Cho hàm số f( x ) = { \begin{array}{l}2x + a\;{\rm{khi}}\;x \ge 1\\3{x^2} + b\;{\rm{khi}}\;x < 1

Ta có \(\int\limits_0^2 {f\left( x \right)dx = } \int\limits_0^1 {f\left( x \right)dx + \int\limits_1^2 {f\left( x \right)dx = } } \)\(\int\limits_0^1 {\left( {3{x^2} + b} \right)dx + \int\limits_1^2 {\left( {2x + a} \right)dx} } \)

\( = \left. {\left( {{x^3} + bx} \right)} \right|_0^1 + \left. {\left( {{x^2} + ax} \right)} \right|_1^2\)\( = 1 + b + 4 + 2a - 1 - a\)\( = a + b + 4\).

Mà \(\int\limits_0^2 {f\left( x \right)dx = 13} \) nên \(a + b = 9\).

Vì \(f\left( x \right)\) liên tục trên \(\mathbb{R}\) nên \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = f\left( 1 \right)\)\( \Leftrightarrow 2 + a = 3 + b \Leftrightarrow a - b = 1\).

Ta có hệ \(\left\{ \begin{array}{l}a + b = 9\\a - b = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 5\\b = 4\end{array} \right.\).

Vậy \(a + b - ab = 5 + 4 - 5.4 = - 11\).