A reference from “introduction to calculus and analysis II/1” p. 89:
$u=f(x,y,z)$, we are able now to define the integral $\int L$ of the linear differential form $$L=A(x,y,z)dx+B(x,y,z)dy+C(x,y,z)dz$$ over a simple oriented arc ${\Gamma}^{*}$. We assume that the coefficients $A,B,C$ of $L$ are continuous in a neighborhood of ${\Gamma}^{*}$. We make the further assumption that the arc ${\Gamma}^{*}$ not only is continuous but sectionally smooth, that is, that it can be represented parameterically by functions $$x=\varphi(t),\;y=\psi(t),\;z=\chi(t);\;a\leq t \leq b, \tag{53}$$ which are sectionally smooth.
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$$\int_{{\Gamma}^{*}}L=\varepsilon\int_{a}^{b}\left(A\:\frac{dx}{dt}+B\:\frac{dy}{dt}+C\:\frac{dz}{dt}\right) dt \tag{56}$$ Here the integrand is the function of the single variable $t$ obtained by substituting for the arguments $x, y, z$ of $A, B, C$ their expressions (53); moreover, $\varepsilon=1$ when ${\Gamma}^{*}$ is oriented positively with respect to $t$ and $\varepsilon=-1$ when oriented negatively.
A reference from “introduction to calculus and analysis II/1” p. 104-p.105:
If the coefficients of the differential form $L=A\;dx + B\;dy + C\;dz$ have continuous first derivatives in a simply connected set $R$ and satisfy the integrability conditions $$B_z - C_y=0, \:C_x-A_z=0,\:A_y-B_x=0,\tag{75a}$$ then $L$ is the total differential of a function $u$ defined in $R$: $$A=u_x,\;B=u_y,\;C=u_z. \tag{75b} $$ For the proof, it is sufficient to show that the integral of $L$ extended over any simple polygonal arc in $R$ with initial point $P’$ and final point $P’’$ has a value that depends only on $P’$ and $P’’$. We represent the two oriented arcs ${C_0}^{*}$ and ${C_1}^{*}$ parametrically by, respectively, $$x=\varphi_{0}(t),\;y=\psi _{0}(t),\;z=\chi _{0}(t),\;0\leq t \leq 1, \tag{76a} $$ and $$x=\varphi_{1}(t),\;y=\psi _{1}(t),\;z=\chi _{1}(t),\;0\leq t \leq 1, \tag{76b} $$ with $t=0$ yielding $P’$ and $t=1$ yielding $P’’$. Using the simple connectivity of $R$, we can “imbed” the paths(75a,b) into a continuous family $$x=\varphi(t,\lambda),\;y=\psi(t ,\lambda),\;z=\chi(t ,\lambda) \tag{76c} $$ reducing to(76a, b) for $\lambda=0,1$ and to $P’$, $P’’$ for $t=0,1$. We have by formula (56), p. 89. $$\int_{{C_1}^{*}}L-\int_{{C_0}^{*}}L \\ =\int_{0}^{1}\left[(Ax_t+By_t+Cz_t)|_{\lambda=1}-(Ax_t+By_t+Cz_t)|_{\lambda=0}\right] dt \tag{76d} $$ where $x,y,z$ are the functions of $t,\lambda$ given by (76c). We assume, to begin with, that those functions have continuous first derivatives with respect to $t,\lambda$ and a continuous mixed second derivative for $0\leq t \leq 1, 0\leq \lambda \leq 1$. Then by (76d) $$\int_{{C_1}^{*}}L-\int_{{C_0}^{*}}L =\int_{0}^{1}dt \int_{0}^{1}(Ax_t+By_t+Cz_t)_{\lambda} \; d\lambda \tag{76e} $$ Now using the chain rule of differentiation and the integrability conditions (76a), we have the identity $$(Ax_t+By_t+Cz_t)_{\lambda} \\ = Ax_{\lambda t} + By_{\lambda t} + Cz_{\lambda t} + A_{x}x_{\lambda}x_{t} + A_{y}y_{\lambda}x_{t}+ A_{z}z_{\lambda}x_{t} \\ + B_{x}x_{\lambda}y_{t} + B_{y}y_{\lambda}y_{t} + B_{z}z_{\lambda}y_{t} \\ + C_{x}x_{\lambda}z_{t} + C_{y}y_{\lambda}z_{t} + C_{z}z_{\lambda}z_{t} \\ = (Ax_{\lambda}+By_{\lambda}+Cz_{\lambda})_{t} \tag{77} $$ ///////////////////////////////////////////////////////////////////////////////////////
The explanation for equation (76e):
Fix $x$, define $v(\lambda)= Ax_t+By_t+Cz_t = A\varphi_{t}(t,\lambda)+B\psi_{t}(t,\lambda)+C\chi_{t}(t,\lambda)\\ v’(\lambda)=\frac{\partial}{\partial \lambda}(Ax_t+By_t+Cz_t )=(Ax_t+By_t+Cz_t )_{\lambda}$, applying the formula $$\int_{\alpha}^{x}f’(x)dx=f(x)-f(\alpha)\implies$$ $$\int_{0}^{1}v’(\lambda)\;d\lambda= \int_{0}^{1} (Ax_t+By_t+Cz_t )_{\lambda} \;d\lambda =v(1)-v(0)\\ (Ax_t+By_t+Cz_t)|_{\lambda=1}-(Ax_t+By_t+Cz_t)|_{\lambda=0} \implies$$ $$\int_{{C_1}^{*}}L-\int_{{C_0}^{*}}L = \\ \int_{0}^{1}\left[(Ax_t+By_t+Cz_t)|_{\lambda=1}-(Ax_t+By_t+Cz_t)|_{\lambda=0}\right] dt \\ = \int_{0}^{1}\left[ \int_{0}^{1} (Ax_t+By_t+Cz_t )_{\lambda} \;d\lambda \right] dt \\ = \int_{0}^{1}dt \int_{0}^{1}(Ax_t+By_t+Cz_t)_{\lambda} \; d\lambda $$
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Please explain the identity (77), thanks.