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प्रश्न
Find the area of the region bounded by 3x – 2y + 6 = 0, x = – 3, x = 1 and x-axis
उत्तर
3x – 2y + 6 = 0
y = `(3x + 6)/2`
To find further limit put y = 0
We get, x = – 2
Area = `int_(-3)^(-2) - y"d"x + int_(-2)^1 y"d"x`
= `(-1)/2 int_(-3)^(-) (3x + 6)"d"x + 1/2 int_(-2)^1 (3x + 6) "d"x`
= `(-1)/2 [(3x^2)/2 + 6x]_(-3)^(-2) + 1/2[(3x^2)/2 + 6x]_(-2)^1`
= `(-1)/2[-6 + 9/2] + 1/2[15/2 + 6]`
= `1/2[6 - 9/2 + 5/2+ 6]`
= `1/2[12 + 3]`
= `15/2`
= 7.5 square units
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