Chọn phương án A.

Ta có $\begin{aligned}[t]
I=\displaystyle\int\limits_0^2\big(x^2+f(3-2x)\big)\mathrm{\,d}x&=\int\limits_0^2x^2\mathrm{\,d}x+\int\limits_0^2f(3-2x)\mathrm{\,d}x\\
&=\dfrac{8}{3}+\displaystyle\int\limits_0^2f(3-2x)\mathrm{\,d}x.
\end{aligned}$

Đặt $t=3-2x\Rightarrow\mathrm{\,d}t=-2\mathrm{\,d}x$ hay $\mathrm{\,d}x=-\dfrac{1}{2}\mathrm{\,d}t$.

• $x=2\Rightarrow t=-1$
• $x=0\Rightarrow t=3$

Khi đó $\begin{aligned}[t]
\displaystyle\int\limits_0^2f(3-2x)\mathrm{\,d}x&=\int\limits_3^{-1}f(t)\left(-\dfrac{1}{2}\right)\mathrm{\,d}t\\
&=\dfrac{1}{2}\displaystyle\int\limits_{-1}^3f(x)\mathrm{\,d}x=\dfrac{1}{2}\big(F(3)-F(-1)\big).
\end{aligned}$

Mặt khác $F(3)-F(-1)=G(3)-G(-1)=\displaystyle\int\limits_{-1}^3f(x)\mathrm{\,d}x$.

Lại có $$\begin{array}{cccc}
&2F(3)+G(3)&=&9+2F(-1)+G(-1)\\
\Leftrightarrow&2\big(F(3)-F(-1)\big)+\big(G(3)-G(-1)\big)&=&9\\
\Leftrightarrow&3\big(F(3)-F(-1)\big)&=&9\\
\Leftrightarrow&F(3)-F(-1)&=&3.
\end{array}$$
Suy ra $\begin{aligned}[t]
I&=\dfrac{8}{3}+\displaystyle\int\limits_0^2f(3-2 x)\mathrm{\,d}x\\
&=\dfrac{8}{3}+\dfrac{1}{2}\big(F(3)-F(-1)\big)\\
&=\dfrac{8}{3}+\dfrac{3}{2}=\dfrac{25}{6}.
\end{aligned}$