Đáp án B

Đặt \(\left\{ \begin{array}{l}u = x + 1\\dv = f'\left( x \right)d{\rm{x}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = d{\rm{x}}\\v = f\left( x \right)\end{array} \right.\) ta có:

\[\int\limits_0^3 {\left( {x + 1} \right)f'\left( x \right)d{\rm{x}}} = \left. {\left( {x + 1} \right)f\left( x \right)} \right|_0^3 - \int\limits_0^3 {f\left( x \right)d{\rm{x}}} = 4f\left( 3 \right) - f\left( 0 \right) - \int\limits_0^3 {f\left( x \right)d{\rm{x}}} = a\].

Mặt khác \(\int\limits_0^1 {\left| {f'\left( x \right)} \right|{\rm{d}}x} = \int\limits_0^1 {f'\left( x \right)dx} = f\left( 1 \right) - f\left( 0 \right) = b \Leftrightarrow f\left( 0 \right) = d - b\).

Lại có \(\int\limits_1^3 {\left| {f'\left( x \right)} \right|{\rm{d}}x} = - \int\limits_1^3 {f'\left( x \right)dx} = f\left( 1 \right) - f\left( 3 \right) = d - f\left( 3 \right) = c \Rightarrow f\left( 3 \right) = d - c\).

Thế vào ta được \(4\left( {d - c} \right) - \left( {d - b} \right) - I = a \Leftrightarrow 3d + b - 4c - a = I.\)