Animated Solution for Mathematics - Definite Integration: The area bounded by the curves $y=\ln x,y=\ln |x|,y=|\ln x|$ and $y=|\ln |x||$ is

The problem asks us to find the area bounded by the four curves: $y=\ln x$, $y=\ln |x|$, $y=|\ln x|$, and $y=|\ln |x||$.

First, we must acknowledge the domain. Because the functions involve $\ln x$ and $\ln |x|$, we know that $x eq 0$. This constraint defines our boundary.

Let us focus on the right half-plane, where $x>0$. In this region, $|x|=x$, which simplifies our curves to $y=\ln x$ and $y=|\ln x|$.

The presence of $|x|$ in the functions $y=\ln |x|$ and $y=|\ln |x||$ indicates that these functions are even. An even function is symmetric about the $y$-axis, meaning the area on the right is perfectly mirrored on the left.

Furthermore, the absolute value applied to the logarithm creates symmetry across the $x$-axis. When we combine these properties, we realize the total area is composed of four identical regions—one in each quadrant. We only need to calculate the area in the first quadrant and multiply by 4.

In the first quadrant, specifically for $x\in (0,1)$, the function $y=\ln x$ is negative. Therefore, the boundary is defined by $y=-\ln x$.

The area $A$ of the region in the first quadrant is given by the integral:

The total area $A_{total}$ is then:

To evaluate the integral, we use Integration by Parts. Let $u=\ln x$ and $dv=dx$. This gives $du=\frac{1}{x}dx$ and $v=x$.

Applying the formula $\int udv=uv-\int vdu$, we obtain:

Now, we evaluate this from $0$ to $1$:

At the upper limit, $1\ln 1-1=0-1=-1$. At the lower limit, the limit ${\lim }_{x\to 0}(x\ln x-x)$ is $0$. Thus, the area of one piece is $-(-1-0)=1$.

Multiplying by 4, we find the total area is 4 square units.