Phương pháp giải:

Sử dụng phương pháp đặt ẩn phụ.

Lời giải chi tiết:

Ta có: \(I = \int\limits_0^2 {\left[ {f\left( {2x} \right) + f\left( {4 - 2x} \right)} \right]dx}  = \int\limits_0^2 {f\left( {2x} \right)dx}  + \int\limits_0^2 {f\left( {4 - 2x} \right)dx} \)

Xét \({I_1} = \int\limits_0^2 {f\left( {2x} \right)dx} \):

Đặt \(t = 2x \Rightarrow \)\({I_1} = \int\limits_0^4 {f\left( t \right)\dfrac{1}{2}dt}  = \dfrac{1}{2}\int\limits_0^4 {f\left( t \right)dt}  = \dfrac{1}{2}\int\limits_0^4 {f\left( x \right)dx = \dfrac{1}{2}.2019 = \dfrac{{2019}}{2}} \)

Xét \({I_2} = \int\limits_0^2 {f\left( {4 - 2x} \right)dx} \):

Đặt \(t = 4 - 2x \Rightarrow \)\({I_1} = \int\limits_4^0 {f\left( t \right)\left( { - \dfrac{1}{2}dt} \right)}  =  - \dfrac{1}{2}\int\limits_4^0 {f\left( t \right)dt}  = \dfrac{1}{2}\int\limits_0^4 {f\left( t \right)dt}  = \dfrac{1}{2}\int\limits_0^4 {f\left( x \right)dx = \dfrac{1}{2}.2019 = \dfrac{{2019}}{2}} \)

\(I = {I_1} + {I_2} = \dfrac{{2019}}{2} + \dfrac{{2019}}{2} = 2019\).

Chọn B.