[proofplan] By definition, the line integral $\int_\gamma \omega$ is the integral over $[a,b]$ of the pullback $1$-form $\gamma^*\omega$. We compute this pullback explicitly: $\gamma^*$ is $\mathbb{R}$-linear, commutes with multiplication by smooth functions via composition with $\gamma$, and acts on each coordinate $1$-form $dx_i$ by $\gamma^*(dx_i) = \gamma_i'(t)\, dt$. Summing the contributions yields a single-variable $1$-form on $[a,b]$ whose integral is the right-hand side of the claimed identity. [/proofplan]

[step:Unfold the definition of the line integral as an integral of the pullback]By definition, the line integral of $\omega$ along $\gamma$ is \begin{align*} \int_\gamma \omega \;:=\; \int_{[a,b]} \gamma^* \omega, \end{align*} where the pullback $\gamma^*\omega$ is the smooth $1$-form on the open interval $(a-\varepsilon, b+\varepsilon)$ (for some $\varepsilon > 0$ to which $\gamma$ extends smoothly, or interpreted on $[a,b]$ via one-sided limits at the endpoints) defined by \begin{align*} \gamma^*: \Omega^1(U) &\to \Omega^1([a,b]), \\ \omega &\mapsto \gamma^*\omega, \quad (\gamma^*\omega)_t(v) := \omega_{\gamma(t)}(d\gamma_t(v)) \text{ for } t \in [a,b],\, v \in T_t[a,b]. \end{align*} Since $\gamma^*\omega \in \Omega^1([a,b])$ is a smooth $1$-form on a one-dimensional manifold, it has the form $\gamma^*\omega = h(t)\, dt$ for a unique smooth function $h: [a,b] \to \mathbb{R}$, and the integral $\int_{[a,b]} \gamma^*\omega$ is by definition the Lebesgue (equivalently, Riemann) integral $\int_a^b h(t)\, d\mathcal{L}^1(t)$. It therefore suffices to identify $h$.[/step]

[guided]The line integral of a $1$-form along a curve is defined intrinsically by pulling the form back to the parameter interval and integrating there. To unpack this: 1. The curve is a smooth map $\gamma: [a,b] \to U$, and its differential at $t \in [a,b]$ is the [linear map](/page/Linear%20Map) $d\gamma_t: T_t[a,b] \to T_{\gamma(t)}U$. With the canonical identification $T_t[a,b] \cong \mathbb{R}$ via the basis vector $\partial_t$, we have $d\gamma_t(\partial_t) = \gamma'(t) \in T_{\gamma(t)}U \cong \mathbb{R}^n$. 2. The pullback $\gamma^*\omega \in \Omega^1([a,b])$ is defined by the universal recipe $(\gamma^*\omega)_t(v) := \omega_{\gamma(t)}(d\gamma_t(v))$. 3. Because $[a,b]$ is one-dimensional, every smooth $1$-form on it has the form $h(t)\, dt$ for a unique smooth $h$. The integral of such a form is the ordinary integral $\int_a^b h(t)\, d\mathcal{L}^1(t)$ — there is no orientation ambiguity beyond the standard orientation of $[a,b]$ from $a$ to $b$. So the entire problem reduces to: **what is the coefficient function $h(t)$ when $\gamma^*\omega$ is written as $h(t)\, dt$?** Once we have $h$, the conclusion is immediate by definition of the integral.[/guided]

[step:Compute the pullback of each coordinate $1$-form $dx_i$]Fix $i \in \{1, \dots, n\}$. We claim that \begin{align*} \gamma^*(dx_i) \;=\; \gamma_i'(t)\, dt \quad \text{in } \Omega^1([a,b]). \end{align*} To verify this, evaluate both sides on the standard basis vector $\partial_t \in T_t[a,b]$. By definition of the pullback, \begin{align*} (\gamma^*(dx_i))_t(\partial_t) \;=\; (dx_i)_{\gamma(t)}(d\gamma_t(\partial_t)) \;=\; (dx_i)_{\gamma(t)}(\gamma'(t)), \end{align*} where we used $d\gamma_t(\partial_t) = \gamma'(t) = \sum_{j=1}^n \gamma_j'(t)\, \partial_{x_j}|_{\gamma(t)} \in T_{\gamma(t)}\mathbb{R}^n$. Since $dx_i$ is the coordinate dual to $\partial_{x_j}$ — i.e., $(dx_i)(\partial_{x_j}) = \delta_{ij}$ — linearity gives \begin{align*} (dx_i)_{\gamma(t)}(\gamma'(t)) \;=\; \sum_{j=1}^n \gamma_j'(t)\, \delta_{ij} \;=\; \gamma_i'(t). \end{align*} Meanwhile $(\gamma_i'(t)\, dt)_t(\partial_t) = \gamma_i'(t)\, dt(\partial_t) = \gamma_i'(t)$. Since the two $1$-forms agree on the basis vector $\partial_t$ of the one-dimensional space $T_t[a,b]$ at every $t \in [a,b]$, they are equal as elements of $\Omega^1([a,b])$.[/step]

