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Find the area of the figure bounded by the curves y = |x – 1| and y = 3 – |x|.
To find the area of the figure bounded by
y = |x – 1|
Y = x – 1 is a straight line passing through A(1,0)
Y = 1 – x is a straight line passing through A(1,0) and cutting y - axis at B(0,1)
y = 3 – |x|
Y = 3 – x is a straight line passing through C(0,3)and O(3,0)
Y = 3 + x is a straight line passing through C(0,3)and D( – 3,0)
Point of intersection for
Y = x – 1
And y = 3 – x
We get
X – 1 = 3 – x
or, 2x – 4 = 0
or, x = 2
or, y = 2 – 1 = 1
Thus, point of intersection for y = x – 1 and y = 3 + x is B(2,1)
Point of intersection for
y = 1 – x
y = 3 + x
or,1 – x = 3 + x
or, 2x = – 2
or, x = – 1
or, y = 1 – ( – 1) = 2
Thus, point of intersection for y = 1 – x and y = 3 + x is D( – 1,2)
These are shown in the graph below:
Required area = Region ABCDA
= Region ABFA + Region AFCEA + Region CDEC
The area of the figure bounded by the curves y = |x – 1| and y = 3 – |x| is 4 sq. units