Answer to Question #156678 in Calculus for Rabiah

Question

find the area of the region bounded by the lines y=x y=3x and x+y=4 using double integral

Answer 1

The intersection between the lines $y=x$ and $y=4-x$ is,

$x=4-x$

$2x=4$

$x=2$

So, the point of intersection is $(2,2)$

The intersection between the lines $y=3x$ and $y=4-x$ is,

$3x=4-x$

$4x=4$

$x=1$

So, the point of intersection is $(1,3)$

The sketch of the bounded region is as shown in the figure below:

The area of the region bounded by the lines is evaluated as,

Area$(A)=\iint_{R_1}+\iint_{R_2}$

$=\int_{x=0}^{1}\int_{y=x}^{3x}dydx+\int_{x=1}^{2}\int_{y=x}^{4-x}dydx$

$=\int_{x=0}^{1}[y]_{y=x}^{3x}dx+\int_{x=1}^{2}[y]_{y=x}^{4-x}dx$

$=\int_{x=0}^{1}(3x-x)dx+\int_{x=1}^{2}(4-x-x)dx$

$=2\int_{x=0}^{1}xdx+\int_{x=1}^{2}(4-2x)dx$

$=2[\frac{x^2}{2}]_{x=0}^{1}+[4x-2(\frac{x^2}{2})]_{x=1}^{2}$

$=1+4(2-1)-(2^2-1^2)$

$=1+4-3$

$=1+1$

$=2$

Therefore, the area of the bounded region is Area$(A)=2$ square units.