Can we say that $\int_{0}^{1} f (x)\ \mathrm {d}x \geq \frac {1} {2}$?

Question

Suppose $f : D \subset \Bbb R \longrightarrow \Bbb R$ be a non-negative non-constant function on $D$ containing $(0,1)$ such that $f (c)=100$ for some $c \in (0,1)$. Then can we say that $$\int_{0}^{1} f (x)\ \mathrm {d}x \geq \frac {1} {2}.$$

I think it is true but can't properly figure out why? Please help me.

Thank you in advance.

Answer 1

$f(x)=0$ if $x\in (0,1-{1\over{400}}]$, $f(x)=80000x+200-80000$ for $x\geq 1-{1\over{400}}$.

Answer 2

If the function is continuous, consider the function (on Desmos)$$100\exp \big({-10^6(x-0.5)^2} \big).$$

Its integral is approximately $0.177$, which is smaller than $\frac{1}{2}$.

We can get this function from $e^{-(x-0.5)^2}$. Since its maximum is $1$, multiplying by $100$ to get $100e^{-(x-0.5)^2}$ guarantees that the function will reach $100$. Now, we can make the integral as small as we want by $100\exp \big({-a(x-0.5)^2} \big)$ choosing an $a$ large enough.