My Calculus book says:

"We know that the area under a curve $y=F(x)$ from $a$ to $b$ is $A=\int_a^b F(x) dx$, where $F(x)\geqslant0$. If the curve is traced out once by parametric equations $x=f(t)$ and $y=g(t)$, $\alpha \leqslant t \leqslant \beta$, then we can calculate an area formula by using the Substitution Rule for Definite Integrals as follows:

$A=\int_a^b y dx =\int_\alpha^\beta g(t)f'(t)dt\,\,\,\,\,\,\,\,\,\,$[or $\int_\beta^\alpha g(t)f'(t)dt$] ".

Also, it never says that if we have $dx/dt=f'(t)$ we can split the ratio of differentials and obtain $dx=f'(t)dt$.

Could someone deduce the formula of the area under a parametric curve without the use of this mysterious rule of "splitting" the ratio of differentials?

The book deduced the Substituition Rule without doing the split, but said that we could use the split as some non-proved equation to remember how to use this rule. But it is not helpful to me to understand how Substituition Rule was used to obtain the area under parametric curve without using this "split trick".


Ok. thank you gimusi. Your notation gave me an idea, and I think I deduced the formula without the use of equations with differentials like $dx=f'(t)dt$, which is what i was looking for. If someone is reading this and think it is right, please let me know.

So, let me try...

Given a parametric curve with $y=f(t), x=g(t),$ and assuming that $y$ could be writen in terms of x, in other words, $y=F(x)$, we have these equalities:


Now, since the Substitution Rule give us the following equation:

$\int f(g(x))g'(x)dx=\int f(u)du\,\,\,\,\,\,\, with\,\,\,\,\,\,\,u=g(x)$

(which can be deduced without the use of equations with differentials like $du=g'(x)dx$), we could rewrite the rule by just changing some letters and have:

$\int F(g(t))g'(t)dt=\int F(x)dx\,\,\,\,\,\,\, with\,\,\,\,\,\,\,x=g(t)$

then, using the case of the parametric curve above, that is, $F(g(t))=f(t)$ and $F(x)=y$, we have:

$\int f(t)g'(t)dt=\int ydx$

That is it. It is enough to satisfy my curiosity, although I think that there is some more steps to prove the relationship between the definite integrals that I firstly asked, which is:

$A=∫^b_aydx=∫^β_αg(t)f′(t)dt\,\,\,\,\,\,\,\,\,[or ∫^α_βg(t)f′(t)dt]$

  • $\begingroup$ Yes that’s right, for the latter doubts refer to some calculus book with an explanation of the substitution rule for integrals. $\endgroup$ – user Jul 31 '18 at 19:22

It is a simple change of varible for integrals, indeed we have


and we set

  • $x=f(t)\implies dx=f'(t)dt$
  • $y=F(x(t))=g(t)$

and then

$$A=\int_a^b F(x) dx=\int_{\alpha}^{\beta} g(t)f'(t) dt$$

If such interpretation don't satisfy your curiosity, for a formal proof you should refer to the theory for the change of variable for integrals.

  • $\begingroup$ It should be mandatory that the person who puts a downvote explains their reasons. $\endgroup$ – Piquito Jul 31 '18 at 18:48
  • $\begingroup$ @Piquito It should be very useful indeed also for me in order to learn something more. $\endgroup$ – user Jul 31 '18 at 19:10
  • $\begingroup$ Apparently, Gimusi, we can "learn" a lot from certain citizens. $\endgroup$ – Piquito Jul 31 '18 at 20:51
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    $\begingroup$ Gimusi, I asked for the proof without the use of "split-trick". But your notation helped me to think how could it be done. Thank you. $\endgroup$ – Guilherme Correa Teixeira Aug 1 '18 at 20:09
  • $\begingroup$ @GuilhermeCorreaTeixeira You are welcome! Bye $\endgroup$ – user Aug 1 '18 at 20:10

HINT.- $ydx=y\cdot\dfrac{dx}{dt}\cdot dt$ and $\dfrac{dx}{dt}=f'(t)$


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