Ta có: $x^2-4|x|+3 \Leftrightarrow \left[\begin{array}{l}x=\pm1\\x=\pm3\end{array}\right.$
Diện tích hình phẳng cần tìm là:
$S=\int\limits_{-3}^3|x^2-4|x|+3|dx$
$=\int\limits_{-3}^{-1}(-x^2-4x-3)dx+\int\limits_{-1}^0(x^2+4x+3)dx+\int\limits_0^1(x^2-4x+3)dx+\int\limits_1^3(-x^2+4x-3)dx$
$=\left(\dfrac{-x^3}{3}-2x^2-3x\right)\left|\begin{array}{l}-1\\-3\end{array}\right.+\left(\dfrac{x^3}{3}+2x^2+3x\right)\left|\begin{array}{l}0\\-1\end{array}\right.+\left(\dfrac{x^3}{3}-2x^2+3x\right)\left|\begin{array}{l}1\\0\end{array}\right.+\left(\dfrac{-x^3}{3}+2x^2-3x\right)\left|\begin{array}{l}3\\1\end{array}\right.$
$=\dfrac{16}{3}$