[tex]f(x) = x-e^x \\ g(x) = f^{-1}(x) \\ \\ \displaystyle \boxed{\int_{a}^bf(x)\, dx +\int_{f(a)}^{f(b)}f^{-1}(x)\, dx = bf(b) - af(a)}\rightarrow \text{Teorema lui Laisant}\\ \\ \int_{-1}^{1-\frac{1}{e}} g(x)\,dx =\int_{-1}^{1-\frac{1}{e}} f^{-1}(x)\, dx\\ \\\\ -1 = f(a) \Rightarrow -1 = a-e^{-a} \Rightarrow a = 0 \\ 1-\frac{1}{e} = f(b) \Rightarrow 1-\dfrac{1}{e} = b-e^{-b} \Rightarrow b = 1[/tex]
[tex]\displaystyle \Rightarrow \int_{0}^1f(x)\, dx+\int_{f(0)}^{f(1)}f^{-1}(x) dx = 1\cdot f(1)-0\cdot f(0) \\ \\ \Rightarrow\int_{0}^1f(x)\, dx+\int_{-1}^{1-\frac{1}{e}}g(x)\,dx = 1\cdot f(1)-0\cdot f(0)\\\int_{0}^1 f(x)\,dx= \dfrac{1}{e} - \dfrac{1}{2}\\ \\ \Rightarrow\dfrac{1}{e}-\dfrac{1}{2}+ \int_{-1}^{1-\frac{1}{e}}g(x)\,dx = 1-\dfrac{1}{e} \Rightarrow \boxed{\int_{-1}^{1-\frac{1}{e}}g(x)\,dx =\dfrac{3}{2}-\dfrac{2}{e}}[/tex]
