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UnkleRhaukus

  • one year ago

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

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  1. anonymous
    • one year ago
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    um let me think so............

  2. ganeshie8
    • one year ago
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    derivative of a constant is 0, so antiderivative of 0 must be a constant ?

  3. Astrophysics
    • one year ago
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    Yes, integrating 0 is C

  4. UnkleRhaukus
    • one year ago
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    (and why does a diagram give the wrong conclusion)

  5. anonymous
    • one year ago
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    what Diagram ?

  6. UnkleRhaukus
    • one year ago
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    |dw:1438851830406:dw| a rectangle of height zero , how can the area be anything other than zero?

  7. anonymous
    • one year ago
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    but it is right that integrating 0 is C

  8. UnkleRhaukus
    • one year ago
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    if it were a definite integral, it will always equal zero |dw:1438851890255:dw|

  9. imqwerty
    • one year ago
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    0

  10. ganeshie8
    • one year ago
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    yes that agrees with fundamental thm of calculus \[\int_{x_1}^{x_2} f(x)~dx = c-c\Bigg|_{x_1}^{x_2} = 0\]

  11. UnkleRhaukus
    • one year ago
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    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?

  12. anonymous
    • one year ago
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    yes

  13. UnkleRhaukus
    • one year ago
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    but why?

  14. ganeshie8
    • one year ago
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    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 : \[\int 0 dx = ? \iff \text{solve } f'(x) = 0, ~f(0)=c_0\]

  15. ganeshie8
    • one year ago
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    evaluating \(\int 0 dx\) is same as finding the general solution of differential equation \(f'(x) = 0\)

  16. ganeshie8
    • one year ago
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    it is easy to see that any constant function satisfies above DE because the slope of any constant function is 0

  17. ganeshie8
    • one year ago
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    `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.

  18. ganeshie8
    • one year ago
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    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 \(\cos x\)

  19. UnkleRhaukus
    • one year ago
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    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

  20. ganeshie8
    • one year ago
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    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

  21. ganeshie8
    • one year ago
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    if you have initial condition, then the solution is a "specific" constant function in the family of constant functions

  22. UnkleRhaukus
    • one year ago
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    \[\sum0\,\Delta x = C?\]

  23. ganeshie8
    • one year ago
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    nope, \(\sum 0 = 0 \)

  24. ganeshie8
    • one year ago
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    that sum is the definition of definite integral, has nothing to do with the indefinite integral (save FTC)

  25. UnkleRhaukus
    • one year ago
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    why dosen't the zero factor outside of the integral \[\int0\,\mathrm dx=0\int\mathrm dx=0\]

  26. ganeshie8
    • one year ago
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    I think pulling out a constant works only for definite integrals, there is no rule like that for antiderivatives

  27. ganeshie8
    • one year ago
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    https://en.wikipedia.org/wiki/Integral#Properties

  28. UnkleRhaukus
    • one year ago
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    so \[\int k\,\mathrm dx\neq k\int\mathrm dx,\qquad k\in\mathbb R\] (in general)

  29. ganeshie8
    • one year ago
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    we have just seen that it breaks for \(k=0\), so i believe so..

  30. UnkleRhaukus
    • one year ago
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    so the calculus rules break linearity, in general

  31. ganeshie8
    • one year ago
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    We can fix it easily by agreeing that the indefinite integral outputs a function that can be known only up to a constant function : \[\int f(x) dx = F(x) + C\]

  32. ganeshie8
    • one year ago
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    \[\int 0 dx = 0 \int dx = 0 + C = C\] we're good

  33. ganeshie8
    • one year ago
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    linearity is preserved with indefinite integrals too, just need to agree that the output is not unique

  34. UnkleRhaukus
    • one year ago
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    \[\int 0\,\mathrm dx = 0 \int\,\mathrm dx = 0 (x+C) = 0\]

  35. ganeshie8
    • one year ago
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    but \(0\) is known upto a constant, because indefinite integral outputs a family of functions...

  36. ganeshie8
    • one year ago
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    maybe i think we shouldn't be pulling out 0 with indefinite integrals haha

  37. UnkleRhaukus
    • one year ago
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    the family x+c, is multiply by the constant, they all go

  38. UnkleRhaukus
    • one year ago
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    zero breaks calculus

  39. ganeshie8
    • one year ago
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    it doesn't, \(0\) is same as \(0+C\) upto a constant

  40. UnkleRhaukus
    • one year ago
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    lol

  41. ganeshie8
    • one year ago
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    indefinite integral promises that it can reveal information upto a constant, so nothing breaks as such..

  42. UnkleRhaukus
    • one year ago
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    0 = 0 + C ( ± C )

  43. ganeshie8
    • one year ago
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    \(0\equiv C\) for the output of indefinite integrals

  44. ganeshie8
    • one year ago
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    \(F \equiv F+C\) for the output of indefinite integrals

  45. ganeshie8
    • one year ago
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    a lil bit more abuse of notation \(F \equiv F+C \pmod{\int f(x)\, dx }\)

  46. ganeshie8
    • one year ago
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    i like vadim's answer here http://math.stackexchange.com/questions/1069664/is-indefinite-integration-non-linear

  47. UnkleRhaukus
    • one year ago
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    a question match!

  48. ganeshie8
    • one year ago
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    real analysis/topology ppl might enjoy this discussion more @zzr0ck3r @eliassaab @oldrin.bataku @Empty

  49. UnkleRhaukus
    • one year ago
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    vadim's answer reminds me vectors/linear algebra

  50. amilapsn
    • one year ago
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    \[\text{For indefinite integrals}:\int kf(x)\mathrm dx=k\int f(x)\mathrm dx+c \\ \ \forall k\in \mathbb R \]

  51. Astrophysics
    • one year ago
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    So (3) in that post is clearly wrong right?

  52. amilapsn
    • one year ago
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    3 doesn't happen all the time.... It's where c=0....

  53. Astrophysics
    • one year ago
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    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 \[\text{For indefinite integrals}:\int kf(x)\mathrm dx=k\int f(x)\mathrm dx+c \\ \ \forall k\in \mathbb R \] \(\color{blue}{\text{End of Quote}}\)

  54. amilapsn
    • one year ago
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    you got it wrong in you 2nd step @Astrophysics

  55. UnkleRhaukus
    • one year ago
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    \[\int 0\,\mathrm dx = 0 \int\,\mathrm dx +C_1= 0 (x+C_2) +C_1= C_1\] right @amilapsn ?

  56. amilapsn
    • one year ago
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    right..

  57. Astrophysics
    • one year ago
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    Can you explain the C1 @UnkleRhaukus

  58. UnkleRhaukus
    • one year ago
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    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"

  59. Astrophysics
    • one year ago
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    Ahh I should've known that, I see it now, thanks @UnkleRhaukus @amilapsn @ganeshie8 and vadim123 haha.

  60. UnkleRhaukus
    • one year ago
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    Reductio ad absurdum.

  61. Astrophysics
    • one year ago
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    Haha, exactly!

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