Let c be the curve which is the union of two line segments, the first going from (0, 0) to (1, -3) and the second going from (1, -3) to (2, 0).
computer the line integral:
(c→)∫ -2 dy + 1 dx.

(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

or

step-by-step explanation:

we need to solve for 'x' the following equation:  

we have that:

⇒

⇒ ⇒

factorizing, we get that:

  ⇒

then, the solutions are: or

nothing extraneous has been found!

i don't know, but that looks pretty bad. my school passed us unless we failed in all the subjects, but idk abt ur school's rules.

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