Cho hàm số \(f(x)\) thỏa mãn \(\int\limits_0^1 {\left( {x + 1} \right)f\left( x \right){\rm{d}}x}  = 10\) và \(2f\left( 1 \right

* Hướng dẫn giải

Đặt \(\left\{ \begin{array}{l}
u = x + 1\\
{\rm{d}}v = f'\left( x \right){\rm{d}}x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{\rm{d}}u = {\rm{d}}x\\
v = f\left( x \right)
\end{array} \right.\). Khi đó \(I = \left. {\left( {x + 1} \right)f\left( x \right)} \right|_0^1 - \int\limits_0^1 {f\left( x \right)dx} \)

Suy ra \(10 = 2f\left( 1 \right) - f\left( 0 \right) - \int\limits_0^1 {f\left( x \right){\rm{d}}x}  \Rightarrow \int\limits_0^1 {f\left( x \right){\rm{d}}x}  =  - 10 + 2 =  - 8\)

Vậy \(\int\limits_0^1 {f\left( x \right){\rm{d}}x}  =  - 8\).