[proofplan] We apply the [Gauss-Green Theorem](/theorems/28) to the product $w := uv$, which relates the integral of $\partial_{x_i} w$ over $U$ to a boundary integral involving the outward unit normal. Expanding $\partial_{x_i} w$ via the product rule produces two interior terms. Rearranging isolates the desired identity: the integral of $(\partial_{x_i} u) \cdot v$ equals minus the integral of $u \cdot \partial_{x_i} v$ plus a boundary correction term. [/proofplan] [step:Define the product $w := uv$ and apply the Gauss-Green theorem] Fix $i \in \{1, \dots, n\}$. Define the function \begin{align*} w: \overline{U} &\to \mathbb{R} \\ x &\mapsto u(x) \, v(x). \end{align*} Since $u, v \in C^1(\overline{U})$, the product $w = uv \in C^1(\overline{U})$. Because $U \subseteq \mathbb{R}^n$ is a bounded open set with $C^1$ [boundary](/pages/1024), the hypotheses of the [Gauss-Green Theorem](/theorems/28) are satisfied. Applying it to $w$ in the $x_i$-direction yields \begin{align*} \int_U \frac{\partial w}{\partial x_i}(x) \, d\mathcal{L}^n(x) = \int_{\partial U} w(x) \, \nu^i(x) \, d\mathcal{H}^{n-1}(x). \end{align*} [guided] Fix $i \in \{1, \dots, n\}$. The strategy is to apply the [Gauss-Green Theorem](/theorems/28) not to $u$ or $v$ individually, but to their product. Why the product? Because integration by parts is fundamentally about redistributing a derivative from one factor to the other, and the product rule is what enables this redistribution. Define the function \begin{align*} w: \overline{U} &\to \mathbb{R} \\ x &\mapsto u(x) \, v(x). \end{align*} Since $u, v \in C^1(\overline{U})$, the product rule for differentiation ensures $w \in C^1(\overline{U})$. The [Gauss-Green Theorem](/theorems/28) requires that (i) $U \subseteq \mathbb{R}^n$ is a bounded open set with $C^1$ boundary, and (ii) the integrand belongs to $C^1(\overline{U})$. Both conditions are satisfied: condition (i) is a hypothesis of the present theorem, and condition (ii) holds because $w = uv \in C^1(\overline{U})$ as just established. Applying the Gauss-Green Theorem to $w$ in the $x_i$-direction gives \begin{align*} \int_U \frac{\partial w}{\partial x_i}(x) \, d\mathcal{L}^n(x) = \int_{\partial U} w(x) \, \nu^i(x) \, d\mathcal{H}^{n-1}(x). \end{align*} This identity converts a volume integral of a derivative into a boundary integral weighted by the $i$-th component of the outward unit normal $\nu(x) = (\nu^1(x), \dots, \nu^n(x))$. [/guided] [/step] [step:Expand $\partial_{x_i} w$ via the product rule] By the product rule for partial derivatives applied to $w = uv$, \begin{align*} \frac{\partial w}{\partial x_i}(x) = \frac{\partial u}{\partial x_i}(x) \, v(x) + u(x) \, \frac{\partial v}{\partial x_i}(x). \end{align*} Substituting this into the left-hand side of the Gauss-Green identity and writing $w = uv$ on the right-hand side gives \begin{align*} \int_U \frac{\partial u}{\partial x_i}(x) \, v(x) \, d\mathcal{L}^n(x) + \int_U u(x) \, \frac{\partial v}{\partial x_i}(x) \, d\mathcal{L}^n(x) = \int_{\partial U} u(x) \, v(x) \, \nu^i(x) \, d\mathcal{H}^{n-1}(x). \end{align*} Here we used the linearity of the [Lebesgue integral](/pages/1152) to split the left-hand side into two integrals. [guided] We now expand the derivative $\partial_{x_i} w$ using the product rule. Since $w(x) = u(x) \, v(x)$ and both $u, v \in C^1(\overline{U})$, the product rule for partial derivatives gives \begin{align*} \frac{\partial w}{\partial x_i}(x) = \frac{\partial u}{\partial x_i}(x) \, v(x) + u(x) \, \frac{\partial v}{\partial x_i}(x). \end{align*} Substituting this expansion into the left-hand side of the Gauss-Green identity from the previous step, and replacing $w(x) = u(x) \, v(x)$ on the right-hand side, we obtain \begin{align*} \int_U \left( \frac{\partial u}{\partial x_i}(x) \, v(x) + u(x) \, \frac{\partial v}{\partial x_i}(x) \right) d\mathcal{L}^n(x) = \int_{\partial U} u(x) \, v(x) \, \nu^i(x) \, d\mathcal{H}^{n-1}(x). \end{align*} By linearity of the Lebesgue integral, the left-hand side splits as \begin{align*} \int_U \frac{\partial u}{\partial x_i}(x) \, v(x) \, d\mathcal{L}^n(x) + \int_U u(x) \, \frac{\partial v}{\partial x_i}(x) \, d\mathcal{L}^n(x) = \int_{\partial U} u(x) \, v(x) \, \nu^i(x) \, d\mathcal{H}^{n-1}(x). \end{align*} Notice how the product rule has produced exactly the two interior terms that appear in the integration-by-parts formula. The derivative that originally acted on the product $w = uv$ has been distributed between $u$ and $v$. [/guided] [/step] [step:Rearrange to isolate the integration-by-parts identity] Subtracting $\int_U u(x) \, \frac{\partial v}{\partial x_i}(x) \, d\mathcal{L}^n(x)$ from both sides of the identity above yields \begin{align*} \int_U \frac{\partial u}{\partial x_i}(x) \, v(x) \, d\mathcal{L}^n(x) = -\int_U u(x) \, \frac{\partial v}{\partial x_i}(x) \, d\mathcal{L}^n(x) + \int_{\partial U} u(x) \, v(x) \, \nu^i(x) \, d\mathcal{H}^{n-1}(x). \end{align*} Since $i \in \{1, \dots, n\}$ was arbitrary, this holds for each coordinate direction, which is the desired identity. [guided] We now perform a simple algebraic rearrangement. The identity from the previous step reads \begin{align*} \int_U \frac{\partial u}{\partial x_i}(x) \, v(x) \, d\mathcal{L}^n(x) + \int_U u(x) \, \frac{\partial v}{\partial x_i}(x) \, d\mathcal{L}^n(x) = \int_{\partial U} u(x) \, v(x) \, \nu^i(x) \, d\mathcal{H}^{n-1}(x). \end{align*} We want the term $\int_U (\partial_{x_i} u) \, v \, d\mathcal{L}^n$ isolated on the left-hand side. Subtracting $\int_U u \, (\partial_{x_i} v) \, d\mathcal{L}^n(x)$ from both sides — which is justified because all three integrals are finite (as $u, v \in C^1(\overline{U})$ and $U$ is bounded, so all integrands are continuous on the compact set $\overline{U}$) — gives \begin{align*} \int_U \frac{\partial u}{\partial x_i}(x) \, v(x) \, d\mathcal{L}^n(x) = -\int_U u(x) \, \frac{\partial v}{\partial x_i}(x) \, d\mathcal{L}^n(x) + \int_{\partial U} u(x) \, v(x) \, \nu^i(x) \, d\mathcal{H}^{n-1}(x). \end{align*} Since $i \in \{1, \dots, n\}$ was chosen arbitrarily at the start, this identity holds for every coordinate direction $i = 1, \dots, n$. This completes the proof of the integration-by-parts formula. [/guided] [/step]