How do I prove that then $f(x) = 1$ for some $x ∈ [0, 1]$?

Question

We are asked the following:

let $f(x)$ be a continuous function on $[0, 1]$ such that $$\int _ {0} ^ {1} \ f(y) dy = 1$$ prove that $f(x) = 1$ for some $x ∈ [0, 1]$

The question asks to take into account Rolle's Theorem and Mean Value Theorem. I'm aware of the definitions, but I am really unsure how to answer this question and my tutors are not really responding till after easter.

Could someone please help? Why is there a "y" involved?

Thanks you

Answer 1

Let $F(x)=\int\limits_0^x f(t)dt$. By the fundamental theorem of calculus this function is continuous and differentiable on $(0,1)$, the derivative at $x$ is $f(x)$.

Now apply the mean value theorem to conclude there is a point $z$ in $(0,1)$ with $F'(z)=1$, of course $F'(z)=f(z)$ so we are done.

Answer 2

Assume there is not x such that $f(x)=1$. Then $f(x)>1$ or $f(x)<1$ for all x . So the integral is either less or greater than 1, contradiction.

Answer 3

Ok, thanks to Jorge, I think I've figured it out. My full answer would be:

We know $f(x)$ is continuous on $[0,1]$.

Set $F(x)$ = $\int\limits_0^x f(y)dy$

The Fundamental Theorem of Calculus shows that the function is differentiable on $(0,1)$. Also it implies $F'(x) = f(x)$.

Using The Mean-Value Theorem:

$F'(x)$ = $\frac{F(1)-F(0)}{1-0}$

Since $F(1)$ = $\int\limits_0^1 f(y)dy$ = $1$ and $F(0) = 0$

$F'(x) = f(x) = 1$ for some $x$ in $[0,1]$

I think this would be the way to answer it.