Scalar Line Integrals:Let x : [a, b] → Rnbe a C1path. f : Rn→ R, a scalar field.∆sk= length of kth segment of path=Rtktk−1||x′(t)||dt = ||x′(t∗∗k)||(tk− tk−1) = ||x′(t∗∗k)||∆tkfor some t∗∗k∈ [tk−1, tk]Rxf ds ∼ Σni=1f(x(t∗k))∆sk= Σni=1f(x(t∗k))||x′(t∗∗k)||∆tkThusRxf ds =Rbaf(x(t))||x′(t)||dtVector Line integrals:Let x : [a, b] → Rnbe a C1path. F : Rn→ Rn, a vectorfield.x′(t∗k) ∼∆xk∆tkRxF · ds ∼ Σni=1F (x(t∗k)) · ∆xk= Σni=1F (x(t∗k)) · x′(t∗k)∆tkThusRxF · ds =RbaF (x(t)) · x′(t)dtOther Formulations of Vector Line integrals:The tangent vector to x at t is T (t) =x′(t)||x′(t)||RxF · ds =RbaF (x(t)) · x′(t)dt =RbaF (x(t)) · x′(t)||x′(t)||||x′(t)||dt =RbaF (x(t)) · T (t)||x′(t)||dt =RxF (x(t)) · T (t)dsNoteRbaF (x(t)) · T (t)ds is a scalar line integral of the scalarfield F · T : Rn→ R over the path x.Note: F ·T is the tangential component of F along the path x.Another notation (differential form):For simplicity, we will work in R2, b ut the following general -izes to any dimension.Let x(t) = (x(t), y(t)). Let F (x, y) = (M (x, y), N(x, y))x = x(t), y = y(t). Also x′(t) =dxdt, y′(t) =dydtZxF · ds =ZbaF (x(t)) · x′(t)dt=Zba(M(x, y), N(x, y)) · (x′(t), y′(t))dt=ZbaM(x, y)x′(t)dt + N(x, y)y′(t)dt=ZxM(x, y)dx + N(x, y)dyDefinitions:A curve is the image of piecewise C1path x : [a, b] → Rn.A curve is simple if it has no self-intersections; that is, x is1:1 on the open interval (a, b)A path i s closed if x(a) = x(b)A curve is closed if x(a) = x(b)RxF · ds is called the circulation of f a l o ng x if x is a closedpath.A parametrization of a curve C is a path whose image is C.Normally we will require a parametrization of a cu rve to be1:1 where possible.A piecewise C1path y : [c, d] → Rnis a reparametrization ofa piecewise C1path x : [a, b] → Rnif there exists a bijectiveC1function u : [c, d] → [a, b] w here the inverse of u is also C1and y = x ◦ u (i.e., y(t) = x(u(t))).Note tha t either1.) u(a) = c and u(b) = d. In this case, we say th at y (and uare or ientation-preservin g OR2.) u(a) = d and u(b) = c. In this case, we say th at y (and uare or ientation-rever sing.Given piecewise C1path, x : [a, b] → Rn, the opposite path isxopp: [a, b] → Rnxopp= x(a + b − t)That is xoppis an orientation- r eversing reparametrization ofx where u[a, b] → [a, b], u(t) = a + b − t.Thm: Let x : [a, b] → Rnbe a piecewise C1path and lety : [c, d] → Rnbe a rep a rametrization of x. Thenif f : Rn→ R is continu o us, th enRyf ds =Rxf dsif F : Rn→ Rnis continuous, thenRyF · ds =RxF · ds if y is orientation -preserving.RyF · ds = −RxF · ds if y is orientation