Using integration,find the area of the region | vedclass

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(A) Let $A(1,0)$,$B(2,2)$,and $C(3,1)$ be the vertices of a triangle $ABC$.The equations of the sides are:Side $AB$: $y - 0 = \frac{2-0}{2-1}(x - 1) \

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(A) Let $A(1,0)$,$B(2,2)$,and $C(3,1)$ be the vertices of a triangle $ABC$.The equations of the sides are:Side $AB$: $y - 0 = \frac{2-0}{2-1}(x - 1) \implies y = 2(x-1)$Side $BC$: $y - 2 = \frac{1-2}{3-2}(x - 2) \implies y - 2 = -1(x-2) \implies y = 4-x$Side $AC$: $y - 0 = \frac{1-0}{3-1}(x - 1) \implies y = \frac{1}{2}(x-1)$The area of $\Delta ABC$ is given by:Area $= \int_{1}^{2} (y_{AB} - y_{AC}) dx + \int_{2}^{3} (y_{BC} - y_{AC}) dx$$= \int_{1}^{2} (2x - 2 - \frac{x-1}{2}) dx + \int_{2}^{3} (4 - x - \frac{x-1}{2}) dx$$= \int_{1}^{2} (\frac{3x-3}{2}) dx + \int_{2}^{3} (\frac{9-3x}{2}) dx$$= \frac{3}{2} [\frac{x^2}{2} - x]_{1}^{2} + \frac{3}{2} [3x - \frac{x^2}{2}]_{2}^{3}$$= \frac{3}{2} [(2 - 2) - (\frac{1}{2} - 1)] + \frac{3}{2} [(9 - 4.5) - (6 - 2)]$$= \frac{3}{2} [0.5] + \frac{3}{2} [4.5 - 4] = \frac{3}{2} (0.5) + \frac{3}{2} (0.5) = 0.75 + 0.75 = 1.5$ square units.

Using integration,find the area of the region bounded by the triangle whose vertices are $(1, 0)$,$(2, 2)$,and $(3, 1)$.

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