UnkleRhaukus
  • UnkleRhaukus
Where \(f(x) = 0\), evaluate the indefinite integral: \[\int f(x)\,\mathrm dx\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
um let me think so............
ganeshie8
  • ganeshie8
derivative of a constant is 0, so antiderivative of 0 must be a constant ?
Astrophysics
  • Astrophysics
Yes, integrating 0 is C

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UnkleRhaukus
  • UnkleRhaukus
(and why does a diagram give the wrong conclusion)
anonymous
  • anonymous
what Diagram ?
UnkleRhaukus
  • UnkleRhaukus
|dw:1438851830406:dw| a rectangle of height zero , how can the area be anything other than zero?
anonymous
  • anonymous
but it is right that integrating 0 is C
UnkleRhaukus
  • UnkleRhaukus
if it were a definite integral, it will always equal zero |dw:1438851890255:dw|
imqwerty
  • imqwerty
0
ganeshie8
  • ganeshie8
yes that agrees with fundamental thm of calculus \[\int_{x_1}^{x_2} f(x)~dx = c-c\Bigg|_{x_1}^{x_2} = 0\]
UnkleRhaukus
  • UnkleRhaukus
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?
anonymous
  • anonymous
yes
UnkleRhaukus
  • UnkleRhaukus
but why?
ganeshie8
  • ganeshie8
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 : \[\int 0 dx = ? \iff \text{solve } f'(x) = 0, ~f(0)=c_0\]
ganeshie8
  • ganeshie8
evaluating \(\int 0 dx\) is same as finding the general solution of differential equation \(f'(x) = 0\)
ganeshie8
  • ganeshie8
it is easy to see that any constant function satisfies above DE because the slope of any constant function is 0
ganeshie8
  • ganeshie8
`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
  • ganeshie8
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 \(\cos x\)
UnkleRhaukus
  • UnkleRhaukus
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
ganeshie8
  • ganeshie8
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
ganeshie8
  • ganeshie8
if you have initial condition, then the solution is a "specific" constant function in the family of constant functions
UnkleRhaukus
  • UnkleRhaukus
\[\sum0\,\Delta x = C?\]
ganeshie8
  • ganeshie8
nope, \(\sum 0 = 0 \)
ganeshie8
  • ganeshie8
that sum is the definition of definite integral, has nothing to do with the indefinite integral (save FTC)
UnkleRhaukus
  • UnkleRhaukus
why dosen't the zero factor outside of the integral \[\int0\,\mathrm dx=0\int\mathrm dx=0\]
ganeshie8
  • ganeshie8
I think pulling out a constant works only for definite integrals, there is no rule like that for antiderivatives
ganeshie8
  • ganeshie8
https://en.wikipedia.org/wiki/Integral#Properties
UnkleRhaukus
  • UnkleRhaukus
so \[\int k\,\mathrm dx\neq k\int\mathrm dx,\qquad k\in\mathbb R\] (in general)
ganeshie8
  • ganeshie8
we have just seen that it breaks for \(k=0\), so i believe so..
UnkleRhaukus
  • UnkleRhaukus
so the calculus rules break linearity, in general
ganeshie8
  • ganeshie8
We can fix it easily by agreeing that the indefinite integral outputs a function that can be known only up to a constant function : \[\int f(x) dx = F(x) + C\]
ganeshie8
  • ganeshie8
\[\int 0 dx = 0 \int dx = 0 + C = C\] we're good
ganeshie8
  • ganeshie8
linearity is preserved with indefinite integrals too, just need to agree that the output is not unique
UnkleRhaukus
  • UnkleRhaukus
\[\int 0\,\mathrm dx = 0 \int\,\mathrm dx = 0 (x+C) = 0\]
ganeshie8
  • ganeshie8
but \(0\) is known upto a constant, because indefinite integral outputs a family of functions...
ganeshie8
  • ganeshie8
maybe i think we shouldn't be pulling out 0 with indefinite integrals haha
UnkleRhaukus
  • UnkleRhaukus
the family x+c, is multiply by the constant, they all go
UnkleRhaukus
  • UnkleRhaukus
zero breaks calculus
ganeshie8
  • ganeshie8
it doesn't, \(0\) is same as \(0+C\) upto a constant
UnkleRhaukus
  • UnkleRhaukus
lol
ganeshie8
  • ganeshie8
indefinite integral promises that it can reveal information upto a constant, so nothing breaks as such..
UnkleRhaukus
  • UnkleRhaukus
0 = 0 + C ( ± C )
ganeshie8
  • ganeshie8
\(0\equiv C\) for the output of indefinite integrals
ganeshie8
  • ganeshie8
\(F \equiv F+C\) for the output of indefinite integrals
ganeshie8
  • ganeshie8
a lil bit more abuse of notation \(F \equiv F+C \pmod{\int f(x)\, dx }\)
ganeshie8
  • ganeshie8
i like vadim's answer here http://math.stackexchange.com/questions/1069664/is-indefinite-integration-non-linear
UnkleRhaukus
  • UnkleRhaukus
a question match!
ganeshie8
  • ganeshie8
real analysis/topology ppl might enjoy this discussion more @zzr0ck3r @eliassaab @oldrin.bataku @Empty
UnkleRhaukus
  • UnkleRhaukus
vadim's answer reminds me vectors/linear algebra
amilapsn
  • amilapsn
\[\text{For indefinite integrals}:\int kf(x)\mathrm dx=k\int f(x)\mathrm dx+c \\ \ \forall k\in \mathbb R \]
Astrophysics
  • Astrophysics
So (3) in that post is clearly wrong right?
amilapsn
  • amilapsn
3 doesn't happen all the time.... It's where c=0....
Astrophysics
  • Astrophysics
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 \[\text{For indefinite integrals}:\int kf(x)\mathrm dx=k\int f(x)\mathrm dx+c \\ \ \forall k\in \mathbb R \] \(\color{blue}{\text{End of Quote}}\)
amilapsn
  • amilapsn
you got it wrong in you 2nd step @Astrophysics
UnkleRhaukus
  • UnkleRhaukus
\[\int 0\,\mathrm dx = 0 \int\,\mathrm dx +C_1= 0 (x+C_2) +C_1= C_1\] right @amilapsn ?
amilapsn
  • amilapsn
right..
Astrophysics
  • Astrophysics
Can you explain the C1 @UnkleRhaukus
UnkleRhaukus
  • 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"
Astrophysics
  • Astrophysics
Ahh I should've known that, I see it now, thanks @UnkleRhaukus @amilapsn @ganeshie8 and vadim123 haha.
UnkleRhaukus
  • UnkleRhaukus
Reductio ad absurdum.
Astrophysics
  • Astrophysics
Haha, exactly!

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