Cho hàm số f(x) có đạo hàm liên tục trên R . Biết f(4)=1

Đáp án D

Đặt \(t = x + 2\) ta có: \(\int\limits_{02}^2 {xf\left( {x + 2} \right)d{\rm{x}}} = 5 \Leftrightarrow \int\limits_0^4 {\left( {t - 2} \right)f\left( t \right)dt} = 5 \Rightarrow \int\limits_0^4 {\left( {x - 2} \right)f\left( x \right)d{\rm{x}}} = 5\)

Đặt \(\left\{ \begin{array}{l}u = {x^2}\\dv = f'\left( x \right)d{\rm{x}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = 2{\rm{xdx}}\\v = f\left( x \right)\end{array} \right.\) suy ra \(\int\limits_0^4 {{x^2}f'\left( x \right)d{\rm{x}}} = \left. {{x^2}f\left( x \right)} \right|_0^4 - \int\limits_0^4 {2{\rm{x}}f\left( x \right)d{\rm{x}}} \)

\( = 16f\left( 4 \right) - 2\int\limits_0^4 {xf\left( x \right)d{\rm{x}}} \)

Do đó \(\int\limits_0^4 {\left[ {{x^2}f'\left( x \right) + 4f\left( x \right)} \right]d{\rm{x}}} = 16 - 2\int\limits_0^2 {\left[ {xf\left( x \right) - 2f\left( x \right)} \right]d{\rm{x}}} = 16 - 2\int\limits_0^4 {\left( {x - 2} \right)f\left( x \right)d{\rm{x}}} \)

\( = 16 - 2.5 = 6\).