Where \(f(x) = 0\), evaluate the indefinite integral: \[\int f(x)\,\mathrm dx\]

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Where \(f(x) = 0\), evaluate the indefinite integral: \[\int f(x)\,\mathrm dx\]

Mathematics
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um let me think so............
derivative of a constant is 0, so antiderivative of 0 must be a constant ?
Yes, integrating 0 is C

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(and why does a diagram give the wrong conclusion)
what Diagram ?
|dw:1438851830406:dw| a rectangle of height zero , how can the area be anything other than zero?
but it is right that integrating 0 is C
if it were a definite integral, it will always equal zero |dw:1438851890255:dw|
0
yes that agrees with fundamental thm of calculus \[\int_{x_1}^{x_2} f(x)~dx = c-c\Bigg|_{x_1}^{x_2} = 0\]
Why is it mathematically erroneous, to conclude that the arbitrary constant of integration C , must be equal to zero for the indefinite integral of the zero function?
yes
but why?
if it was not for fundamental thm of calculus, there is no need to assume that "definite integral" is related to antiderivatives the indefinite integral of 0 is a constant function, it is not 0, sometimes it can be 0 too depending on the initial conditions of IVP : \[\int 0 dx = ? \iff \text{solve } f'(x) = 0, ~f(0)=c_0\]
evaluating \(\int 0 dx\) is same as finding the general solution of differential equation \(f'(x) = 0\)
it is easy to see that any constant function satisfies above DE because the slope of any constant function is 0
`Why is it mathematically erroneous, to conclude that the arbitrary constant of integration C , must be equal to zero for the indefinite integral of the zero function?` If I interpret it correctly, I think saying \(\int 0 dx= 0\) leaves out other family of constant functions that are not 0, so this doesn't work.
that is like saying \(\int \cos x dx = \sin x\) but what about \(\sin x+1, ~\sin x+2\) etc... they are also antiderivatives of \(\cos x\)
the initial conditions are defined implicitly by the function \[f(x) = 0 \implies f(x_1) = 0,\quad f(x_2) = 0. \] (this is independent of the choice of \(x_1\), \(x_2\) ) there are no other member of the family, for this (zero) function, that satisfy its definition
definite integral is a different story; area is just one interpretation of definite integral, it is not the only definition of definite integral. suppose we don't have fundamental theorem of calculus. then i think the proper definition definite integral uses limits. \[\int\limits_{x_1}^{x_2} 0\, dx=\lim\limits_{n\to\infty}\sum\limits_{i=1}^n f(x_i)\Delta x = 0+0+\cdots = 0\] which trivially gave 0 as expected
if you have initial condition, then the solution is a "specific" constant function in the family of constant functions
\[\sum0\,\Delta x = C?\]
nope, \(\sum 0 = 0 \)
that sum is the definition of definite integral, has nothing to do with the indefinite integral (save FTC)
why dosen't the zero factor outside of the integral \[\int0\,\mathrm dx=0\int\mathrm dx=0\]
I think pulling out a constant works only for definite integrals, there is no rule like that for antiderivatives
https://en.wikipedia.org/wiki/Integral#Properties
so \[\int k\,\mathrm dx\neq k\int\mathrm dx,\qquad k\in\mathbb R\] (in general)
we have just seen that it breaks for \(k=0\), so i believe so..
so the calculus rules break linearity, in general
We can fix it easily by agreeing that the indefinite integral outputs a function that can be known only up to a constant function : \[\int f(x) dx = F(x) + C\]
\[\int 0 dx = 0 \int dx = 0 + C = C\] we're good
linearity is preserved with indefinite integrals too, just need to agree that the output is not unique
\[\int 0\,\mathrm dx = 0 \int\,\mathrm dx = 0 (x+C) = 0\]
but \(0\) is known upto a constant, because indefinite integral outputs a family of functions...
maybe i think we shouldn't be pulling out 0 with indefinite integrals haha
the family x+c, is multiply by the constant, they all go
zero breaks calculus
it doesn't, \(0\) is same as \(0+C\) upto a constant
lol
indefinite integral promises that it can reveal information upto a constant, so nothing breaks as such..
0 = 0 + C ( ± C )
\(0\equiv C\) for the output of indefinite integrals
\(F \equiv F+C\) for the output of indefinite integrals
a lil bit more abuse of notation \(F \equiv F+C \pmod{\int f(x)\, dx }\)
i like vadim's answer here http://math.stackexchange.com/questions/1069664/is-indefinite-integration-non-linear
a question match!
real analysis/topology ppl might enjoy this discussion more @zzr0ck3r @eliassaab @oldrin.bataku @Empty
vadim's answer reminds me vectors/linear algebra
\[\text{For indefinite integrals}:\int kf(x)\mathrm dx=k\int f(x)\mathrm dx+c \\ \ \forall k\in \mathbb R \]
So (3) in that post is clearly wrong right?
3 doesn't happen all the time.... It's where c=0....
That's weird, I guess I just don't get it lol, \[\int\limits 0 dx = 0 \int\limits dx = 0(x+C) = 0\] this makes sense to me and seems to follow \(\color{blue}{\text{Originally Posted by}}\) @amilapsn \[\text{For indefinite integrals}:\int kf(x)\mathrm dx=k\int f(x)\mathrm dx+c \\ \ \forall k\in \mathbb R \] \(\color{blue}{\text{End of Quote}}\)
you got it wrong in you 2nd step @Astrophysics
\[\int 0\,\mathrm dx = 0 \int\,\mathrm dx +C_1= 0 (x+C_2) +C_1= C_1\] right @amilapsn ?
right..
Can you explain the C1 @UnkleRhaukus
when factoring out a constant factor from the integrand of an integral, you always have to add a constant of integration, because the constant factor might be zero which would imply that an indefinite integral could be evaluated exactly; to a precision, beyond "up to a constant"
Ahh I should've known that, I see it now, thanks @UnkleRhaukus @amilapsn @ganeshie8 and vadim123 haha.
Reductio ad absurdum.
Haha, exactly!

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