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

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(A) We know that $\ln x$ is defined for $x > 0$ and $\ln |x|$ is defined for all $x \in \mathbb{R} - \{0\}$.Also,$|\ln x| \ge 0$ and $|\ln |x|| \ge 0$

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$4$

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(A) We know that $\ln x$ is defined for $x > 0$ and $\ln |x|$ is defined for all $x \in \mathbb{R} - \{0\}$.Also,$|\ln x| \ge 0$ and $|\ln |x|| \ge 0$.The region bounded by these curves is symmetric about both the $x$-axis and the $y$-axis.The area is given by $4 	imes \int_{0}^{1} |\ln x| \, dx$.Since for $x \in (0, 1)$,$\ln x Area $= -4 \int_{0}^{1} \ln x \, dx$.Using integration by parts,$\int \ln x \, dx = x \ln x - x$.Area $= -4 [x \ln x - x]_{0}^{1} = -4 [(1 \ln 1 - 1) - (\lim_{x 	o 0^+} x \ln x - 0)]$.Since $\lim_{x 	o 0^+} x \ln x = 0$,the area $= -4 [0 - 1 - 0] = -4(-1) = 4 \, sq. \, units$.

The area bounded by the curves $y = \ln x$,$y = \ln |x|$,$y = |\ln x|$ and $y = |\ln |x||$ is ......... $sq. \,unit$.

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