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

* Hướng dẫn giải

Đáp án C.

Đặt $\left\{\begin{array}{l}u=f\left(x\right)\\ dv=\left(x+1\right)e^{x}dx\end{array}\right.\iff \left\{\begin{array}{l}du=f\text{'}\left(x\right)dx\\ v=x{e}^x\end{array}\right.$, khi đó ${\int }_0^1\left(x+1\right)e^x.f\left(x\right)dx$

$={\overline{)x{e}^x.f\left(x\right)}}_0^1-{\int }_0^{1}x{e}^x.f\text{'}\left(x\right)dx$

$=e.f\left(1\right)-{\int }_0^{1}x{e}^x.f\text{'}\left(x\right)dx\iff {\int }_0^{1}x{e}^x.f\text{'}\left(x\right)dx=-{\int }_0^1\left(x+1\right)e^x.f\left(x\right)dx=\frac{1-e^2}{4}$. 

Xét tích phân ${\int }_0^1{\left[f\text{'}\left(x\right)+k.x{e}^x\right]}^{2}dx={\int }_0^1{\left[f\text{'}\left(x\right)\right]}^{2}dx+2k.{\int }_0^{1}x{e}^x.f\text{'}\left(x\right)dx+k^2.{\int }_0^1{x}^2{e}^{2x}dx=0$ 

$\iff \frac{e^2-1}{4}+2k.\frac{1-e^2}{4}+k^2.\frac{e^2-1}{4}=0\Rightarrow k^2-2k+1=0\iff k=1\Rightarrow f\text{'}\left(x\right)=-x.e^x.$ 

Do đó $f\left(x\right)=\int f\text{'}\left(x\right)dx=-\int x.e^{x}dx=\left(1-x\right)e^x+C$ mà $f\left(1\right)=0\Rightarrow C=0$. 

Vậy $I={\int }_0^{1}f\left(x\right)dx={\int }_0^1(1-x)e^{x}dx\stackrel{casio}{\to }I=e-2.$