Area of the region bounded by the curve $y = e^{x}$ and lines $x = 0$ and $y = e$ is

$\text{curve: } y = e^{x}$ $\text{lines: } x = 0 \quad \text{and} \quad y = e$ $y=e^{x} \cap y=e \Rightarrow x=1; \quad y=e^{x} \cap x=0 \Rightarrow (0,1)$ $\text{To find: area of the region with corners } (0,1),\ (0,e),\ (1,e)$ $A = \displaystyle\int_{0}^{1} (e - e^{x})\,dx$ $A = \Big[ex - e^{x}\Big]_{0}^{1} = (e - e) - (0 - 1) = 1$ $\displaystyle\int_{0}^{1} e^{x}\,dx = e - 1$ $(3)\ e - (e-1) = 1 \quad \checkmark$ $A = \displaystyle\int_{1}^{e} \ln y\,dy = \Big[y\ln y - y\Big]_{1}^{e} = 1$ $\displaystyle\int_{1}^{e} \ln(e+1-y)\,dy \ \xrightarrow{\ t=e+1-y\ } \ \int_{1}^{e} \ln t\,dt = 1$ $(1)\ e - 1 \approx 1.718 \neq 1 \quad \times$ $A = 1 \ \text{square unit}$ $\boxed{\text{Answer: (2), (3), (4)}}$