Find the area bounded by the curves y = ||x - | vedclass

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(B) The given curves are $y = ||x - 1| - 2|$ and $y = 2$.To find the intersection points,we set $||x - 1| - 2| = 2$.This implies $|x - 1| - 2 = 2$ or

Vedclass · Accepted Answer

$8$

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(B) The given curves are $y = ||x - 1| - 2|$ and $y = 2$.To find the intersection points,we set $||x - 1| - 2| = 2$.This implies $|x - 1| - 2 = 2$ or $|x - 1| - 2 = -2$.Case $1$: $|x - 1| = 4 \implies x - 1 = 4$ or $x - 1 = -4$,so $x = 5$ or $x = -3$.Case $2$: $|x - 1| = 0 \implies x = 1$.Looking at the graph,the region is bounded between $x = -3$ and $x = 5$ with the upper boundary $y = 2$ and lower boundary $y = ||x - 1| - 2|$.The area $A$ is given by $\int_{-3}^{5} (2 - ||x - 1| - 2|) \, dx$.Due to symmetry about $x = 1$,we can calculate $2 	imes \int_{1}^{5} (2 - ||x - 1| - 2|) \, dx$.For $1 \le x \le 5$,$|x - 1| = x - 1$,so $y = |x - 1 - 2| = |x - 3|$.Area $= 2 \int_{1}^{5} (2 - |x - 3|) \, dx = 2 \left[ \int_{1}^{3} (2 - (3 - x)) \, dx + \int_{3}^{5} (2 - (x - 3)) \, dx ight]$.Area $= 2 \left[ \int_{1}^{3} (x - 1) \, dx + \int_{3}^{5} (5 - x) \, dx ight] = 2 \left[ \left( \frac{x^2}{2} - x ight)_{1}^{3} + \left( 5x - \frac{x^2}{2} ight)_{3}^{5} ight]$.Area $= 2 \left[ (4.5 - 3) - (0.5 - 1) + (25 - 12.5) - (15 - 4.5) ight] = 2 [2 + 2] = 8$ square units.

Find the area bounded by the curves $y = ||x - 1| - 2|$ and $y = 2$.

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