Answer :

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


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