The area of the region enclosed by the curves $y = e^x$, $y = |e^x-1|$ and $y$-axis is:
Step-by-Step Solution
Key Concept: Find where $e^x = 1-e^x$ (i.e., $x = \ln\tfrac{1}{2} < 0$); for $x\in[\ln\tfrac{1}{2},0]$ the enclosed strip is $e^x - (1-e^x) = 2e^x-1$.
For $x<0$: $|e^x-1|=1-e^x$. Intersection with $y=e^x$: $e^x = 1-e^x \Rightarrow x = \ln\tfrac{1}{2}$.
$$\text{Area} = \int_{\ln(1/2)}^0\!(e^x-(1-e^x))dx = \int_{\ln(1/2)}^0(2e^x-1)dx = \left[2e^x-x\right]_{\ln(1/2)}^0$$
$$= (2-0)-(1+\ln 2) = 1-\ln 2.$$
Correct Answer: 1