Let f be a real polynomial of degree n such that f(x) = f'(x)...

Let the degree of the polynomial $f(x)$ be $n$.

The degree of $f'(x)$ is $n-1$ and the degree of $f''(x)$ is $n-2$.

Since $f(x) = f'(x) f''(x)$, equating the degrees on both sides gives:

$n = (n-1) + (n-2) \Rightarrow n = 3$

Let $f(x) = ax^3 + bx^2 + cx + d$.

Given $f(0) = 0$, we get $d = 0$. Thus, $f(x) = ax^3 + bx^2 + cx$.

Differentiating $f(x)$ with respect to $x$:

$f'(x) = 3ax^2 + 2bx + c$

$f''(x) = 6ax + 2b$

Substituting these into the given equation $f(x) = f'(x) f''(x)$:

$ax^3 + bx^2 + cx = (3ax^2 + 2bx + c)(6ax + 2b)$

$ax^3 + bx^2 + cx = 18a^2x^3 + 18abx^2 + (4b^2 + 6ac)x + 2bc$

Comparing the coefficients of corresponding powers of $x$:

For $x^3$: $a = 18a^2 \Rightarrow a = \dfrac{1}{18}$ (since $a \neq 0$ for a cubic polynomial)

For the constant term: $0 = 2bc$

For $x$: $c = 4b^2 + 6ac$

Substituting $a = \dfrac{1}{18}$ into the $x$ coefficient equation:

$c = 4b^2 + \dfrac{c}{3} \Rightarrow \dfrac{2c}{3} = 4b^2 \Rightarrow c = 6b^2$

From $0 = 2bc$, either $b = 0$ or $c = 0$. In either case, since $c = 6b^2$, we get $b = 0$ and $c = 0$.

Thus, the polynomial is $f(x) = \dfrac{1}{18}x^3$.

Now, finding the required values:

$f'(x) = \dfrac{1}{6}x^2 \Rightarrow f'(2) = \dfrac{4}{6} = \dfrac{2}{3}$

$f''(x) = \dfrac{1}{3}x \Rightarrow f''(2) = \dfrac{2}{3}$

$\int_0^2 f(x)\,dx = \int_0^2 \dfrac{1}{18}x^3\,dx = \dfrac{1}{18} \left[ \dfrac{x^4}{4} \right]_0^2 = \dfrac{16}{72} = \dfrac{2}{9}$

Substituting these into the given expression:

$36\left(f'(2) + f''(2) + \int_0^2 f(x)\,dx\right) = 36\left(\dfrac{2}{3} + \dfrac{2}{3} + \dfrac{2}{9}\right)$

$= 36\left(\dfrac{4}{3} + \dfrac{2}{9}\right) = 36\left(\dfrac{14}{9}\right) = 4 \times 14 = 56$

Answer: $56$