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Question
Find the area of the region bounded by 2x – y + 1 = 0, y = – 1, y = 3 and y-axis
Solution
2x – y + 1 = 0
x = `(y - 1)/2`
To find further limit put x = 0, we get y = 1
Area = `int_(-1)^1 - x "d"xy + int_(-1)^3 x "d"y`
= `(-1)/2 int_(-1)^1 (y - 1) "d"y + 1/2 int_1^3 (y - 1) "d"y`
= `(-1)/2[y^2/2 - y]_(-1)^1 + 1/2[y^2/2 - y]_1^3`
= `(-1)/2[(1/2 - 1) - (1/2 + 1)] + 1/2[(9/2 - 3) - (1/2 - 1)]`
= `(-1)/2 - [1/2 - 3/2] + 1/2[3/2 + 1/2]`
= `1/2[1/2 + 3/2 + 3/2 + 1/2]`
= `1/2[4]`
= 2 square units
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