Suppose $f:\mathbb R\to \mathbb R$ is a differentiable map satisfying $f(0)=0$ and $f(1)=1$. For all real $x$, we have $|f'(x)|\leq2$. What is the maximum possible value of $\int_{0}^{1}f(x)dx$, over all such possible functions
Now here is what I tried.
We have $|f'(x)|\leq2\Rightarrow-2\leq f'(x)\leq 2$
$$\Rightarrow -2t\leq \int_{0}^{t}f'(x)dx\leq 2t\space \text{or}\space -2t\leq f(t)\leq 2t$$ $$\Rightarrow -1\leq \int_{0}^{1}f(t)dt=\int_{0}^{1}f(x)dx\leq1$$
Thus $\int_{0}^{1}f(x)dx$ cannot exceed $1$.
Now in first sight, it looks like $f(x)=2x$ gives us the maximum value of the integral. But $f(1)=2$ here which does not satisfy the given condition that $f(1)=1$
Now assuming $f(x)=x^{1/n}$. This gives us $f(0)=0$ and $f(1)=1$. It is also differentiable.
$$\int_{0}^{1}f(x)dx=\int_{0}^{1}x^{1/n}dx=\dfrac{n-1}{n}$$
Observe that as $n\to \infty$, $\int_{0}^{1}f(x)dx=\dfrac{n-1}{n}\to 1$.
Thus if we assume any $k<1$ to be the maximum value of $\int_{0}^{1}f(x)dx$, then we can find a sufficiently large $n\in N$ such that setting $f(x)=x^{1/n}$ gives $\int_{0}^{1}f(x)dx=\dfrac{n-1}{n}>k$, thus contradicting the maximality of $k$.
$\therefore$ Maximum possible value of $\int_{0}^{1}f(x)dx$ is $1$
But on further examination, I found out that if $f(x)=x^{1/n}$, then $f'(x)=\dfrac{1}{nx^{n-1/n}}$. We should have $f'(x)\leq2$, so $\dfrac{1}{nx^{n-1/n}}\leq2\Rightarrow x^{n-1/n}\geq\dfrac{1}{2n}$ which need not be true for all $x\in R$.
Thus this definition of function violates the condition that $|f'(x)|\leq2\space\forall\ x\in \mathbb R$
Here I tried two definitions both of which failed and now I'm stuck. Please help me out.
THANKS