Question

Using the method of integration find the area bounded by the curve [ Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x – y = 11]

Answer 1

The given curve is | x |+| y |=1. The area of the region bounded by the given curve is represented by the shaded region ABCD.

The point of intersection of lines −x+y=1 and x+y=1 is ( 0,1 ).

The point of intersection of lines x−y=1 and x+y=1 is ( 1,0 ).

The point of intersection of lines −x−y=1 and x−y=1 is ( 0,−1 ).

The point of intersection of lines −x+y=1 and −x−y=1 is ( −1,0 ).

So, the curve intersect the axes at points A( 0,1 ),B( 1,0 ),C( 0,−1 ) and D( −1,0 ).

Since the values of x and y correspond to each other, then it is observed that the given curve is symmetrical about x-axis and y-axis.

The area of the shaded region is,

Area ABCD=4×Area OABO

Consider the line AB which gives the x-value from 0 to 1 with the equation,

x+y=1 y=1−x

So the area of OBAO is,

Area OBAO= ∫ 0 1 ( 1−x )dx

This gives the area of ABCD as,

Area of ABCD=4 ∫ 0 1 ( 1−x )dx =4 ( x− x 2 2 ) 0 1 =4[ 1− 1 2 ] =2

Therefore, the area of bounded region is 2 sq. units.