The area of the region bounded by the curves | vedclass

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(D) The required area is given by the integral $\int_{1}^{3} |x - 2| \, dx$.Since $|x - 2| = -(x - 2)$ for $x < 2$ and $|x - 2| = (x - 2)$ for $x \geq

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$1$

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(D) The required area is given by the integral $\int_{1}^{3} |x - 2| \, dx$.Since $|x - 2| = -(x - 2)$ for $x $	ext{Area} = \int_{1}^{2} -(x - 2) \, dx + \int_{2}^{3} (x - 2) \, dx$$= \int_{1}^{2} (2 - x) \, dx + \int_{2}^{3} (x - 2) \, dx$$= \left[ 2x - \frac{x^2}{2} ight]_{1}^{2} + \left[ \frac{x^2}{2} - 2x ight]_{2}^{3}$$= \left( (4 - 2) - (2 - 0.5) ight) + \left( (4.5 - 6) - (2 - 4) ight)$$= (2 - 1.5) + (-1.5 - (-2))$$= 0.5 + 0.5 = 1$.Thus,the area is $1$ square unit.

The area of the region bounded by the curves $y = |x - 2|$,$x = 1$,$x = 3$ and the $x$-axis is

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