Cho hàm số f(x) có đồ thị trên đoạn \[\left[ { - 3;3} \right]\;\]là đường gấp khúc ABCD như hình vẽ.

Tính \[\mathop \smallint \limits_{ - 3}^3 f\left( x \right)dx\]

A.\[506\,\,\left( {c{m^2}} \right)\]

B. \[747\,\,\left( {c{m^2}} \right)\]

C. \[507\,\,\left( {c{m^2}} \right)\]

D. \[746\,\,\left( {c{m^2}} \right)\]

A.\[\frac{9}{2}\]

B.\[\frac{{18}}{5}\]

C.4

D.5

A.\[S = \mathop \smallint \nolimits_{ - 2}^2 f(x)dx\]

B. \[S = \mathop \smallint \nolimits_1^{ - 2} f(x)dx + \mathop \smallint \nolimits_1^2 f(x)dx\]

C. \[S = \mathop \smallint \nolimits_{ - 2}^1 f(x)dx + \mathop \smallint \nolimits_1^2 f(x)dx\]

D. \[S = \mathop \smallint \nolimits_{ - 2}^1 f(x)dx - \mathop \smallint \nolimits_1^2 f(x)dx\]

A.\[\frac{{107}}{6}\]

B. \[\frac{{109}}{6}\]

C. \[\frac{{109}}{7}\]

D. \[\frac{{109}}{8}\]

A.\[S = \left| {\mathop \smallint \nolimits_{ - 1}^1 \left( {3x - {x^3}} \right)dx} \right|\]

B. \[S = \mathop \smallint \nolimits_{ - 1}^0 \left( {3x - {x^3}} \right)dx + \mathop \smallint \nolimits_0^1 \left( {{x^3} - 3x} \right)dx\]

C. \[S = \mathop \smallint \nolimits_{ - 1}^1 \left( {3x - {x^3}} \right)dx\]

D. \[S = \mathop \smallint \nolimits_{ - 1}^0 \left( {{x^3} - 3x} \right)dx + \mathop \smallint \nolimits_0^1 \left( {3x - {x^3}} \right)dx\]

A.\[S = \mathop \smallint \limits_0^e \left| {{e^x} + x} \right|dx\]

B. \[S = \mathop \smallint \limits_0^e \left| {{e^x} - x} \right|dx\]

C. \[S = \mathop \smallint \limits_e^0 \left| {{e^x} - x} \right|dx\]

D. \[S = \mathop \smallint \limits_e^0 \left| {{e^x} + x} \right|dx\]