Ta có: $|x^2-4x+3|=x+3 \Leftrightarrow \left[\begin{array}{l}x=0\\x=5\end{array}\right.$
Diện tích hình phảng cần tìm là:
$S=\int\limits_0^5(x+3-|x^2-4x+3|)dx$
$=\int\limits_0^1(x+3-(x^2-4x+3))dx+\int\limits_1^3(x+3+(x^2-4x+3))dx+\int\limits_3^5(x+3-(x^2-4x+3))dx$
$=\int\limits_0^1(-x^2+5x)dx+\int\limits_1^3(x^2-3x+6)dx+\int\limits_0^1(-x^2+5x)dx$
$=\left(-\dfrac{x^3}{3}+\dfrac{5x^2}{2}\right)\left|\begin{array}{l}1\\0\end{array}\right.+\left(\dfrac{x^3}{3}-\dfrac{3x^2}{2}+6x\right)\left|\begin{array}{l}3\\1\end{array}\right.+\left(-\dfrac{x^3}{3}+\dfrac{5x^2}{2}\right)\left|\begin{array}{l}5\\3\end{array}\right.=\dfrac{109}{6}$