Answer:
67/4
Step-by-step explanation:
Let C, D be the line segments that join (-1,0) with (0,8) and (0,8) with (2,8) respectively.
The work required by the force field F(x,y)=(x,y) is the curve integral
[tex]\large \int_{C+D}^{} F=\int_{C}^{}F+\int_{D}^{}F [/tex]
the line segments that joins (-1,0) with (0,8) can be parameterized as
r(t) = t(0,8)+(1-t)(-1,0)=(t-1,8t) with 0≤ t≤ 1
So
[tex]\large \int_{C}^{}F=\int_{0}^{1}F(r(t))\bullet r'(t)dt=\int_{0}^{1}(t-1,8t)\bullet (1,8)dt=\int_{0}^{1}(t-1+64t)dt=\\\\=\int_{0}^{1}(65t-1)dt=65\int_{0}^{1}tdt-\int_{0}^{1}dt=\frac{65}{2}-1=\frac{63}{2} [/tex]
the line segments that joins (0,8) with (2,8) can be parameterized as
r(t) = t(2,8)+(1-t)(0,8)=(2t,8) with 0≤ t≤ 1
hence
[tex]\large \int_{D}^{}F=\int_{0}^{1}F(r(t))\bullet r'(t)dt=\int_{0}^{1}(2t,8)\bullet (2,0)dt=\int_{0}^{1}4tdt=4\int_{0}^{1}tdt=2 [/tex]
and we have the work required by F to move a particle alon the path is
[tex]\large \int_{C}^{}F+\int_{D}^{}F=\frac{63}{2}+2=\frac{67}{4}[/tex]
