The area of the figure bounded by the curves | vedclass

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(C) To find the area bounded by $y = |x - 1|$ and $y = 3 - |x|$,we first find the intersection points of the curves.For $x \ge 1$,$y = x - 1$ and $y =

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$4$

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(C) To find the area bounded by $y = |x - 1|$ and $y = 3 - |x|$,we first find the intersection points of the curves.For $x \ge 1$,$y = x - 1$ and $y = 3 - |x|$. If $x \ge 1$,then $x - 1 = 3 - x \Rightarrow 2x = 4 \Rightarrow x = 2$. At $x = 2$,$y = 1$.For $x For $0 \le x The required area is given by the integral:$A = \int_{-1}^{2} (3 - |x| - |x - 1|) dx$We split the integral based on the definitions of the absolute values:$A = \int_{-1}^{0} ((3 + x) - (1 - x)) dx + \int_{0}^{1} ((3 - x) - (1 - x)) dx + \int_{1}^{2} ((3 - x) - (x - 1)) dx$$A = \int_{-1}^{0} (2 + 2x) dx + \int_{0}^{1} (2) dx + \int_{1}^{2} (4 - 2x) dx$$A = [2x + x^2]_{-1}^{0} + [2x]_{0}^{1} + [4x - x^2]_{1}^{2}$$A = (0 - (-2 + 1)) + (2 - 0) + ((8 - 4) - (4 - 1))$$A = 1 + 2 + (4 - 3) = 1 + 2 + 1 = 4 	ext{ sq. units}$.Thus,the correct option is $C$.

The area of the figure bounded by the curves $y = |x - 1|$ and $y = 3 - |x|$ is ....... $sq. \text{ unit}$.

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