Đáp án:
\[\int\limits_1^2 {f\left( x \right)dx} = f\left( 2 \right) - a\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\int\limits_1^2 {\left( {x - 1} \right)f'\left( x \right)dx} = a\\
\left\{ \begin{array}{l}
u = x - 1\\
v' = f'\left( x \right)
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 1\\
v = f\left( x \right)
\end{array} \right.\\
\Rightarrow \int\limits_1^2 {\left( {x - 1} \right)f'\left( x \right)dx} = \mathop {\left. {\left( {x - 1} \right).f\left( x \right)} \right|}\nolimits_1^2 - \int\limits_1^2 {1.f\left( x \right)dx} \\
\Leftrightarrow a = 1.f\left( 2 \right) - 0.f\left( 1 \right) - \int\limits_1^2 {f\left( x \right)dx} \\
\Rightarrow \int\limits_1^2 {f\left( x \right)dx} = f\left( 2 \right) - a
\end{array}\)
