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

- UnkleRhaukus

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

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- schrodinger

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- anonymous

um let me think so............

- ganeshie8

derivative of a constant is 0, so antiderivative of 0 must be a constant ?

- Astrophysics

Yes, integrating 0 is C

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## More answers

- UnkleRhaukus

(and why does a diagram give the wrong conclusion)

- anonymous

what Diagram ?

- UnkleRhaukus

|dw:1438851830406:dw|
a rectangle of height zero , how can the area be anything other than zero?

- anonymous

but it is right that integrating 0 is C

- UnkleRhaukus

if it were a definite integral,
it will always equal zero
|dw:1438851890255:dw|

- imqwerty

0

- 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

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

yes

- UnkleRhaukus

but why?

- 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

evaluating \(\int 0 dx\) is same as finding the general solution of differential equation \(f'(x) = 0\)

- ganeshie8

it is easy to see that any constant function satisfies above DE because the slope of any constant function is 0

- 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

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

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

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

if you have initial condition, then the solution is a "specific" constant function in the family of constant functions

- UnkleRhaukus

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

- ganeshie8

nope, \(\sum 0 = 0 \)

- ganeshie8

that sum is the definition of definite integral,
has nothing to do with the indefinite integral (save FTC)

- UnkleRhaukus

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

- ganeshie8

I think pulling out a constant works only for definite integrals, there is no rule like that for antiderivatives

- ganeshie8

https://en.wikipedia.org/wiki/Integral#Properties

- UnkleRhaukus

so \[\int k\,\mathrm dx\neq k\int\mathrm dx,\qquad k\in\mathbb R\] (in general)

- ganeshie8

we have just seen that it breaks for \(k=0\), so i believe so..

- UnkleRhaukus

so the calculus rules break linearity, in general

- 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

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

- ganeshie8

linearity is preserved with indefinite integrals too, just need to agree that the output is not unique

- UnkleRhaukus

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

- ganeshie8

but \(0\) is known upto a constant,
because indefinite integral outputs a family of functions...

- ganeshie8

maybe i think we shouldn't be pulling out 0 with indefinite integrals haha

- UnkleRhaukus

the family x+c, is multiply by the constant, they all go

- UnkleRhaukus

zero breaks calculus

- ganeshie8

it doesn't, \(0\) is same as \(0+C\) upto a constant

- UnkleRhaukus

lol

- ganeshie8

indefinite integral promises that it can reveal information upto a constant, so nothing breaks as such..

- UnkleRhaukus

0 = 0 + C ( ± C )

- ganeshie8

\(0\equiv C\)
for the output of indefinite integrals

- ganeshie8

\(F \equiv F+C\)
for the output of indefinite integrals

- ganeshie8

a lil bit more abuse of notation
\(F \equiv F+C \pmod{\int f(x)\, dx }\)

- ganeshie8

i like vadim's answer here
http://math.stackexchange.com/questions/1069664/is-indefinite-integration-non-linear

- UnkleRhaukus

a question match!

- ganeshie8

real analysis/topology ppl might enjoy this discussion more
@zzr0ck3r @eliassaab @oldrin.bataku @Empty

- UnkleRhaukus

vadim's answer reminds me vectors/linear algebra

- amilapsn

\[\text{For indefinite integrals}:\int kf(x)\mathrm dx=k\int f(x)\mathrm dx+c \\ \ \forall k\in \mathbb R \]

- Astrophysics

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

- amilapsn

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

- 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

you got it wrong in you 2nd step @Astrophysics

- UnkleRhaukus

\[\int 0\,\mathrm dx = 0 \int\,\mathrm dx +C_1= 0 (x+C_2) +C_1= C_1\]
right @amilapsn ?

- amilapsn

right..

- Astrophysics

Can you explain the C1 @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

Ahh I should've known that, I see it now, thanks @UnkleRhaukus @amilapsn @ganeshie8
and vadim123 haha.

- UnkleRhaukus

Reductio ad absurdum.

- Astrophysics

Haha, exactly!

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