Đáp án đúng: A
Giải chi tiết:Ta có:
\(\begin{array}{l}\int\limits_{ - 3}^4 {f\left( {\left| {x + 1} \right|} \right)dx}  = \int\limits_{ - 3}^{ - 1} {f\left( { - x - 1} \right)dx}  + \int\limits_{ - 1}^4 {f\left( {x + 1} \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =  - \int\limits_{ - 3}^{ - 1} {f\left( { - x - 1} \right)d\left( { - x - 1} \right)}  + \int\limits_{ - 1}^4 {f\left( {x + 1} \right)d\left( {x + 1} \right)} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =  - \int\limits_2^0 {f\left( t \right)dt}  + \int\limits_0^5 {f\left( u \right)du} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int\limits_0^2 {f\left( x \right)dx}  + \int\limits_0^5 {f\left( x \right)dx} \end{array}\)
Ta có:
\(\begin{array}{l}\int\limits_0^2 {f\left( x \right)dx}  = \int\limits_0^1 {f\left( x \right)dx}  + \int\limits_1^2 {f\left( x \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {S_1} - {S_2} = 3 - 10 =  - 7\end{array}\)
\(\begin{array}{l}\int\limits_0^5 {f\left( x \right)dx}  = \int\limits_0^1 {f\left( x \right)dx}  + \int\limits_1^2 {f\left( x \right)dx}  + \int\limits_2^3 {f\left( x \right)dx}  + \int\limits_3^4 {f\left( x \right)dx}  + \int\limits_4^5 {f\left( x \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {S_1} - {S_2} + {S_3} - {S_4} + {S_5} = 3 - 10 + 5 - 6 + 16 = 8\end{array}\)
Vậy \(\int\limits_{ - 3}^4 {f\left( {\left| {x + 1} \right|} \right)dx}  =  - 7 + 8 = 1\).
Chọn A