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Question
Draw a rough sketch of the given curve y = 1 + |x +1|, x = –3, x = 3, y = 0 and find the area of the region bounded by them, using integration.
Solution
Given equations are y = 1 + |x + 1|
x = –3 and x = 3
y = 0
Taking y = 1 + |x + 1|
⇒ y = 1 + x + 1
⇒ y = x + 2
And y = 1 – x – 1
⇒ y = –x
On solving we get x = –1
Area of the required regions = `int_(-3)^(-1) -x "d"x + int_(-1)^3 (x + 2) "d"x`
= `-[x^2/2]_-3^-1 + [x^2/2 + 2x]_1^3`
= `-[1/2 - 9/2] + [(9/2 + 6) - (1/2 - 2)]`
= `-(-4) + [21/2 + 3/2]`
= 4 + 12
= 16 sq.units
Hence, the required area = 16 sq.units
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