Câu 47: Cho hàm số \(f\left(x\right)\) nhận giá trị dương và có đạo hàm liên tục trên \(\left[0;1\right]\) sao cho \(f\left(1\right)=1\) và \(f\left(x\right).f\left(1-x\right)=e^{x^2-x}\) \(\forall x\in\left[0;1\right]\) . Tính \(I=\int\limits^1_0\dfrac{\left(2x^3-3x^2\right)f'\left(x\right)}{f\left(x\right)}dx.\)
A. \(\dfrac{-3}{10}\) B. \(\dfrac{-1}{10}\) C. \(\dfrac{-2}{10}\) D. \(\dfrac{-1}{5}\)
\(f\left(x\right).f\left(1-x\right)=e^{x^2-x}\Rightarrow f\left(0\right).f\left(1\right)=1\)
\(\Rightarrow f\left(0\right)=1\)
Đặt \(\left\{{}\begin{matrix}u=2x^3-3x^2\\dv=\dfrac{f'\left(x\right)}{f\left(x\right)}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=6x^2-6x\\x=ln\left[f\left(x\right)\right]\end{matrix}\right.\)
\(\Rightarrow I=ln\left[f\left(x\right)\right].\left(2x^3-3x^2\right)|^1_0-\int\limits^1_0\left(6x^2-6x\right).ln\left[f\left(x\right)\right]dx\)
\(=0-\int\limits^1_0\left(6x^2-6x\right).ln\left[f\left(x\right)\right]dx\)
Đặt \(u=1-x\Rightarrow x=1-u\Rightarrow dx=-du\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=1\Rightarrow u=0\end{matrix}\right.\)
\(x-x^2=x\left(1-x\right)=\left(1-u\right)u\)
\(\Rightarrow I=6\int\limits^0_1u\left(1-u\right).ln\left[f\left(1-u\right)\right].\left(-du\right)=-6\int\limits^1_0\left(u^2-u\right).ln\left[f\left(1-u\right)\right]du\)
\(=-6\int\limits^1_0\left(x^2-x\right)ln\left[f\left(1-x\right)\right]dx\)
\(=-6\int\limits^1_0\left(x^2-x\right)ln\left[\dfrac{e^{x^2-x}}{f\left(x\right)}\right]dx\)
\(=-6\int\limits^1_0\left(x^2-x\right).\left[lne^{x^2-x}-ln\left[f\left(x\right)\right]\right]dx\)
\(=-6\int\limits^1_0\left(x^2-x\right)^2dx+6\int\limits^1_0\left(x^2-x\right)ln\left[f\left(x\right)\right]dx\)
\(=-6\int\limits^1_0\left(x^2-x\right)^2dx-I\)
\(\Rightarrow2I=-6\int\limits^1_0\left(x^2-x\right)^2dx=-\dfrac{1}{5}\)
\(\Rightarrow I=-\dfrac{1}{10}\)
