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When calculating arc-length in parametric equation, stewart's book showed me a way to alter the arc length formula:

to substitute the dy/sx with the chain rule version

I understand why this work and we are making a function of x into a function of t so we should change the definitive upper/lower bound and the change dx into dt according to the substitution rule. But I don't understand how the (dx/dt)dt in the right hand side come from. it is the result of the substitution rule but I just don't understand how the substitution rule give us this, please tell me how to perform this substitution and the reason behind it. I'm feeling like figuring out this question could level up my mathematics understanding dramatically, please help me out! thank you very much

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  • $\begingroup$ It is called "inverse substitution". Part of the substitution rule is that $dx$ has to change to $x'(t) \, dt$. Think of it as doing $u$-substitution backwards. When you do $u$-sub, you look for an expression like $u'(t) \, dt$ and just change it into $du$. Here you are just doing that in the other direction. $\endgroup$
    – Nick
    Dec 14, 2019 at 4:43

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The substitution rule comes from the Chain Rule. Concretely, you have $(f(g(x))'=f'(g(x))\,g'(x)$. Then $$\tag1 f(g(\beta))-f(g(\alpha))=\int_\alpha^\beta f'(g(x))\,g'(x)\,dx. $$ Taking $a=g(\alpha)$, $b=g(\beta)$, we also have $$\tag2 f(g(\beta))-f(g(\alpha))=f(b)-f(a)=\int_a^bf'(x)\,dx $$ Comparing $(1)$ and $(2)$, and renaming $f'$ as $f$, we get the Substitution Rule: $$\tag3 \int_a^bf(x)\,dx=\int_\alpha^\beta f(g(x))\,g'(x)\,dx, $$ where $g(\alpha)$, $b=g(\beta)$. The equality $(3)$ is precisely what happens in your picture. If you write $x=x(t)$, then $$ \int_a^bf(x)\,dx=\int_\alpha^\beta f(x(t))\,x'(t)\,dt. $$

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