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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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### Video Transcript

Let's try toe derive this again. We first right R t is given by given by this except the wild T It's truly anyth I mentioned, but for simplicity are right Should I mention if you wantto do in St Amon changes at another component The derivative should be ex private t y proud t So I will be this integration. This integration will be from a to B the dot product off them plus why off t y prompt e However what is this? This one If you observe that or you can just observe from the final from not that ex off t square If you take the derivative with respect with tea, you should get exactly X off t. Well, you have a chewing from ex prompt he by the general. So this is nothing but or or even just observing something by by U substitution. And so the first part below the antiterrorist tiff, Opie with respect to Teo bx and sub t square over too crosswise of t square over two from a to B and this is not on x X square, plus y square is just are absolutely off our square. So this is our square over too, from a to B And this one after you probably be in a you should guess this.

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Vector Calculus

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