The area enclosed by the curves y = ln x,y = | vedclass

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(B) Let us analyze the given curves:$1$. $y = \ln x$ is defined for $x > 0$.$2$. $y = \ln |x|$ is defined for $x 
eq 0$.$3$. $y = |\ln x|$ is defined

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$4$

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(B) Let us analyze the given curves:$1$. $y = \ln x$ is defined for $x > 0$.$2$. $y = \ln |x|$ is defined for $x 
eq 0$.$3$. $y = |\ln x|$ is defined for $x > 0$.$4$. $y = |\ln |x||$ is defined for $x 
eq 0$.The region enclosed by these curves is symmetric about the $y$-axis. The curves form four identical regions in the four quadrants defined by the lines $x = 1, x = -1, x = 0$ and the $x$-axis.Each region is bounded by the curve $y = |\ln |x||$ and the $x$-axis between $x = 0$ and $x = 1$ (or $x = -1$ and $x = 0$).The area of one such region is given by $\int_0^1 |\ln x| \, dx$.Since $\ln x Area $= \int_0^1 -\ln x \, dx = -[x \ln x - x]_0^1 = -((1 \cdot 0 - 1) - (0)) = 1$.Since there are four such symmetric regions,the total area is $4 	imes 1 = 4$.

The area enclosed by the curves $y = \ln x$,$y = \ln |x|$,$y = |\ln x|$,and $y = |\ln |x||$ is equal to:

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