The area of the region enclosed by the curves $y=\mathrm{e}^x, y=\left|\mathrm{e}^x-1\right|$ and $y$-axis is:

Question

Aicharya · Accepted Answer

$\begin{aligned} & e^x=1-e^x \Rightarrow 2 e^x=1 \\ & \Rightarrow e^x=\frac{1}{2} \\ & \Rightarrow x=\ln \frac{1}{2} \\ & \int_{\ln (1 / 2)}^0\left[e^x-\left(1-e^x\right)\right] d x \\ & =\int_{\ln 2}^0\left(2 e^x-1\right) d x=2 e^x-\left.x\right|_{-\ln 2} ^0 \\ & =2-(1+\ln 2) \\ & =1-\log _e 2\end{aligned}$

The area of the region enclosed by the curves $y=\mathrm{e}^x, y=\left|\mathrm{e}^x-1\right|$ and $y$-axis is:

Solution