The area of the region bounded by y = |x - 1| | vedclass

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(B) The given curves are $y = |x - 1|$ and $y = 1$.To find the intersection points,set $|x - 1| = 1$,which gives $x - 1 = 1$ or $x - 1 = -1$. Thus,$x

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$1$

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(B) The given curves are $y = |x - 1|$ and $y = 1$.To find the intersection points,set $|x - 1| = 1$,which gives $x - 1 = 1$ or $x - 1 = -1$. Thus,$x = 2$ or $x = 0$.The region is bounded between $x = 0$ and $x = 2$.The area $A$ is given by $\int_{0}^{2} (1 - |x - 1|) dx$.Since $|x - 1| = 1 - x$ for $x $A = \int_{0}^{1} (1 - (1 - x)) dx + \int_{1}^{2} (1 - (x - 1)) dx$$A = \int_{0}^{1} x dx + \int_{1}^{2} (2 - x) dx$$A = \left[ \frac{x^2}{2} \right]_{0}^{1} + \left[ 2x - \frac{x^2}{2} \right]_{1}^{2}$$A = \left( \frac{1}{2} - 0 \right) + \left( (4 - 2) - (2 - \frac{1}{2}) \right)$$A = \frac{1}{2} + (2 - \frac{3}{2}) = \frac{1}{2} + \frac{1}{2} = 1$.Thus,the area is $1$ square unit.

The area of the region bounded by $y = |x - 1|$ and $y = 1$ is

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