[guided]We want to compute $\gamma^*(dx_i)$. Since $[a,b]$ is one-dimensional, it is enough to evaluate the resulting $1$-form on a single basis vector — namely $\partial_t$ — at each point $t$. **Step 1: Apply the definition of pullback.** By definition, \begin{align*} (\gamma^*(dx_i))_t(\partial_t) \;=\; (dx_i)_{\gamma(t)}\bigl(d\gamma_t(\partial_t)\bigr). \end{align*} This is just the definition of pullback of a $1$-form: push the vector forward by $d\gamma_t$, then evaluate the form at the pushed-forward vector. **Step 2: Compute the pushforward $d\gamma_t(\partial_t)$.** In the chart on $[a,b]$ with coordinate $t$ and on $U$ with coordinates $x_1, \dots, x_n$, the Jacobian matrix of $\gamma$ is the column vector $(\gamma_1'(t), \dots, \gamma_n'(t))^\top$. Acting on $\partial_t$ (i.e., the column $(1)$ in the one-dimensional source) gives \begin{align*} d\gamma_t(\partial_t) \;=\; \sum_{j=1}^n \gamma_j'(t)\, \partial_{x_j}\big|_{\gamma(t)}. \end{align*} This is just the velocity vector $\gamma'(t)$ expressed in the coordinate basis. **Step 3: Apply $dx_i$ to the pushforward.** The defining property of the coordinate $1$-form $dx_i$ is $(dx_i)(\partial_{x_j}) = \delta_{ij}$ — it is the linear functional that reads off the $i$-th component in the coordinate basis. By linearity, \begin{align*} (dx_i)_{\gamma(t)}\!\left(\sum_{j=1}^n \gamma_j'(t)\, \partial_{x_j}\big|_{\gamma(t)}\right) \;=\; \sum_{j=1}^n \gamma_j'(t)\, \delta_{ij} \;=\; \gamma_i'(t). \end{align*} **Step 4: Compare to $\gamma_i'(t)\, dt$.** The $1$-form $\gamma_i'(t)\, dt$ on $[a,b]$, evaluated on $\partial_t$ at the point $t$, returns $\gamma_i'(t) \cdot dt(\partial_t) = \gamma_i'(t) \cdot 1 = \gamma_i'(t)$. So both $1$-forms have the same value on the basis vector $\partial_t$ at every $t \in [a,b]$. Since $T_t[a,b]$ is one-dimensional, agreement on $\partial_t$ forces agreement as $1$-forms, and we conclude $\gamma^*(dx_i) = \gamma_i'(t)\, dt$. (Equivalently, this calculation is the special case of the [Coordinate Formula for the Pullback of a Differential Form](/theorems/3570) applied to the smooth map $\gamma$ and the $1$-form $dx_i$.)[/guided]

[step:Assemble the pullback of $\omega$ using $\mathbb{R}$-linearity and the pullback-of-coefficient identity] The pullback operator $\gamma^*: \Omega^\bullet(U) \to \Omega^\bullet([a,b])$ is $\mathbb{R}$-linear and satisfies $\gamma^*(f \cdot \alpha) = (f \circ \gamma) \cdot \gamma^*\alpha$ for every smooth function $f \in C^\infty(U)$ and every smooth form $\alpha \in \Omega^\bullet(U)$ (these are the standard naturality properties of the pullback; see the [Coordinate Formula for the Pullback of a Differential Form](/theorems/3570)). Applying these to $\omega = \sum_i P_i\, dx_i$ and using the previous step, \begin{align*} \gamma^*\omega \;=\; \gamma^*\!\left(\sum_{i=1}^n P_i\, dx_i\right) \;=\; \sum_{i=1}^n (P_i \circ \gamma)\, \gamma^*(dx_i) \;=\; \sum_{i=1}^n P_i(\gamma(t))\, \gamma_i'(t)\, dt. \end{align*} Thus the coefficient function $h$ identified at the end of Step 1 is \begin{align*} h: [a,b] &\to \mathbb{R}, \\ t &\mapsto \sum_{i=1}^n P_i(\gamma(t))\, \gamma_i'(t), \end{align*} which is smooth as a finite sum of products of compositions of smooth functions with the smooth curve $\gamma$ and the smooth derivatives $\gamma_i'$. [/step]

[step:Integrate over $[a,b]$ to obtain the recovery formula] By the definition of the line integral recorded in Step 1 and the identification of $h$ in Step 3, \begin{align*} \int_\gamma \omega \;=\; \int_{[a,b]} \gamma^*\omega \;=\; \int_a^b h(t)\, d\mathcal{L}^1(t) \;=\; \int_a^b \sum_{i=1}^n P_i(\gamma(t))\, \gamma_i'(t)\, d\mathcal{L}^1(t). \end{align*} The interchange of the finite sum with the integral on the right-hand side is justified because the sum has finitely many terms and each summand is continuous (hence integrable) on the compact interval $[a,b]$. This establishes the claimed identity and completes the proof. [/step]