Evaluate I= int(sinx+4y)dx + (4x+y) dy for the nonclosed path ABCD

Question

A = (0,0)

B = (4,4)

C = (4,8)

D = (0,12)

Answer 1

Integral along AB(y=x) (x=0 to x=4), dy=dx + Integral [along BC(x=4, dx=0, from y=4 to y=8] +Integral along CD ( y=-x+12, dy = -dx, from x=4 to x=0] + Integral along DA(x=0), dx =0 from y=12 to y=0]

1. Int. along AB =Int.[sinx + 4x +4x + 5x]dx = Int[sinx +13x]dx(x= 0to 4) = [-cosx + (13/2)x^2] = [(-Cos4 + 1) +(104-0)] =-cos 4 + 105
2. Int along BC = Int [ 16+5y]dy (from y = 4 to y=8) = (16y + 2.5y^2 )= 64+120 = 184
3. Int along CD = Int.[ (sinx + 4x) dx- {-4x-5(-x+12)}]dx = [-cosx+2.5x^2-60x] (from x=4 to x=0]

cos 4x -199

4.Int.along DA(x=0, dx=0] = Int. [5ydy (from y= 12 to y=0) = -360

Int 1 + Int 2 + Int 3 + Int 4 = 105+184-199-360 = -270

=

Answer 2

I = ∫₀⁴(sinx + 9x)dx + ∫₄⁸(16 + y)dy + ∫₄⁰(sinx + 36 - 7x)dx = (- cosx + 9x²/2)₀⁴ + (16y + y²/2)₄⁸ + (- cosx + 36x - 7x²/2)₄⁰ = - cos4 + 73 + 128 + 32 - 64 - 8 - 1 + cos4 - 144 + 56 = 72.

AB: y = x; dy = dx; 0 ≤ x ≤ 4.

BC: x = 4; dx = 0; 4 ≤ y ≤ 8

CD ; y = 12 - x, dy = - dx; x from 4 to 0