Answer:
(C→)∫ -2 dy + 1 dx=2
Step-by-step explanation:
From Exercise we have :
(0, 0) to (1, -3) and (1, -3) to (2, 0).
(C→)∫ -2 dy + 1 dx
We will divide the given curve into two curves.
We find the parametric equation for the points (0,0) and (1,-3), we get
r(t)=(0,0)+t((1,-3)-(0,0))
r(t)=t(1,-3)
r(t)=(t,-3t) for 0≤t≤1.
We get
x=t ⇒ dx=dt
y=-3t ⇒ dy=-3dt
Now, we get
∫_{C_1} -2 dy + 1 dx=\int\limits^1_0 {-2·(-3)} \, dt +\int\limits^1_0 {1} \, dt =
=6\int\limits^1_0 {1} \, dt+\int\limits^1_0 {1} \, dt
=7\int\limits^1_0 {1} \, dt
=7· [t]^1_0
=7
We find the parametric equation for the points (1,-3) and (2,0), we get
r(t)=(1,-3)+t((2,0)-(1,-3))
r(t)=(1,-3)+t(1,3)
r(t)=(1,-3)+(t,3t)
r(t)=(1+t,-3+3t) for 0≤t≤1.
We get
x=1+t ⇒ dx=dt
y=-3+3t ⇒ dy=3dt
Now, we get
∫_{C_2} -2 dy + 1 dx=\int\limits^1_0 {-2·(3)} \, dt +\int\limits^1_0 {1} \, dt =
=-6\int\limits^1_0 {1} \, dt+\int\limits^1_0 {1} \, dt
=-5\int\limits^1_0 {1} \, dt
=-5 [t]^1_0
=-5
Therefore, we get
(C→)∫ -2 dy + 1 dx=∫_{C_1} -2 dy + 1 dx+∫_{C_2} -2 dy + 1 dx
(C→)∫ -2 dy + 1 dx=7-5
(C→)∫ -2 dy + 1 dx=2
