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I need help with this problem:

Let $f:S\subset\mathbb{R}^n\rightarrow\mathbb{R}$ be continuous on $S$, and let $\alpha:[a,b]\subset\mathbb{R}\rightarrow\mathbb{R}^n$ and $\beta:[c,d]\subset\mathbb{R}\rightarrow\mathbb{R}^n$ be piecewise $C^1$ (continuously differentiable) paths in $S$, such that $\alpha(b)=\beta(c)$. Prove the following integral formulae concerning the inverse path $\alpha^-$ and the product path $\alpha\beta$: $$\int_{\alpha^-}f\ ds=\int_{\alpha}f\ ds$$ $$\int_{\alpha\beta} f \ ds=\int_{\alpha}f \ ds\ + \int_{\beta}f \ ds$$

I don't know if I'm correct, but I think that $\alpha^-(t)=\alpha(-t)$, so $\alpha^-:[-b,-a]\rightarrow\mathbb{R}^n$. Using this I tried to rpove the first one like this: $$\int_{\alpha^-}f\ ds=\int_{-b}^{-a} f(\alpha^-(t))\Vert \alpha{^-}'(t)\Vert dt=\int_{-b}^{-a} f(\alpha(-t))\vert(-1)\vert\Vert\alpha(-t)\Vert dt$$ let $u=-t$, thus $du=-dt$ $$-\int_a^b-f(\alpha(u))\Vert\alpha'(u)\Vert du=\int_a^bf(\alpha(u))\Vert\alpha'(u)\Vert dt=\int_\alpha f\ ds$$ Am I correct? For the second one, I don't know how to prove it. I think that $\alpha\beta$ woudl be the path that goess from $a$ to $d$, right?

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The first one looks correct, for the second, you have to join the paths $\alpha, \beta$ into one parameterisation $\gamma$. One way to do this is by letting,

$\gamma : [a, b + d - c] \to \mathbb{R}^n$ with,

$$\gamma(t) = \begin{cases}\alpha(t), \ t \in [a,b] \\ \beta(t - b + c), \ t \in [b, b + d- c] \end{cases}$$

See here how I have combined the two paths into one interval from $a$ to $b + d -c$. I have not changed the paths at all here, so that $\gamma = \alpha \beta$. That means,

$$\int_\gamma f \ \mathrm{d}s = \int_a^{b+d-c} f(\gamma(t)) || \gamma'(t) || \ \mathrm{d}t = \int_a^b f(\gamma(t)) || \gamma'(t) || \ \mathrm{d}t + \int_b^{b+d-c} f(\gamma(t)) || \gamma'(t) || \ \mathrm{d}t = \dots$$

I'll let you finish the proof, at this point you have to use the definition of $\gamma$. You may also want to verify that $\gamma$ is in fact $C^1$ at every point except at $t = b$, which won't affect the integral since we only require $\gamma$ be piecewise $C^1$.

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  • $\begingroup$ I don't understand why the interval goes from $a$ to $b+d-c$. Also, why $gamma$ is not $C^1$ at $t=b$? $\endgroup$ – davidllerenav Apr 20 at 3:22

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