The area of the region enclosed by the curves $y = e^x$, $y = |e^x - 1|$, and the $y$ -axis is:

Step 1: Analyze the curves and region

We need to find the area enclosed by three curves: $y = e^x$, $y = |e^x - 1|$, and the $y$ -axis ($x=0$). First, let's understand the behavior of the function $y = |e^x - 1|$.

Step 2: Define the absolute value function

The absolute value function $y = |e^x - 1|$ can be split into two cases. When $e^x - 1 \ge 0$, which means $x \ge 0$, the function is $y = e^x - 1$. When $e^x - 1 < 0$, which means $x < 0$, the function is $y = -(e^x - 1) = 1 - e^x$.

Step 3: Identify intersection points

We need to find the intersection points of the curves. The $y$ -axis is $x=0$. At $x=0$, $y=e^0=1$ and $y=|e^0-1|=|1-1|=0$. The curves $y=e^x$ and $y=1-e^x$ intersect when $e^x = 1-e^x$, which gives $2e^x=1$, so $e^x=1/2$. This means $x = \ln(1/2) = -\ln 2$. At this point, $y=1/2$. The curves $y=e^x$ and $y=e^x-1 do$ not intersect.

Step 4: Set up the integral for the area

The region is bounded by $x=0$ on the right. The intersection point of $y=e^x$ and $y=1-e^x$ is at $x = -\ln 2$. For $x < 0$, $e^x > 1-e^x$. For $x > 0$, $e^x > e^x-1$. The area is split into two parts: from $x = -\ln 2$ to $x = 0$. In the interval $[-\ln 2, 0]$, the upper curve is $y=e^x$ and the lower curve is $y=1-e^x$. For $x \ge 0$, the curve $y=|e^x-1|$ becomes $y=e^x-1$. The region is bounded by $y=e^x$, $y=e^x-1$ and $x=0$. However, $y=e^x$ is always above $y=e^x-1$, and the $y$ -axis is $x=0$. The area for $x \ge 0$ is not enclosed by the three given curves. Thus, the area is only from $x = -\ln 2$ to $x = 0$.

Step 5: Evaluate the integral

Now we evaluate the definite integral. We integrate $(2e^x - 1)$ from $x = -\ln 2$ to $x = 0$. The antiderivative of $2e^x - 1$ is $2e^x - x$. Plugging in the limits, we get $(2e^0 - 0) - (2e^{-\ln 2} - (-\ln 2))$. Since $e^0 = 1$ and $e^{-\ln 2} = e^{\ln(1/2)} = 1/2$, the expression simplifies to $(2 - 0) - (2 \cdot 1/2 + \ln 2) = 2 - (1 + \ln 2) = 1 - \ln 2$.

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