Tính diện tích hình phẳng giới hạn bởi hai đường:

Ta có\[\mid {x^2} - 4x + 3\mid = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = 1}\\{x = 3}\end{array}} \right.\]

Ta có:

\[\left| {{x^2} - 4{\rm{x}} + 3} \right| = x + 3 \Leftrightarrow {{\rm{x}}^4} - 8{{\rm{x}}^3} + 22{{\rm{x}}^2} - 24{\rm{x}} + 9 = {x^2} + 6{\rm{x}} + 9\]

\[ \Leftrightarrow {x^4} - 8{x^3} + 21{x^2} - 30x = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = 0}\\{x = 5}\end{array}} \right.\]

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

Có

\[\begin{array}{l}S = \int\limits_0^5 {\mid \mid x2 - 4x + 3\mid - x - 3\mid dx} \\ = \int\limits_0^1 {\left[ {x + 3 - (x2 - 4x + 3)} \right]} dx + \int\limits_1^3 {\left[ {x + 3 - ( - x2 + 4x - 3)} \right]} dx + \int\limits_3^5 {\left[ {x + 3 - (x2 - 4x + 3)} \right]} dx\end{array}\]

\( = \int\limits_0^1 {[ - x2 + 5x]dx + \int\limits_1^3 {[x2 - 3x + 6]dx + \int\limits_3^5 {[ - x2 + 5x]dx} } } \)

\( = \left( { - \frac{{{x^3}}}{3} + 5.\frac{{{x^2}}}{2}} \right)\left| {_0^1} \right. + \left( {\frac{{{x^3}}}{2} - 3.\frac{{{x^2}}}{2} + 6x} \right)\left| {_1^3} \right. + \left( { - \frac{{{x^3}}}{3} + 5.\frac{{{x^2}}}{2}} \right)\left| {_3^5} \right.\)

\( = - \frac{1}{3} + \frac{5}{4} + \frac{{27}}{2} - 3.\frac{9}{2} + 18 - \frac{1}{2} + \frac{3}{2} - 6 - \frac{{125}}{3} + \frac{{125}}{2} + \frac{{27}}{3} - \frac{{5.9}}{2} = \frac{{109}}{6}\)