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I've always taken integration by substitution for granted but recently I've learned that differentials can't fully be treated as variables and that the process of integration by substitution is really just a shorthand. Now, writing something like $du=u'(x)dx$ during substitution feels weird and empty.

I understood the justification for $\int f(g(x))g'(x)\,dx = \int f(u)\,du$ (given $u = g(x)$) from this page, but how do I apply this process to prove $\int f(x)f'(x)\,dx = \int v\,dv$ (given $v=f(x)$)?

Sorry that I don't have any work to show for myself, it's because I don't even know how to approach this problem.

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3 Answers 3

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This follows from the fundamental theorem of calculus (FTC) and the chain rule.

By the chain rule $(\frac{f(x)^2}2)'=f(x)\cdot f'(x)$.

Then by (FTC) $\int f(x)\cdot f'(x)\operatorname{dx}= \frac{f(x)^2}2+C$.

To see the last part, write $g(x)=\frac{f(x)^2}2$. Then we have $\int g'(x)\operatorname{dx}=g(x)+C$.

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I'll flip the $f$ and $g$ around, i.e. lets say that we can assume $$ \int f(g(x))g'(x)\,dx = \int f(u)\,du$$ and we want to prove $$ \int g(x) g'(x)\, dx = \int u\, du $$ We can write this as $\int f(g(x))g'(x)\,dx $ using $f(x)=x$, so that the substitution $u=g(x)$ gives the result.

Alternatively, since $(g^2)' = 2gg'$,$$ \int g(x) g'(x)\, dx =\int \left(\frac{g^2}2\right)' \, dx = \frac{g^2}2 + C$$

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  • $\begingroup$ Exactly what I wanted, thank you $\endgroup$ Commented Oct 25, 2018 at 9:58
  • $\begingroup$ (there was a typo in my alternative solution, forgot to bring down the 2) $\endgroup$ Commented Oct 25, 2018 at 10:14
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$$I=\int f(x)f'(x)dx$$ let $u=f(x)$. Therefore $du=f'(x)dx$. This substitution gives $$I=\int udu$$ $$I=\frac{u^2}2+C$$ $$I=\frac{f(x)^2}{2}+C$$

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  • $\begingroup$ I think you mistook my question, the title was not my whole question. What I really wanted to know is how substitution works behind the scenes in proving the integral at hand. Writing "$du = f'(x)dx$" would just be a circular argument $\endgroup$ Commented Oct 25, 2018 at 19:23

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