The value of the integration for a non closed path abcd will be -88
Cyclic integration is the integration over a closed loop.Close the path by connecting D to A. Then by Green's theorem, the integral over the closed path ABCDA - which I'll just abbreviate C - is
[tex]\oint_c(Sin(x)+9y)dx+(4x+y)dy\\\\\\[/tex]
[tex]=\int\int \dfrac{\partial (4x+y)}{\partial x} -\dfrac{\partial (Sin(x)+9y)}{\partial y}dxdy[/tex]
[tex]=-5\int\int dxdy[/tex]
(where int(C ) denotes the region interior to the path C )
The remaining double integral is -5 times the area of the trapezoid, which is
[tex]-5 \int\int dxdy=\dfrac{-5}{2}\times (12+4)\times 4=-160[/tex]
To get the line integral you want, just subtract the integral taken over the path DA. On this line segment, we have x = 0 and dx = 0, so this integral reduces to
[tex]\int _{DA} ydy=\int\limits^0_{12} y dy=-\int\limits^{12}_0ydy =-72[/tex]
Then
[tex]\int_{ABCD}9Sin(x)+9y)dx+(4x+y)dy=-160-(-72)=-88[/tex]
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