The area bounded by the curves y=|x-1|+|x-2| | vedclass

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(B) The function $y = |x-1| + |x-2|$ can be defined as:$y = \begin{cases} -(x-1) - (x-2) = -2x+3, & 	ext{if } x < 1 \ (x-1) - (x-2) = 1, & 	ext{if

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$4$

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(B) The function $y = |x-1| + |x-2|$ can be defined as:$y = \begin{cases} -(x-1) - (x-2) = -2x+3, & 	ext{if } x  2 \end{cases}$To find the intersection with $y=3$:For $x For $x > 2$: $2x-3 = 3 \implies x=3$.The region is a trapezoid bounded by $x=0$ to $x=3$ with height $h=3-1=2$ (since the minimum value of the curve is $1$ at $x \in [1, 2]$).The area is the integral of $(3 - (|x-1| + |x-2|))$ from $x=0$ to $x=3$.Alternatively,the area is the area of the rectangle formed by $x=0$ to $x=3$ and $y=0$ to $y=3$ minus the area under the curve $y=|x-1|+|x-2|$.Area $= \int_{0}^{3} 3 \, dx - \int_{0}^{3} (|x-1| + |x-2|) \, dx = 9 - [\int_{0}^{1} (-2x+3) \, dx + \int_{1}^{2} 1 \, dx + \int_{2}^{3} (2x-3) \, dx] = 9 - [2 + 1 + 2] = 9 - 5 = 4$.

The area bounded by the curves $y=|x-1|+|x-2|$ and $y=3$ is equal to

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