Câu hỏi 21

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

Tính \(I = \int\limits_0^{\dfrac{\pi }{4}} {f\left( {2x} \right)dx} \).

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

Tính \(J = \int\limits_0^{\dfrac{\pi }{4}} {\sin x} dx =  - \left. {\cos x} \right|_0^{\dfrac{\pi }{4}} =  - \left( {\dfrac{{\sqrt 2 }}{2} - 1} \right) = 1 - \dfrac{{\sqrt 2 }}{2}\).

Vậy \(I - J = 2 - \left( {1 - \dfrac{{\sqrt 2 }}{2}} \right) = 1 + \dfrac{{\sqrt 2 }}{2}\) hay \(\int\limits_0^{\dfrac{\pi }{4}} {\left[ {f(2x) - \sin x} \right]} dx = 1 + \dfrac{{\sqrt 2 }}{2}\).

Chọn D