Line integral?

Question

∫cy^3dx+x^3dy, where c is the right-angle curve from (-4,1) to (-4,-2), to (2, -2)

Answer 1

Line integral ∫(y^3dx+x^3dy) over C can be decomposed into sum of two line integrals over C1: x = -4 and C2: y = -2.

Hence the reqd. integral is
∫c1( ) + ∫c2( )
= ∫{y^3*0+(-4)^3dy} + ∫{(-2)^3dx+x^3*0}
= ∫(-4)^3dy + ∫(-2)^3dx
= -64∫dy - 8∫dx [first integral has limits: 1 to -2 and second one: -4 to 2]
= 192 - 48
= 144

Answer 2

Two steps:

1. From (-4,1) to (-4,-2), x = -4 = constant, dx = 0
∫cy^3dx+x^3dy
= ∫(-4)^3dy[y:1...-2] dx
= -64(-3)
= 192

2. From (-4,-2) to (2, -2), y = -2 = constant, dy = 0
∫cy^3dx+x^3dy
= ∫(-2)^3dy[x:-4...2] dx
= -8(6)
= -48

Combine the two steps:
∫cy^3dx+x^3dy
= 192-48
= 144