Answer to Question #156460 in Calculus for nurul

Question #156460

Find the area of region bounded by the lines y=x,y=3x,x+y=4 using double integral.


1
Expert's answer
2021-01-19T18:16:36-0500

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)={3xforx<14xforx>1f(z) = \left\{ \begin{array}{rcl} {3x} & {for}& x<1\\ {4-x} & {for} & x>1 \end{array}\right.

The inner area of the triangle can be obtained from expression

A=02xf(x)dydx=01x3xdydx+12x4xdydx=012xdx+12(42x)dx=2A=\int{}^2_0\int{}^{f(x)}_xdy dx = \int{}^1_0\int{}^{3x}_xdy dx + \int{}^2_1\int{}^{4-x}_xdy dx = \int{}^1_02xdx+\int{}^2_1(4-2x)dx=2


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