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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

(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

but why?

\[\sum0\,\Delta x = C?\]

nope, \(\sum 0 = 0 \)

why dosen't the zero factor outside of the integral
\[\int0\,\mathrm dx=0\int\mathrm dx=0\]

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

\[\int 0 dx = 0 \int dx = 0 + C = C\]
we're good

\[\int 0\,\mathrm dx = 0 \int\,\mathrm dx = 0 (x+C) = 0\]

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

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
\(F \equiv F+C \pmod{\int f(x)\, dx }\)

a question match!

vadim's answer reminds me vectors/linear algebra

So (3) in that post is clearly wrong right?

3 doesn't happen all the time.... It's where c=0....

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

Reductio ad absurdum.

Haha, exactly!