Cho hàm số f(x) có đạo hàm liên tục trên [0;1] thỏa mãn điều kiện

* Hướng dẫn giải

$\Rightarrow A=x{e}^{x}f\left(x\right)\overline{)\begin{array}{c}1\\ 0\end{array}-{\int }_0^{1}x{e}^{x}f\text{'}\left(x\right)dx}=-{\int }_0^{1}x{e}^{x}f\text{'}\left(x\right)dx=\frac{1-e^2}{4}$

Xét ${\int }_0^1{x}^2{e}^{2x}dx=e^{2x}\left(\frac{1}{2}{x}^2-\frac{1}{2}x+\frac{1}{4}\right)\overline{)\begin{array}{c}1\\ 0\end{array}=\frac{e^2-1}{4}}$

Ta có $$\begin{gathered} {\int }_0^1{\left[f\text{'}\left(x\right)\right]}^{2}dx+2{\int }_0^{1}x{e}^{x}f\text{'}\left(x\right)dx+{\int }_0^1{x}^2{e}^{2x}f\left(x\right)dx=0 \\ \iff {\int }_0^1{\left(f\text{'}\left(x\right)+x.e^x\right)}^{2}dx=0 \\ f\text{'}\left(x\right)+x.e^x=0,\forall x\in \left[0;1\right]\left(do {\left(f\text{'}\left(x\right)+x.e^x\right)}^2\ge 0, \forall x\in \left[0;1\right]\right) \\ \Rightarrow f\text{'}\left(x\right)=-x{e}^x\Rightarrow f\left(x\right)=\left(1-x\right)e^x+C \\ f\left(1\right)=0\Rightarrow f\left(x\right)=\left(1-x\right)e^x \\ \Rightarrow I={\int }_0^{1}f\left(x\right)dx={\int }_0^1\left(1-x\right)e^{x}dx=\left(2-x\right)e^x\overline{)\begin{array}{c}1\\ 0\end{array}=e-2} \end{gathered}$$