VC4.3
From Exampleproblems
Since the integration is performed in xy-plane, we have 
The path of integration C consists of the straight line OA and AB. Now, we have
![\int_{C} F\cdot\,dr=\int_{C} [(2x+y)i+(3y-x)j]\cdot(dx i+dy j)\,](/wiki/images/math/e/a/7/ea7e95f07c7509b8733542844c1d83e6.png)
=![\int_{C} [(2x+y)dx+(3y-x)dy]\,](/wiki/images/math/1/c/4/1c45ccc03cfd0c102e285a518567664b.png)
--(1)
The equation of the straight line AB is 
i.e 
Therefore,on OA,y=0,dy=0 and x varies from 0 to 2, and on AB,
and x varies from 2 to 3. Then from the equation (1),
![\int_{C} F\cdot\,dr=\int_{0}^{2} 2x\,dx+\int_{2}^{3}[(2x+2x-4)dx+(6x-12-x)\cdot\,2dx]\,](/wiki/images/math/1/8/3/183a831cc4fcd12ed4c7f9238fc6aa5d.png)
=![[x^2]_{0}^{2}+\int_{2}^{3} (14x-28)dx=4+14\int_{2}^{3}(x-2)dx=4+14[\frac{(x-2)^{2}}{2}]_{2}^{3}=4+7=11\,](/wiki/images/math/e/2/9/e2977147c69bb0cb69a9a084d39f00c0.png)
