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]\displaystyle \oint_C (\sin(x)+9y)\,\mathrm dx + (4x+y)\,\mathrm dy \\\\ = \iint_{\mathrm{int}(C)}\frac{\partial(4x+y)}{\partial x} - \frac{\partial(\sin(x)+9y)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = -5\iint_{\mathrm{int}(C)}\mathrm dx\,\mathrm dy[/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]\displaystyle -5\iint_{\mathrm{int}(C)}\mathrm dx\,\mathrm dy = -\frac52\times(12+4)\times4=-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]\displaystyle\int_{DA}y\,\mathrm dy = \int_{12}^0y\,\mathrm dy = -\int_0^{12}y\,\mathrm dy = -72[/tex]
Then
[tex]\displaystyle \int_{ABCD} (\sin(x)+9y)\,\mathrm dx + (4x+y)\,\mathrm dy = -160 - (-72) = \boxed{-88}[/tex]
