(Sở Hà Tĩnh 2022) Cho hàm số \(f\left( x \right)\) liên tục trên \(\mathbb{R}\) và \(f\left( 4 \right) = 2\), \(\int\limits_0^4 {f\left( x \right){\rm{d}}x = 4} \). Tính \(\int\limits_0^2 {xf’\left( {2x} \right){\rm{d}}x} \)

A. \(I = 17\).

B. \(I = 1\).

C. \(I = 12\).

D. \(I = 4\).

Lời giải:

Chọn B

Xét \(I = \int\limits_0^2 {xf’\left( {2x} \right){\rm{d}}x} \).

Đặt \(\left\{ \begin{array}{l}u = x\\dv = f’\left( {2x} \right){\rm{d}}x\end{array} \right.\)\( \Rightarrow \left\{ \begin{array}{l}{\rm{d}}u = {\rm{d}}x\\v = \frac{1}{2}f\left( {2x} \right)\end{array} \right.\).

\(I = \frac{1}{2}\left. {x.f\left( {2x} \right)} \right|_0^2 – \frac{1}{2}\int\limits_0^2 {f\left( {2x} \right){\rm{d}}x} \)\( = f\left( 4 \right) – \frac{1}{2}\int\limits_0^2 {f\left( {2x} \right){\rm{d}}x} = 2 – \frac{1}{2}\int\limits_0^2 {f\left( {2x} \right){\rm{d}}x} \)

Đặt \(t = 2x \Rightarrow {\rm{d}}t = 2{\rm{d}}x\).

Đổi cận \(x = 0 \Rightarrow t = 0;x = 2 \Rightarrow t = 4\).

\( \Rightarrow I = 2 – \frac{1}{4}\int\limits_0^4 {f\left( t \right){\rm{d}}t} = 2 – \frac{1}{4}\int\limits_0^4 {f\left( x \right){\rm{d}}x = 2} – 1 = 1\).