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The area of the region enclosed by the curves $$y=e^x,\; y=|e^x-1|$$ and the $$y$$ -axis is:
The area is $$\int_{\ln(1/2)}^{0} (e^x - (1-e^x)) dx = \int_{\ln(1/2)}^{0} (2e^x - 1) dx = [2e^x - x]_{\ln(1/2)}^{0} $$
$$= (2-0) - (2(1/2) - \ln(1/2)) = 2 - 1 - \ln 2 = \mathbf{1 - \ln 2}$$ (Option A)
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