Timeline for Dummy variable rule for indefinite integrals?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| 14 hours ago | answer | added | invictus | timeline score: 0 | |
| 23 hours ago | history | edited | ryang | CC BY-SA 4.0 |
deleted 197 characters in body
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| yesterday | history | became hot network question | |||
| yesterday | vote | accept | Samuel Ho | ||
| yesterday | answer | added | ryang | timeline score: 5 | |
| yesterday | answer | added | Gamer Rossi | timeline score: 2 | |
| yesterday | comment | added | Bowei Tang | @JonathanZ Maybe we shall ask when $\int g(x)\ dx$ is equal to $\int g(u)\ du$. Set $u=f(x)$. If $\int g(x)\ dx=\int g(u)\ du$, then $$\int g(x)\ dx=\int g(f(x)) f'(x)\ dx.$$ So we should have $g(x)=g(f(x))\cdot f'(x)$. | |
| yesterday | comment | added | JonathanZ | I think you're onto something pointing out that the assumption that $\int g(x)\ dx=\int g(u)\ du$ might be wrong, but I'm having the devil of a time saying why. A lot of calculus notation is a bit iffy, and when pushed to the limit even basic function notation breaks, (like is $f(x)$ the function or the evaluation of the function at some point)? | |
| yesterday | answer | added | JonathanZ | timeline score: 3 | |
| yesterday | comment | added | Bowei Tang | It's true that $\int g(x)\ dx=-\int g(u)\ du$, but you implicitly assume that $\int g(x)\ dx=\int g(u)\ du$, which is not the case. For example, if $g(x)=x^2$, then $\int g(x)\ dx=\frac{1}{3}x^3+C$ and $\int g(u)\ du=\frac{1}{3}u^3+C=-\frac{1}{3}x^3+C.$ | |
| yesterday | history | asked | Samuel Ho | CC BY-SA 4.0 |