The area bounded by the curve x=log (|y|),the | vedclass

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(D) Given the curve is $x = \log |y|$.This can be rewritten as $|y| = e^x$,which implies $y = \pm e^x$.The curve is symmetric about the $x$-axis.The a

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$2\left(1-e^{-1}\right)$

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(D) Given the curve is $x = \log |y|$.This can be rewritten as $|y| = e^x$,which implies $y = \pm e^x$.The curve is symmetric about the $x$-axis.The area bounded by the curve,the lines $x = -1$ and $x = 0$ is given by the integral of $y$ with respect to $x$ from $-1$ to $0$.Since the curve is symmetric,the total area is twice the area above the $x$-axis:$	ext{Area} = 2 \int_{-1}^{0} |y| \, dx = 2 \int_{-1}^{0} e^x \, dx$Evaluating the integral:$	ext{Area} = 2 \left[ e^x ight]_{-1}^{0}$$= 2 (e^0 - e^{-1})$$= 2 (1 - e^{-1})$Thus,the required area is $2(1 - e^{-1})$ square units.

The area bounded by the curve $x=\log (|y|)$,the lines $x=-1$ and $x=0$ is

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