Let \(f\) be a real polynomial of degree \(n\), \(f(\mathrm{x})={f}^{'}(\mathrm{x}){f}^{''}(\mathrm{x}),\) for all \(x\i…

Let degree of \(f(x)\) be \(n\)

Since \(f(x)=f^{\prime}(x) f^{\prime \prime}(x)\),

comparing degrees gives \(n=(n-1)+(n-2)\), so \(n=3\)

Let \(f(x)=a x^3+b x^2+c x+d\)
Since \(f(0)=0, d=0\), so \(f(x)=a x^3+b x^2+c x\)
Now \(f^{\prime}(x)=3 a x^2+2 b x+c\) and \(f^{\prime \prime}(x)=6 a x+2 b\)
Using \(f(x)=f^{\prime}(x) f^{\prime \prime}(x)\):

\(a x^3+b x^2+c x=\left(3 a x^2+2 b x+c\right)(6 a x+2 b)\)
So \(a x^3+b x^2+c x=18 a^2 x^3+18 a b x^2+\left(6 a c+4 b^2\right) x+2 b c\)
Comparing coefficients:

\(a=18 a^2\)

Hence \(a=\frac{1}{18}\)

Also \(b c=0\) and \(c=6 a c+4 b^2\)

Using \(a=\frac{1}{18}, c=\frac{c}{3}+4 b^2\), so \(c=6 b^2\)

Since \(b c=0\), we get \(b=0\), hence \(c=0\)

Therefore \(f(x)=\frac{x^3}{18}\)

So \(f^{\prime}(x)=\frac{x^2}{6}\) and \(f^{\prime \prime}(x)=\frac{x}{3}\)

Hence \(f^{\prime}(2)=\frac{2}{3}, f^{\prime \prime}(2)=\frac{2}{3}\),

and \(\int_0^2 f(x) d x=\int_0^2 \frac{x^3}{18} d x=\left.\frac{x^4}{72}\right|_0 ^2=\frac{2}{9}\)

Therefore \(36\left(f^{\prime}(2)+f^{\prime \prime}(2)+\int_0^2 f(x) d x\right)=36\left(\frac{2}{3}+\frac{2}{3}+\frac{2}{9}\right)=56\)