[proofplan] The integral of a 1-form along a curve is defined via pullback to the parameter interval $[0,1]$. Writing $\alpha = df$ and using the naturality identity $\gamma^* df = d(f \circ \gamma)$, the pulled-back form becomes the differential of the scalar function $f \circ \gamma: [0,1] \to \mathbb{R}$. The integral then collapses to a one-dimensional integral of a total derivative, which the [Fundamental Theorem of Calculus](/theorems/632) evaluates as $f(\gamma(1)) - f(\gamma(0))$. The closedness hypothesis $\gamma(1) = \gamma(0)$ kills this boundary term. [/proofplan] [step:Unfold the definition of the line integral as a pullback integral] By the [definition of integration of differential forms](/theorems/1529) along a smooth curve, the line integral of $\alpha$ along $\gamma$ is given by pulling $\alpha$ back to the parameter interval and integrating: \begin{align*} \int_\gamma \alpha := \int_{[0,1]} \gamma^* \alpha, \end{align*} where $\gamma^* \alpha \in \Omega^1([0,1])$ is the pullback of $\alpha$ under the smooth map $\gamma: [0,1] \to M$. Concretely, for each $t \in [0,1]$ the value of $\gamma^*\alpha$ at $t$ is the linear functional on $T_t [0,1] \cong \mathbb{R}$ defined by \begin{align*} (\gamma^*\alpha)_t: T_t[0,1] &\to \mathbb{R}, \\ v &\mapsto \alpha_{\gamma(t)}(d\gamma_t(v)). \end{align*} [guided] We must reduce the manifold-level object $\int_\gamma \alpha$ to a concrete one-dimensional integral. By the [definition of integration of differential forms](/theorems/1529) along a smooth curve $\gamma: [0,1] \to M$, the line integral is **defined** as \begin{align*} \int_\gamma \alpha := \int_{[0,1]} \gamma^* \alpha, \end{align*} where $\gamma^* \alpha$ is the pullback. Why does this definition make sense? The form $\alpha \in \Omega^1(M)$ lives on $M$, but $[0,1]$ is the natural place to integrate — it carries the Lebesgue measure $\mathcal{L}^1$, and a 1-form on $[0,1]$ is exactly the kind of object that can be integrated against $\mathcal{L}^1$. The pullback $\gamma^* \alpha$ is the 1-form on $[0,1]$ produced by transporting $\alpha$ via $\gamma$: at each $t$, it acts on a tangent vector $v \in T_t[0,1]$ by first pushing $v$ forward to $T_{\gamma(t)} M$ via the differential $d\gamma_t$, then evaluating $\alpha$ there. Explicitly, \begin{align*} (\gamma^*\alpha)_t: T_t[0,1] &\to \mathbb{R}, \\ v &\mapsto \alpha_{\gamma(t)}(d\gamma_t(v)). \end{align*} This is the standard pullback construction; it depends only on the smoothness of $\gamma$ and the smoothness of $\alpha$. [/guided] [/step] [step:Apply the exactness hypothesis $\alpha = df$ and use $\gamma^* d = d \gamma^*$] By the exactness hypothesis, fix $f \in C^\infty(M)$ with $\alpha = df$. The [exterior derivative](/theorems/1525) is natural with respect to pullback by smooth maps: for any smooth map $F: N \to M$ and any $g \in C^\infty(M) = \Omega^0(M)$, we have $F^* (dg) = d(F^* g) = d(g \circ F)$. Applying this with $F = \gamma$ and $g = f$, \begin{align*} \gamma^* \alpha = \gamma^* (df) = d(f \circ \gamma). \end{align*} The composite $f \circ \gamma: [0,1] \to \mathbb{R}$ is smooth because $\gamma$ and $f$ are smooth. Identifying $\Omega^1([0,1])$ with $C^\infty([0,1])\, dt$ via the global coordinate $t$, we obtain \begin{align*} \gamma^* \alpha = (f \circ \gamma)'(t)\, dt, \end{align*} where $(f \circ \gamma)'(t)$ denotes the ordinary one-dimensional