Phương trình hoành độ giao điểm:
\(\left(e+1\right)x=\left(1+e^x\right)x\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Trên \(\left[0;1\right]\) ta có \(\left(e+1\right)x\ge\left(1+e^x\right)x\)
Do đó:
\(S=\int\limits^1_0\left[\left(e+1\right)x-\left(1+e^x\right)x\right]dx=\int\limits^1_0e.xdx-\int\limits^1_0x.e^xdx=I_1-I_2\)
\(I_1=\int\limits^1_0e.xdx=\dfrac{1}{2}e.x^2|^1_0=\dfrac{e}{2}\)
Xét \(I_2\), đặt \(\left\{{}\begin{matrix}u=x\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I_2=x.e^x|^1_0-\int\limits^1_0e^xdx=\left(x-1\right)e^x|^1_0=1\)
\(\Rightarrow S=\dfrac{e}{2}-1\)
