If $\int x^2 \cdot e^x dx = e^x f(x) + c$, then the minimum value of $f(x)$ is ...

Using integration by parts:

$\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx$

$= x^2 e^x - 2(x e^x - \int e^x dx)$

$= x^2 e^x - 2x e^x + 2e^x + c$

$= e^x (x^2 - 2x + 2) + c$

Comparing with the given expression, we get $f(x) = x^2 - 2x + 2$.

To find the minimum value of $f(x)$, we can complete the square:

$f(x) = (x - 1)^2 + 1$

Since $(x - 1)^2 \geq 0$ for all real $x$, the minimum value of $f(x)$ is $1$.

Answer: $1$