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

1. anonymous

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

2. ganeshie8

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

3. Astrophysics

Yes, integrating 0 is C

4. UnkleRhaukus

(and why does a diagram give the wrong conclusion)

5. anonymous

what Diagram ?

6. UnkleRhaukus

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

7. anonymous

but it is right that integrating 0 is C

8. UnkleRhaukus

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

9. imqwerty

0

10. ganeshie8

yes that agrees with fundamental thm of calculus $\int_{x_1}^{x_2} f(x)~dx = c-c\Bigg|_{x_1}^{x_2} = 0$

11. 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?

12. anonymous

yes

13. UnkleRhaukus

but why?

14. 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 : $\int 0 dx = ? \iff \text{solve } f'(x) = 0, ~f(0)=c_0$

15. ganeshie8

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

16. ganeshie8

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

17. 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.

18. 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 $$\cos x$$

19. 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

20. 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

21. ganeshie8

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

22. UnkleRhaukus

$\sum0\,\Delta x = C?$

23. ganeshie8

nope, $$\sum 0 = 0$$

24. ganeshie8

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

25. UnkleRhaukus

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

26. ganeshie8

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

27. ganeshie8
28. UnkleRhaukus

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

29. ganeshie8

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

30. UnkleRhaukus

so the calculus rules break linearity, in general

31. 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 : $\int f(x) dx = F(x) + C$

32. ganeshie8

$\int 0 dx = 0 \int dx = 0 + C = C$ we're good

33. ganeshie8

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

34. UnkleRhaukus

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

35. ganeshie8

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

36. ganeshie8

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

37. UnkleRhaukus

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

38. UnkleRhaukus

zero breaks calculus

39. ganeshie8

it doesn't, $$0$$ is same as $$0+C$$ upto a constant

40. UnkleRhaukus

lol

41. ganeshie8

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

42. UnkleRhaukus

0 = 0 + C ( ± C )

43. ganeshie8

$$0\equiv C$$ for the output of indefinite integrals

44. ganeshie8

$$F \equiv F+C$$ for the output of indefinite integrals

45. ganeshie8

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

46. ganeshie8

47. UnkleRhaukus

a question match!

48. ganeshie8

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

49. UnkleRhaukus

50. amilapsn

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

51. Astrophysics

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

52. amilapsn

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

53. 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 $\text{For indefinite integrals}:\int kf(x)\mathrm dx=k\int f(x)\mathrm dx+c \\ \ \forall k\in \mathbb R$ $$\color{blue}{\text{End of Quote}}$$

54. amilapsn

you got it wrong in you 2nd step @Astrophysics

55. UnkleRhaukus

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

56. amilapsn

right..

57. Astrophysics

Can you explain the C1 @UnkleRhaukus

58. 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"

59. Astrophysics

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

60. UnkleRhaukus