Tính diện tích hình phẳng giới hạn bởi hai đường: $y = \left| {{x^2} - 4x + 3} \right|\,\,\,;\,\,y = x + 3$

Ta có $\left| {{x^2} - 4{\rm{x}} + 3} \right| = 0 \Leftrightarrow \left[ \begin{array}{l}x = 1\\x = 3\end{array} \right.$

Ta có: $\left| {{x^2} - 4{\rm{x}} + 3} \right| = x + 3 \Leftrightarrow {{\rm{x}}^4} - 8{{\rm{x}}^3} + 22{{\rm{x}}^2} - 24{\rm{x}} + 9 = {x^2} + 6{\rm{x}} + 9$

$\Leftrightarrow {{\rm{x}}^4} - 8{{\rm{x}}^{^3}} + 21{{\rm{x}}^2} - 30{\rm{x}} = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = 5\end{array} \right.$

Với $0 \le x \le 5$ thì $\left| {{x^2} - 4{\rm{x}} + 3} \right| \le x + 3$

Có

$\begin{array}{l}S=\int_0^5 {\left| {\left| {{x^2} - 4x + 3} \right| - x - 3} \right|dx}  = \int_0^1 {\left[ {x + 3 - \left( {{x^2} - 4{\rm{x}} + 3} \right)} \right]dx}  \\ + \int_1^3 {\left[ {x + 3 - \left( { - {x^2} + 4x - 3} \right)} \right]dx}  + \int_3^5 {\left[ {x + 3 - \left( {{x^2} - 4{\rm{x}} + 3} \right)} \right]dx} \\ = \int_0^1 {\left[ { - {x^2} + 5x} \right]dx}  + \int_1^3 {\left[ {{x^2} - 3x + 6} \right]dx}  + \int_3^5 {\left[ { - {x^2} + 5x} \right]dx}  \\ = \left. {\left( { - \dfrac{{{x^3}}}{3} + 5.\dfrac{{{x^2}}}{2}} \right)} \right|_0^1 + \left. {\left( {\dfrac{{{x^3}}}{2} - 3.\dfrac{{{x^2}}}{2} + 6x} \right)} \right|_1^3 + \left. {\left( { - \dfrac{{{x^3}}}{3} + 5.\dfrac{{{x^2}}}{2}} \right)} \right|_3^5\\ =  - \dfrac{1}{3} + \dfrac{5}{2} + \dfrac{{27}}{2} - 3.\dfrac{9}{2} + 18 - \dfrac{1}{2} + \dfrac{3}{2} - 6 - \dfrac{{125}}{3} + \dfrac{{125}}{2} + \dfrac{{27}}{3} - \dfrac{{5.9}}{2} = \dfrac{{109}}{6}\end{array}$