VC4.9
From Exampleproblems
Let C1 denote the curve made up of the straight lines AB,BC and CA. Since the integration is performed in the xy-plane,we have 
. Then the required integral around the triangle ABC is given by
![\int_{C_1}F\cdot\,dr=\int_{C_1} [(2x^2+y^2)i+(3y-4x)j]\cdot (dxi+dyj)\,](/wiki/images/math/d/4/c/d4cfc56f663487fbc4a5f6d2374d8517.png)
=![\int_{C_1} [(2x^2+y^2)dx+(3y-4x)dy]\,](/wiki/images/math/4/7/c/47c425f5314c8342a33aba123e4c150c.png)
=![\int_{AB}[(2x^2+y^2)dx+(3y-4x)dy]+\int_{BC}[(2x^2+y^2)dx+(3y-4x)dy]+\int_{CA}[(2x^2+y^2)dx+(3y-4x)dy]\,](/wiki/images/math/3/c/6/3c626f43e0bb1f2989c15552424ca30e.png)
Here on AB,y=0,dy=0 and x varies from 0 to 2.
On BC,x=2,dx=0 and y varies from 0 to 1.
Now,equation of the line CA is 
Therefore,on CA,
and y varies from 1 to 0. Using the above,
![\int_{C_1}F\cdot\,dr=\int_{0}^{2}2x^2dx+\int_{0}^{1}(3y-8)dy+\int_{0}^{1}[2(2y)^2+y^2]2dy+(3y-8ydy)\,](/wiki/images/math/2/5/2/25283580f2f42cb2defba2bacaa5ca07.png)
=![2[\frac{x^3}{3}]_{0}^{2}+[\frac{3y^2}{2}-8y]_{0}^{1}+\int_{1}^{0}(18y^2-5y)dy\,](/wiki/images/math/8/e/a/8ea3ee2918af67ed0c9528e919cd1b5e.png)
=![\frac{16}{3}+\frac{3}{2}-8+[6y^3-\frac{5y^2}{2}]_{1}^{0}=\frac{16}{3}+\frac{3}{2}-8-6+\frac{5}{2}=-\frac{14}{3}\,](/wiki/images/math/4/4/d/44d0f0c23a0c5f3ac5e5af3b8991582c.png)
