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  • Using the method of integration find the area of the region bounded by lines: 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0

Q : 14 Using the method of integration find the area of the region bounded by lines: \small 2x+y=4,3x-2y=6\hspace{1mm}and\hspace{1mm}x-3y+5=0.

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best_answer

Applications of integrals

We have to find the area of the shaded region ABC

ar(ABC)=ar(ACLM)-ar(ALB)-ar(BMC)

The lines intersect at points (1,2), (4,3) and (2,0)

\\x-3y=-5\\ y=\frac{x}{3}+\frac{5}{3}

\\ar(ACLM)=\int_{1}^{4}(\frac{x}{3}+\frac{5}{3})dx\\ =[\frac{x^{2}}{6}+\frac{5x}{3}]_{1}^{4}\\ =(\frac{4^{2}}{6}+\frac{5\times 4}{3})-(\frac{1}{6}+\frac{5}{3})\\ =\frac{15}{2}\ units

\\2x+y=4\\ y=4-2x

\\ar(ALB)=\int_{1}^{2}(4-2x)dx\\ =[4x-x^{2}]_{1}^{2}\\ =(8-4)-(4-1)\\ =1\ unit

\\3x-2y=6\\ y=\frac{3x}{2}-3

\\ar(BMC)=\int_{2}^{4}(\frac{3x}{2}-3)dx\\ =[\frac{3x^{2}}{4}-3x]_{2}^{4}\\ =(12-12)-(3-6)\\ =3\ units

\\ ar(ABC)=\frac{15}{2}-1-3\\ =\frac{7}{2}\ units

Area of the region bounded by the lines is 3.5 units

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Sayak

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