The area bounded by the curve |y| + frac{1}{2 | vedclass

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(A) The given equation is $|y| + \frac{1}{2} = e^{-|x|}$.Since the equation involves $|x|$ and $|y|$,the curve is symmetric about both the $x$-axis an

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$2(1 - \ln 2)$

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(A) The given equation is $|y| + \frac{1}{2} = e^{-|x|}$.Since the equation involves $|x|$ and $|y|$,the curve is symmetric about both the $x$-axis and the $y$-axis.We can write $|y| = e^{-|x|} - \frac{1}{2}$.For $y$ to be defined,we must have $e^{-|x|} - \frac{1}{2} \ge 0$,which implies $e^{-|x|} \ge \frac{1}{2}$,or $|x| \le \ln 2$.In the first quadrant $(x \ge 0, y \ge 0)$,the equation becomes $y = e^{-x} - \frac{1}{2}$.The area in the first quadrant is $\int_0^{\ln 2} (e^{-x} - \frac{1}{2}) dx$.Evaluating the integral: $\int_0^{\ln 2} e^{-x} dx - \int_0^{\ln 2} \frac{1}{2} dx = [-e^{-x}]_0^{\ln 2} - [\frac{1}{2}x]_0^{\ln 2} = (-e^{-\ln 2} - (-e^0)) - \frac{1}{2}\ln 2 = (-\frac{1}{2} + 1) - \frac{1}{2}\ln 2 = \frac{1}{2} - \frac{1}{2}\ln 2 = \frac{1}{2}(1 - \ln 2)$.Since the curve is symmetric in all four quadrants,the total area is $4 	imes \frac{1}{2}(1 - \ln 2) = 2(1 - \ln 2)$.

The area bounded by the curve $|y| + \frac{1}{2} = e^{-|x|}$ is

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