Verified answer

C₁ = Line from A(0,0) to B(1,1)

x = t, y = t, t = 0 to 1

dx = dt, dy = dt

∫c₁ (sinx + 3y) dx + (2x + y) dy

= ∫ [0 to 1] (sin t + 3t) dt + (2t + t) dt

= ∫ [0 to 1] (sin t + 6t) dt

= −cos t + 3t² | [0 to 1]

= 4 − cos 1

C₂ = Line from B(1,1) to C(1,2)

x = 1, y = 1+t, t = 0 to 1

dx = 0, dy = dt

∫c₂ (sinx + 3y) dx + (2x + y) dy

= ∫ [0 to 1] (sin 1 + 3(1+t)) 0 + (2 + (1+t)) dt

= ∫ [0 to 1] (3 + t) dt

= 3t + t²/2 | [0 to 1]

= 7/2

C₃ = Line from C(1,2) to D(0,3)

x = 1−t, y = 2+t, t = 0 to 1

dx = −dt, dy = dt

∫c₃ (sinx + 3y) dx + (2x + y) dy

= ∫ [0 to 1] (sin(1−t) + 3(2+t)) (−dt) + (2(1−t) + (2+t)) dt

= ∫ [0 to 1] (−sin(1−t) − 2 − 4t) dt

= (−cos(1−t) − 2t − 2t²) | [0 to 1]

= cos 1 − 5

∫c (sinx + 3y) dx + (2x + y) dy = (4 − cos 1) + 7/2 + (cos 1 − 5) = 5/2

——————————————————————————————

Alternate method using Green's Method.

We usually Green's Method for a closed path.

However, in the case of this non-closed path, we can use Green's Method, then subtract along path from D to A.

First we make sure that closed curve C* = ABCD has positive orientation (counter-clockwise) --- it does.

Region enclosed by path is bounded above by y = 3−x, below by y = x, for x between 0 and 1

Green's theorem: ∫c P dx + Q dy = ∫∫d (∂Q/∂x − ∂P/∂y) dA

Q = 2x + y ----> ∂Q/∂x = 2

P = sinx + 3y ---> ∂P/∂y = 3

∫c* (sinx + 3y) dx + (2x + y) dy

= ∫ [0 to 1] ∫ [x to 3−x] (2−3) dydx

= ∫ [0 to 1] −y | [x to 3−x] dx

= ∫ [0 to 1] (2x−3) dx

= (x²−3x) | [0 to 1]

= −2

C₄ = Line from D(0,3) to A(0,0)

x = 0, y = 3−3t, t = 0 to 1

dx = 0, dy = −3dt

∫c₄ (sinx + 3y) dx + (2x + y) dy

= ∫ [0 to 1] (0 + 3(3−3t)) 0 + (0 + (3−3t)) * −3dt

= ∫ [0 to 1] (9t − 9) dt

= (9t²/2 − 9t) | [0 to 1]

= −9/2

∫c* (sinx + 3y) dx + (2x + y) dy

= ∫c (sinx + 3y) dx + (2x + y) dy + ∫c₄ (sinx + 3y) dx + (2x + y) dy

∫c (sinx + 3y) dx + (2x + y) dy

= ∫c* (sinx + 3y) dx + (2x + y) dy − ∫c₄ (sinx + 3y) dx + (2x + y) dy

= −2 − (−9/2)

= 5/2