Non negative function at a point is less the the integral of the function

Solution 1:

Your question can be answered by many counter-examples from probability; just because the probability integrates to $1$ doesn't mean that the density function must be less than one. Consider for example the pdf of the $\chi^2_1$ distribution:

$$f(x)=\begin{cases}0&x\le0\\\frac{\exp\left(-\frac{1}{2}x\right)}{\sqrt{2\pi x}}&x\gt 0\end{cases}$$

We have that:

$$\int_{\Bbb R}f(x)\,\mathrm{d}x=1$$

But $f\to+\infty$ as $x\to 0^+$, and attains values greater than $1$; e.g., $f(10^{-4})\approx40\gt1$.

With regards to intuition, I'll say that a very thin slice is integrated to a very small value - our function here blows up to $\infty$, yet in integration the region where it is large is so slim that the integral is small...