Using integration,find the area of the region | vedclass

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(A) $BL$ and $CM$ are drawn perpendicular to the $x$-axis.It can be observed in the figure that,Area $(\Delta ABC) = 	ext{Area}(ALBA) + 	ext{Area}(B

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$4$ sq. units

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(A) $BL$ and $CM$ are drawn perpendicular to the $x$-axis.It can be observed in the figure that,Area $(\Delta ABC) = 	ext{Area}(ALBA) + 	ext{Area}(BLMCB) - 	ext{Area}(AMCA)$ ...... $(1)$The equation of the line segment $AB$ passing through $(-1, 0)$ and $(1, 3)$ is:$y - 0 = \frac{3 - 0}{1 - (-1)}(x - (-1)) \implies y = \frac{3}{2}(x + 1)$$	ext{Area}(ALBA) = \int_{-1}^{1} \frac{3}{2}(x + 1) dx = \frac{3}{2} \left[ \frac{x^2}{2} + x ight]_{-1}^{1} = \frac{3}{2} \left[ (\frac{1}{2} + 1) - (\frac{1}{2} - 1) ight] = \frac{3}{2} [2] = 3$ sq. units.The equation of the line segment $BC$ passing through $(1, 3)$ and $(3, 2)$ is:$y - 3 = \frac{2 - 3}{3 - 1}(x - 1) \implies y - 3 = -\frac{1}{2}(x - 1) \implies y = -\frac{1}{2}x + \frac{7}{2}$$	ext{Area}(BLMCB) = \int_{1}^{3} (-\frac{1}{2}x + \frac{7}{2}) dx = \left[ -\frac{x^2}{4} + \frac{7x}{2} ight]_{1}^{3} = (-\frac{9}{4} + \frac{21}{2}) - (-\frac{1}{4} + \frac{7}{2}) = (\frac{33}{4}) - (\frac{13}{4}) = \frac{20}{4} = 5$ sq. units.The equation of the line segment $AC$ passing through $(-1, 0)$ and $(3, 2)$ is:$y - 0 = \frac{2 - 0}{3 - (-1)}(x - (-1)) \implies y = \frac{2}{4}(x + 1) \implies y = \frac{1}{2}(x + 1)$$	ext{Area}(AMCA) = \int_{-1}^{3} \frac{1}{2}(x + 1) dx = \frac{1}{2} \left[ \frac{x^2}{2} + x ight]_{-1}^{3} = \frac{1}{2} \left[ (\frac{9}{2} + 3) - (\frac{1}{2} - 1) ight] = \frac{1}{2} [\frac{15}{2} + \frac{1}{2}] = \frac{1}{2} [8] = 4$ sq. units.Therefore,from equation $(1)$,$	ext{Area}(\Delta ABC) = 3 + 5 - 4 = 4$ sq. units.

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

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