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My book claims the following: $$ x = f(t) $$ $$y = g(t)$$

then, by substitution rule $$ \int y \ dx = \int g(t)f'(t) \ dt$$

I cannot find a way to obtain this result. Could someone show all the steps to obtain it?

Lucas Alanis
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2 Answers2

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If $x = f(t)$, then $\frac{dx}{dt} = \frac{df(t)}{dt} = f'(t)$, so $dx = f'(t)dt$. Alternatively, one can immediately obtain $dx = f'(t)dt$ by chain rule.

The rest is simply substituting $g(t)$ where you see $y$ and $f'(t)dt$ where you see $dx$.

Alex Wertheim
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  • How can you multiply both sides of $\frac {dx}{dt} = f'(t)$ by dt? I thought it was just a notation for derivative. – Lucas Alanis May 09 '13 at 03:19
  • Perhaps this question (and answer by Arturo Magidin) might be of use. http://math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio/21209#21209 – Alex Wertheim May 09 '13 at 03:38
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Let's break this down. We have: $$ x = f(t) $$ $$y = g(t)$$ And we're looking to express in terms of those functions the following integral:$$ \int y \ dx$$ Let's substitute only $y$ to begin with. So the integral becomes:
$$\int g(t)dx$$

Now let's look at substituting the $dx$ and rearranging.

If $x=f(t)$, then $\frac{dx}{dt} = \frac{df(t)}{dt}.$
$i.e. \frac{dx}{dt} = f'(t).$
$\therefore {dx} = f'(t)dt.$

Now to complete the substitution, substitute $dx$ with our result: $$\int g(t)dx = \int g(t)f'(t)dt$$

xisk
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