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In order to make the question simpler, I won't give a specific example, but rather a general one:

Suppose I have a DEFINITE integral with boundaries $a$ and $b$ and I use partial integration to get this casual form:

$$\int_a^b udv = u v - \int vdu.$$

As I have boundaries $a$ and $b$, in the end, I have to compute them for the left side( $u*v$ )as well as for the right side ($\int vdu$).

But what if I use substitution when computing $\int vdu$? That should make the boundaries change, according to what is being substituted. Do the boundaries change then for the left side as well or do no changes happen at all? After comparing my solution to a specific task with the solution of an online integral calculator, it seems that no boundaries should change at all. Is that really how it works and why?

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  • $\begingroup$ You should give the example where the book answer is different than yours... $\endgroup$ – Isham Oct 9 '17 at 17:39
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Formally you have $$ \int_{x=a}^{x=b} u(x) dv(x) = u(b)v(b) - u(a)v(a) - \int_{x=a}^{x=b} v(x) du(x) $$ and if the RHS integral needs to be evaluated by substitution, you can do it in the usual way, changing the boundaries on it.

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  • $\begingroup$ That's exactly what I did, but the symbolab online calculator gives a different answer, as if no boundaries changed at all. It could be wrong though. Thanks for clarifying! $\endgroup$ – Infecto Oct 9 '17 at 15:31
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The boundaries for $[uv]|_a^b$ do not change at all.

$\int v\,du$ is in the end just a number. You can calculate it however you please.

If you use substitution in $y=f(x)$ to change the calculation of that integral, you indeed have to use $f(a)$ and $f(b)$ as the limits in the new (presumedly easier) integral.

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  • $\begingroup$ Is there a formatting error here? Did you mean $\int$? $\endgroup$ – John Doe Oct 9 '17 at 15:33

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