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Step-by-Step Solution
Step 1: Identify the Curves
The question deals with four curves:
1. $y = \ln x$
2. $y = \ln \lvert x \rvert$
3. $y = \lvert \ln x \rvert$
4. $y = \lvert \ln \lvert x \rvert \rvert.$
For $x>0$, $\ln\lvert x\rvert = \ln x$, and thus $| \ln\lvert x\rvert| = |\ln x|$. These expressions lead to symmetrical regions above and below the x-axis when considering the absolute values, creating a bounded region between 0 and 1 on the x-axis.
Step 2: Visualize the Graphs
Below are the illustrative graphs (as provided):
Reflecting the negative part above the x-axis (because of the absolute value) creates a symmetric region whose total area can be calculated by focusing on one relevant part and multiplying appropriately.
The final bounded region is shown:
Step 3: Set Up the Integral
Consider the region from $x=0$ to $x=1$, where $y = - \ln x$ (since $\ln x$ is negative in $(0,1)$, its absolute value is $-\ln x$). Because of the symmetry, the total area is 4 times the integral from 0 to 1:
$$
\text{Required Area} = 4 \int_{0}^{1} -\ln x \, dx.
$$
Step 4: Evaluate the Integral
Compute $I = \int_{0}^{1} -\ln x \, dx$:
Use integration by parts or the known result
$$
\int \ln x \, dx = x \ln x - x + C.
$$
Thus,
$$
\int -\ln x \, dx = - ( x \ln x - x ) + C = -x \ln x + x + C.
$$
Evaluate from $0$ to $1$:
$$
\left[ -x \ln x + x \right]_0^1.
$$
At $x = 1$, $- (1)\ln(1) + 1 = 0 + 1 = 1.$
As $x \to 0$, $x \ln x \to 0,$ so the expression goes to $0.$
Hence,
$$
I = 1 - 0 = 1.
$$
Therefore,
$$
\int_{0}^{1} -\ln x \, dx = 1.
$$
Step 5: Calculate the Final Area
Multiply by 4 to account for the symmetrical regions:
$$
\text{Required Area} = 4 \times 1 = 4 \text{ sq. units}.
$$
