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If $ C $ is a smooth curve given by a vector function $ \textbf{r}(t) $, $ a \leqslant t \leqslant b $, show that $$ \int_C \textbf{r} \cdot d \textbf{r} = \frac{1}{2} \bigl[ | \textbf{r}(b) |^2 - | \textbf{r}(a) |^2 \bigr] $$

$$

\begin{array}{c}{\int_{C} \mathbf{r} \cdot d \mathbf{r}=\int_{a}^{b} \frac{1}{d t}\left[x^{2}(t)+y^{2}(t)+z^{2}(t)\right] d t} \\ {\int_{C} \mathbf{r} \cdot d \mathbf{r}=\frac{1}{2}\left[x^{2}(t)+y^{2}(t)+z^{2}(t)\right]_{a}^{b}} \\ {\int_{C} \mathbf{r} \cdot d \mathbf{r}=\frac{1}{2}\left[x^{2}(b)+y^{2}(b)+z^{2}(b)-x^{2}(a)-y^{2}(a)-z^{2}(a)\right]}\end{array}

$$

Vector Calculus

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Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

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