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  "body": "\n## Statement\n\nLet $U \\subseteq \\mathbb{R}^n$ be open, let $\\phi \\colon U \\to \\mathbb{R}$ be of class $C^1$, and let $\\gamma \\colon [a, b] \\to U$ be a piecewise $C^1$ curve.\nThen\n$$\\int_\\gamma \\nabla \\phi \\cdot d\\mathbf{r} = \\phi(\\gamma(b)) - \\phi(\\gamma(a)).$$\n\n## Remarks\n\nProof for a $C^1$ curve: by the [[chain rule along a curve|0A8TJ0VHC7]], $(\\phi \\circ \\gamma)'(t) = \\nabla \\phi(\\gamma(t)) \\cdot \\gamma'(t)$. Then the [[parametric formula|0W3W1VYWBP]] and the fundamental theorem of calculus give the result. For a piecewise $C^1$ curve, add the pieces.\nSo the line integral of a gradient field depends only on the end points. Around a closed curve it is $0$.\nExample: for $\\phi(x, y) = x^2 y$ and any path from $(0, 0)$ to $(1, 2)$, $\\int_\\gamma (2 x y, x^2) \\cdot d\\mathbf{r} = 2$.\nThis is the case of curves in the [[generalised Stokes theorem|0HPHKCGEQ0]]: the boundary of a curve consists of its two end points.\n\n## Explanations\n\n### L1 human\n\nThis is the fundamental theorem of calculus for curves.\nTo integrate the gradient of $\\phi$ along a path, you only need the values of $\\phi$ at the end and at the start.\n",
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