Cho hàm số y=f(x) có đạo hàm cấp hai liên tục trên đoạn [0;1]

* Hướng dẫn giải

Đáp án D

$\begin{array}{l}\underset{0}{\overset{1}{\int }}{e}^{x}f\left(x\right)dx=\underset{0}{\overset{1}{\int }}{e}^x{f}^{\text{'}}\left(x\right)dx\\ =\underset{0}{\overset{1}{\int }}{e}^x{f^{\text{'}}}^{\text{'}}\left(x\right)dx=k\ne 0\end{array}$

Đặt 

$\begin{array}{l}\left\{\begin{array}{l}u=e^x\\ dv=f^{\text{'}}\left(x\right)dx\end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}du=e^{x}dx\\ v=f\left(x\right)\end{array}\right.\\ \Rightarrow \underset{0}{\overset{1}{\int }}{e}^x{f}^{\text{'}}\left(x\right)dx\\ ={\left.e^{x}f\left(x\right)\right|}_0^1-\underset{0}{\overset{1}{\int }}{e}^{x}f\left(x\right)dx\end{array}$

$\begin{array}{l}\Rightarrow k=e.f\left(1\right)-f\left(0\right)-k\\ \Rightarrow ef\left(1\right)-f\left(0\right)=2k.\end{array}$

Đặt 

$\begin{array}{l}\left\{\begin{array}{l}u=e^x\\ dv={f^{\text{'}}}^{\text{'}}\left(x\right)dx\end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}du=e^{x}dx\\ v=f^{\text{'}}\left(x\right)\end{array}\right.\\ \Rightarrow \underset{0}{\overset{1}{\int }}{e}^x{f^{\text{'}}}^{\text{'}}\left(x\right)dx\\ ={\left.e^x{f}^{\text{'}}\left(x\right)\right|}_0^1-\underset{0}{\overset{1}{\int }}{e}^x{f}^{\text{'}}\left(x\right)dx\end{array}$

$\begin{array}{l}\Rightarrow k=e.f^{\text{'}}\left(1\right)-f^{\text{'}}\left(0\right)-k\\ \Rightarrow e.f^{\text{'}}\left(1\right)-f^{\text{'}}\left(0\right)=2k.\end{array}$

Vậy $\frac{e.f^{\text{'}}\left(1\right)-f^{\text{'}}\left(0\right)}{e.f\left(1\right)-f\left(0\right)}=\frac{2k}{2k}=1$