Cho hàm số \(f(x)\) có đạo hàm liên tục trên \(\left[ {0;1} \right],\) thỏa \(2f\left( x \right) + 3f\left( {1 - x} \right) =

* Hướng dẫn giải

Ta có \(\int\limits_0^1 {f'\left( x \right){\rm{d}}x}  = f\left( x \right)\left| {\begin{array}{*{20}{c}}
1\\
0
\end{array}} \right. = f\left( 1 \right) - f\left( 0 \right).\)

\(2f\left( x \right) + 3f\left( {1 - x} \right) = \sqrt {1 - {x^2}}  \to \left\{ \begin{array}{l}
2f\left( 0 \right) + 3f\left( 1 \right) = 1\\
2f\left( 1 \right) + 3f\left( 0 \right) = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
f\left( 0 \right) =  - \frac{2}{5}\\
f\left( 1 \right) = \frac{3}{5}
\end{array} \right..\)

\(I = \int\limits_0^1 {f'\left( x \right){\rm{d}}x}  = f\left( 1 \right) - f\left( 0 \right) = \frac{3}{5} + \frac{2}{5} = 1.\)