\(f\left(x\right)=x^2-3x+2\int\limits^1_0f\left(x\right).f'\left(x\right)dx\)
\(\Leftrightarrow f\left(x\right)=x^2-3x+f^2\left(x\right)|^1_0\)
\(\Leftrightarrow f\left(x\right)=x^2-3x+f^2\left(1\right)-f^2\left(0\right)=x^2-3x+C\)
Với \(C=f^2\left(1\right)-f^2\left(0\right)\)
\(f\left(1\right)=C-2\) ; \(f\left(0\right)=C\Rightarrow f^2\left(1\right)-f^2\left(0\right)=\left(C-2\right)^2-C^2=C\)
\(\Rightarrow-4C+4=C\Rightarrow C=\dfrac{4}{5}\)
\(\Rightarrow f\left(x\right)=x^2-3x+\dfrac{4}{5}\)
\(\Rightarrow\int\limits^a_0f\left(x\right)dx=\int\limits^a_0\left(x^2-3x+\dfrac{4}{5}\right)dx=\left(\dfrac{1}{3}x^3-\dfrac{3}{2}x^2+\dfrac{4}{5}x\right)|^a_0\)
\(=\dfrac{a^3}{3}-\dfrac{3a^2}{2}+\dfrac{4a}{5}=\dfrac{4a}{5}\)
\(\Rightarrow\dfrac{a^3}{3}-\dfrac{3a^2}{2}=0\Rightarrow a=\dfrac{9}{2}\)
