a) Ta có \({S_1} = \int\limits_1^3 {\left| { - {x^2} + 4x} \right|dx} \)\( = \int\limits_1^3 {\left( { - {x^2} + 4x} \right)dx} \)\( = \left. {\left( { - \frac{{{x^3}}}{3} + 2{x^2}} \right)} \right|_1^3\)\( = 9 - \frac{5}{3} = \frac{3}\).

\({S_2} = \int\limits_1^3 {\left| x \right|} dx\)\( = \int\limits_1^3 x dx\)\( = \left. {\frac{{{x^2}}}{2}} \right|_1^3 = \frac{9}{2} - \frac{1}{2} = 4\).

Do đó S = S₁ – S₂ = \(\frac{3} - 4 = \frac{3}\).

b) \(\int\limits_1^3 {\left| {f\left( x \right) - g\left( x \right)} \right|} dx\)\( = \int\limits_1^3 {\left| { - {x^2} + 4x - x} \right|} dx\)\( = \int\limits_1^3 {\left| { - {x^2} + 3x} \right|} dx\)\( = \int\limits_1^3 {\left( { - {x^2} + 3x} \right)} dx\)

\( = \left. {\left( { - \frac{{{x^3}}}{3} + 3.\frac{{{x^2}}}{2}} \right)} \right|_1^3\)\( = \frac{9}{2} - \frac{7}{6} = \frac{3}\).

Vậy \(S = \int\limits_1^3 {\left| {f\left( x \right) - g\left( x \right)} \right|} dx\).