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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0um let me think so............

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3derivative of a constant is 0, so antiderivative of 0 must be a constant ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yes, integrating 0 is C

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1(and why does a diagram give the wrong conclusion)

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1dw:1438851830406:dw a rectangle of height zero , how can the area be anything other than zero?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but it is right that integrating 0 is C

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1if it were a definite integral, it will always equal zero dw:1438851890255:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3yes that agrees with fundamental thm of calculus \[\int_{x_1}^{x_2} f(x)~dx = cc\Bigg_{x_1}^{x_2} = 0\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1Why 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?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3if 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 : \[\int 0 dx = ? \iff \text{solve } f'(x) = 0, ~f(0)=c_0\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3evaluating \(\int 0 dx\) is same as finding the general solution of differential equation \(f'(x) = 0\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3it is easy to see that any constant function satisfies above DE because the slope of any constant function is 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3`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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3that is like saying \(\int \cos x dx = \sin x\) but what about \(\sin x+1, ~\sin x+2\) etc... they are also antiderivatives of \(\cos x\)

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1the 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3definite 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3if you have initial condition, then the solution is a "specific" constant function in the family of constant functions

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[\sum0\,\Delta x = C?\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3nope, \(\sum 0 = 0 \)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3that sum is the definition of definite integral, has nothing to do with the indefinite integral (save FTC)

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1why dosen't the zero factor outside of the integral \[\int0\,\mathrm dx=0\int\mathrm dx=0\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I think pulling out a constant works only for definite integrals, there is no rule like that for antiderivatives

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1so \[\int k\,\mathrm dx\neq k\int\mathrm dx,\qquad k\in\mathbb R\] (in general)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3we have just seen that it breaks for \(k=0\), so i believe so..

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1so the calculus rules break linearity, in general

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3We can fix it easily by agreeing that the indefinite integral outputs a function that can be known only up to a constant function : \[\int f(x) dx = F(x) + C\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[\int 0 dx = 0 \int dx = 0 + C = C\] we're good

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3linearity is preserved with indefinite integrals too, just need to agree that the output is not unique

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[\int 0\,\mathrm dx = 0 \int\,\mathrm dx = 0 (x+C) = 0\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3but \(0\) is known upto a constant, because indefinite integral outputs a family of functions...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3maybe i think we shouldn't be pulling out 0 with indefinite integrals haha

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1the family x+c, is multiply by the constant, they all go

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1zero breaks calculus

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3it doesn't, \(0\) is same as \(0+C\) upto a constant

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3indefinite integral promises that it can reveal information upto a constant, so nothing breaks as such..

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.10 = 0 + C ( ± C )

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(0\equiv C\) for the output of indefinite integrals

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(F \equiv F+C\) for the output of indefinite integrals

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3a lil bit more abuse of notation \(F \equiv F+C \pmod{\int f(x)\, dx }\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3i like vadim's answer here http://math.stackexchange.com/questions/1069664/isindefiniteintegrationnonlinear

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1a question match!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3real analysis/topology ppl might enjoy this discussion more @zzr0ck3r @eliassaab @oldrin.bataku @Empty

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1vadim's answer reminds me vectors/linear algebra

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1\[\text{For indefinite integrals}:\int kf(x)\mathrm dx=k\int f(x)\mathrm dx+c \\ \ \forall k\in \mathbb R \]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0So (3) in that post is clearly wrong right?

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.13 doesn't happen all the time.... It's where c=0....

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0That'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 \[\text{For indefinite integrals}:\int kf(x)\mathrm dx=k\int f(x)\mathrm dx+c \\ \ \forall k\in \mathbb R \] \(\color{blue}{\text{End of Quote}}\)

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1you got it wrong in you 2nd step @Astrophysics

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[\int 0\,\mathrm dx = 0 \int\,\mathrm dx +C_1= 0 (x+C_2) +C_1= C_1\] right @amilapsn ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Can you explain the C1 @UnkleRhaukus

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1when 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"

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Ahh I should've known that, I see it now, thanks @UnkleRhaukus @amilapsn @ganeshie8 and vadim123 haha.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1Reductio ad absurdum.
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