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Question Number 2163 by Filup last updated on 06/Nov/15
For: y=f(x) → x=g(y)  Therefore:   { ((x(t)=t)),((y(t)=f(t))) :}  let r(t)=⟨x(t), y(t)⟩    ∴∫rdt=⟨∫tdt, ∫f(t)dt⟩    Does:  ∫x(t)dt=∫g(y)dy  and  ∫y(t)dt=∫f(x)dx
$$\mathrm{For}:\:{y}={f}\left({x}\right)\:\rightarrow\:{x}={g}\left({y}\right) \\ $$$$\mathrm{Therefore}: \\ $$$$\begin{cases}{{x}\left({t}\right)={t}}\\{{y}\left({t}\right)={f}\left({t}\right)}\end{cases} \\ $$$$\mathrm{let}\:\boldsymbol{{r}}\left({t}\right)=\langle{x}\left({t}\right),\:{y}\left({t}\right)\rangle \\ $$$$ \\ $$$$\therefore\int\boldsymbol{{r}}{dt}=\langle\int{tdt},\:\int{f}\left({t}\right){dt}\rangle \\ $$$$ \\ $$$$\mathrm{Does}: \\ $$$$\int{x}\left({t}\right){dt}=\int{g}\left({y}\right){dy} \\ $$$$\mathrm{and} \\ $$$$\int{y}\left({t}\right){dt}=\int{f}\left({x}\right){dx} \\ $$
Commented by prakash jain last updated on 06/Nov/15
g(y)=x=t  y=f(x)=f(t)  dy= f ′(t)dt  ∫x(t)dt=∫tdt  ∫g(y)dy=∫tf ′(t)dt≠∫tdt
$${g}\left({y}\right)={x}={t} \\ $$$${y}={f}\left({x}\right)={f}\left({t}\right) \\ $$$${dy}=\:{f}\:'\left({t}\right){dt} \\ $$$$\int{x}\left({t}\right){dt}=\int{tdt} \\ $$$$\int{g}\left({y}\right){dy}=\int{tf}\:'\left({t}\right){dt}\neq\int{tdt} \\ $$
Answered by prakash jain last updated on 06/Nov/15
Second part   ∫y(t)dt=∫f(x)dx is correct since y(t)=f(t)  ∫f(t)dt=∫f(x)dx ∵x=t
$$\mathrm{Second}\:\mathrm{part}\: \\ $$$$\int{y}\left({t}\right){dt}=\int{f}\left({x}\right){dx}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{since}\:{y}\left({t}\right)={f}\left({t}\right) \\ $$$$\int{f}\left({t}\right){dt}=\int{f}\left({x}\right){dx}\:\because{x}={t} \\ $$
Commented by Filup last updated on 06/Nov/15
Thanks
$$\mathcal{T}{hanks} \\ $$

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