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I know that there's a fundamental theorem for line integrals. That is, suppose $C$ is a smooth curve given by $r(t)$, $a \leq t \leq b$ and suppose $\nabla f$ is continuous on $C$. Then $$\int_{C}\nabla f\cdot dr = f(r(b)) - f(r(a)).$$ Is there something similar for integrals of the form $$\int_{C}f(s)\, ds = \int_{a}^{b}f(x(t), y(t))\sqrt{x'(t)^{2} + y'(t)^{2}}\, dt?$$

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Isn't this just the normal fundamental theorem? Ultimately you're going to get a 1D integral out of that. – Robert Mastragostino Oct 4 '12 at 19:21

There is an inequality that can easily be proved using Schwarz' inequality:

$$\bigl|f(r(b))-f(r(a))\bigr| \leq\int_a^b \bigl|\nabla f(r(t))\bigr|\ \bigl|r'(t)\bigr|\ dt = \int_C |\nabla f|\ ds\ .$$

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