# Talk:Integration by Substitution/Definite Integral

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In the theorem statement, should the symbol $t$ in the left-hand side integrand be replaced with $x$ to match the statement "where $x = \phi(u)$"? Also, for the substitution to be valid, shouldn't we also impose that $\phi$ be bijective? (Otherwise we might be "summing" over the same values of $x$ "multiple times" in the right hand side, right?) --Plammens (talk) 19:10, 1 March 2021 (UTC)

- Good call on $x \to t$, changed as appropriate as needed.

- Not sure I'm onside as regards $\phi$ being a bijection. Nothing in any of the sources I have saying anything about that. Care to check your own sources? Or find a counterexample? --prime mover (talk) 20:30, 1 March 2021 (UTC)

- You're right, I've looked at it again and, in fact, the proof doesn't require anything of the sort. I think my confusion was due to coming from another context where bijectivity is needed. The intuitive answer to my question above is that, if we think of $\phi$ as a path on the real line, even if on its way to $\phi(b)$ from $\phi(a)$ it "turns around" and passes over a previously traversed segment again but in the opposite direction (meaning it's not injective), the negative sign of $\phi'$ will essentially mean that that part of the "sum" (integral) will cancel out with the previous sum over the same segment, and moreover since $\phi$ is continuous, it will be forced to pass a third time through that but this time in the positive direction (towards $\phi(b)$), which overall will produce the same outcome as if it had just traversed it once. Although now that I've reread this explanation it seems I'm not able to explain this in words :) --Plammens (talk) 11:44, 6 March 2021 (UTC)

- Usually I would not pay attention to such details, but recently I had a smiliar problem in physics. I had to integrate a certain function along a trajectory of a body orbinting a central mass. Integration wrt time from $r_1 = \map r 0$ to $r_2 = \map r t$ with $r_2 > r_1$ is straightforward, because time always increases. But if we change the variable to $r$, suddenly we have two options: $r_1 \to r_2$ and $r_1 \to r_0 \to r_1 \to r_2$ with $r_0 : r_0 < r_1$ being the turning point. Furthermore, the part between $r_0$ and $r_1$ does not cancel out and has to be included twice. The choice of trajectory is dictated by the sign of $\ds \frac {\d r} {\d t}$ at $t = 0$. Basically, to solve for $\map r t$ initially one is given an equation in the form $\map {\dot r^2} t = \map f t$, and based on initial conditions the correct branch has to be chosen. Hence, I understand the confusion and would like to understand this statment in greater detail. Maybe some other time.--Julius (talk) 15:07, 6 March 2021 (UTC)