The area bounded by the curve y=|x-2|,x=1,x=3 | vedclass

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(C) The required area is given by the integral $\int_{1}^{3} |x-2| dx$. Since the function $|x-2|$ changes its definition at $x=2$,we split the integr

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$1 	ext{ sq. units}$

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(C) The required area is given by the integral $\int_{1}^{3} |x-2| dx$. Since the function $|x-2|$ changes its definition at $x=2$,we split the integral: $	ext{Area} = \int_{1}^{2} -(x-2) dx + \int_{2}^{3} (x-2) dx$. Evaluating the first part: $\int_{1}^{2} (2-x) dx = [2x - \frac{x^2}{2}]_{1}^{2} = (4 - 2) - (2 - \frac{1}{2}) = 2 - 1.5 = 0.5$. Evaluating the second part: $\int_{2}^{3} (x-2) dx = [\frac{x^2}{2} - 2x]_{2}^{3} = (4.5 - 6) - (2 - 4) = -1.5 - (-2) = 0.5$. Adding both parts: $0.5 + 0.5 = 1 	ext{ sq. unit}$.

The area bounded by the curve $y=|x-2|$,$x=1$,$x=3$ and $X$-axis is

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