derivative. [guided] The exactness hypothesis gives a function $f \in C^\infty(M)$ with $\alpha = df$. We want to push this structure through the pullback. The key fact is the **naturality of the [exterior derivative](/theorems/1525)**: for any smooth map $F: N \to M$ and any smooth differential form $\omega$ on $M$, \begin{align*} F^*(d\omega) = d(F^* \omega). \end{align*} We use this in degree zero, where forms are just smooth functions and pullback is precomposition: $F^* g = g \circ F$ for $g \in C^\infty(M) = \Omega^0(M)$. So with $F = \gamma$ and $\omega = f$, \begin{align*} \gamma^* \alpha = \gamma^* (df) = d(\gamma^* f) = d(f \circ \gamma). \end{align*} Why is this useful? Because $f \circ \gamma$ is a smooth real-valued function on the one-dimensional manifold $[0,1]$, where the [exterior derivative](/theorems/1525) reduces to the ordinary single-variable derivative times $dt$. The interval $[0,1]$ has a global coordinate $t$, so every 1-form on it can be written $h(t)\, dt$ for a unique $h \in C^\infty([0,1])$. For the function $f \circ \gamma$, the [exterior derivative](/theorems/1525) is \begin{align*} d(f \circ \gamma) = (f \circ \gamma)'(t)\, dt, \end{align*} where $(f \circ \gamma)'(t)$ is the ordinary derivative from one-variable calculus. Combining, $\gamma^* \alpha = (f \circ \gamma)'(t)\, dt$. [/guided] [/step] [step:Reduce to a one-dimensional integral and apply the Fundamental Theorem of Calculus] Substituting into the definition from Step 1, \begin{align*} \int_\gamma \alpha = \int_{[0,1]} \gamma^* \alpha = \int_0^1 (f \circ \gamma)'(t)\, d\mathcal{L}^1(t). \end{align*} The function $f \circ \gamma: [0,1] \to \mathbb{R}$ is smooth and hence continuously differentiable on $[0,1]$. The [Fundamental Theorem of Calculus](/theorems/632) applies to the continuously differentiable function $f \circ \gamma$ and yields \begin{align*} \int_0^1 (f \circ \gamma)'(t)\, d\mathcal{L}^1(t) = (f \circ \gamma)(1) - (f \circ \gamma)(0) = f(\gamma(1)) - f(\gamma(0)). \end{align*} [guided] We now have an honest one-variable integral to evaluate. Substituting the expression $\gamma^* \alpha = (f \circ \gamma)'(t)\, dt$ from Step 2 into the definition of the line integral from Step 1, \begin{align*} \int_\gamma \alpha = \int_{[0,1]} \gamma^* \alpha = \int_0^1 (f \circ \gamma)'(t)\, d\mathcal{L}^1(t). \end{align*} We want to apply the [Fundamental Theorem of Calculus](/theorems/632), which requires that the integrand be the derivative of a continuously differentiable function on $[0,1]$. We verify this hypothesis: $\gamma: [0,1] \to M$ is smooth and $f: M \to \mathbb{R}$ is smooth, so the composite $f \circ \gamma: [0,1] \to \mathbb{R}$ is smooth, and in particular $C^1$. Hence the FTC applies and yields \begin{align*} \int_0^1 (f \circ \gamma)'(t)\, d\mathcal{L}^1(t) = (f \circ \gamma)(1) - (f \circ \gamma)(0) = f(\gamma(1)) - f(\gamma(0)). \end{align*} The line integral is now reduced to a difference of two [real numbers](/page/Real%20Numbers) — the values of $f$ at the two endpoints of $\gamma$. [/guided] [/step] [step:Use the closedness of $\gamma$ to conclude] The hypothesis that $\gamma$ is a closed curve means $\gamma(1) = \gamma(0)$. Hence $f(\gamma(1)) = f(\gamma(0))$, and combining with Step 3, \begin{align*} \int_\gamma \alpha = f(\gamma(1)) - f(\gamma(0)) = 0. \end{align*} This is the desired conclusion. [/step]