral as you can only have equality at 2 isolated points and you have strict inequality otherwise.
Define $g:x\mapsto f(x) - (x+1)^{2017}$, which is continuous and satisfies $\int_0^1 g(x)\,{\rm d}x = 0$.
You want to prove that there exists $a\in(0,1)$ such that $g(a)=0$. So, what happen if $g(x)\neq 0$ for all $x\in(0,1)$ ?