Find the area of region bounded by the lines y=x,y=3x,x+y=4 using double integral.
Solution
Lines are shown on the plot.
The points of intersection of lines can be obtained from the equations:
a) y=x, y=3x: x = 3x => x=0, y=0
b) y=3x, x+y=4: 3x=4-x => x=1, y=3
c) y=x, x+y=4: x=4-x => x=2, y=2
Let
"f(z) = \\left\\{ \\begin{array}{rcl}\n{3x} & {for}& x<1\\\\\n{4-x} & {for} & x>1\n\\end{array}\\right."
The inner area of the triangle can be obtained from expression
"A=\\int{}^2_0\\int{}^{f(x)}_xdy dx =\n\\int{}^1_0\\int{}^{3x}_xdy dx + \\int{}^2_1\\int{}^{4-x}_xdy dx =\n\\int{}^1_02xdx+\\int{}^2_1(4-2x)dx=2"
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