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PFuser1232
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Argument A
##∫ 0 dx = 0x + C = C##
Argument B
##∫ 0 dx = ∫ (0)(1) dx = 0 ∫ 1 dx = 0(x+C) = 0##
Discuss.
##∫ 0 dx = 0x + C = C##
Argument B
##∫ 0 dx = ∫ (0)(1) dx = 0 ∫ 1 dx = 0(x+C) = 0##
Discuss.
Mark44 said:The "Discuss" part makes me think this is a homework problem.
Mark44 said:Argument B is silly, as it factors 0 into 0 times 1, and then moves the 0 outside the integral.
There's no need to write 0x in the above.MohammedRady97 said:Argument A
##∫ 0 dx = 0x + C = C##
##\int 0 dx = 0\int dx = 0x + C = C##MohammedRady97 said:Argument B
##∫ 0 dx = ∫ (0)(1) dx = 0 ∫ 1 dx = 0(x+C) = 0##
MohammedRady97 said:Discuss.
Can you elaborate on this? I'm not following you. How does the domain enter into this problem, given that we're working with indefinite integrals?pwsnafu said:Further, ##\int 0 \, dx## being constant is wrong. If for example the domain is ##\mathbb{R}\setminus\{0\}## then you would have two constants, not necessarily equal.
MohammedRady97 said:Argument A
##∫ 0 dx = 0x + C = C##
Argument B
##∫ 0 dx = ∫ (0)(1) dx = 0 ∫ 1 dx = 0(x+C) = 0##
Discuss.
Mark44 said:Can you elaborate on this? I'm not following you. How does the domain enter into this problem, given that we're working with indefinite integrals?
pwsnafu said:The indefinite integral does not calculate a function. It calculates an equivalence class of function. All constant functions are in the same equivalence class.
Further, ##\int 0 \, dx## being constant is wrong. If for example the domain is ##\mathbb{R}\setminus\{0\}## then you would have two constants, not necessarily equal.
PeroK said:For what it's worth, I think this is an interesting question. What do you think? A or B?
Hint: one answer is clearly wrong, but it's not so easy to explain exactly why.
MohammedRady97 said:My gut tells me it's A.
PeroK said:Your head should tell you that as well. In general, you have to be careful with the constant of integration. It's an informal way of specifying an equivalence class of functions (as pointed out above). So, that's where the multiplication by 0 breaks down. If you do things more formally, you still have an equivalence class after multiplication by 0 - so you don't get rid of the constant of integration.
And, yes, an indefinite integral is actually a set of functions.
No. You work with equivalence classes of functions, not with a constant of integration.MohammedRady97 said:And how exactly do I do things more formally?
Does ##∫a f(x) dx = a ∫ f(x) dx## only apply for nonzero a?
PeroK said:No. You work with equivalence classes of functions, not with a constant of integration.
An analogy would be ##0 \cdot x(mod \ n) = 0 (mod \ n) ≠ 0##
Yes.MohammedRady97 said:So what you're saying is I should treat the indefinite integral as a set of functions ## {F(x) + C | C ∈ ℝ} ##, but not as a single function?
MohammedRady97 said:What about the rule ##∫a f(x) dx = a ∫ f(x) dx##?
Mark44 said:Yes.
If a function f has two distinct antiderivatives F1 and F2 (IOW F1] = f and F2] = f) then F1(x) - F2(x) ≡ C.
An example in the same vein is this:
$$\int sin(x)cos(x)dx$$
One student uses substitution to evaluate this integral, using u = sin(x), so du = cos(x)dx.
Then integral becomes ##\int udu = (1/2)u^2 = (1/2)sin^2(x)##.
Another student also uses substitution, but with u = cos(x), du = -sin(x)dx
With this substitution, the integral becomes ##-\int udu = -(1/2)u^2 = -(1/2)cos^2(x)##.
(Notice that both students omitted the constant of integration.)
Here we have two distinct antiderivatives for the same integrand. As it turns out, the two antiderivatives differ by a constant: (1/2)sin2(x) = -(1/2)cos2(x) + 1/2, independent of the value of x.
Stephen Tashi said:As PeroK mentioned, such rules are not about symbols representing unique functions. The symbol [itex] \int f(x) [/itex] does not represent a unique function. Such rules don't work if you interpret the anti-derivative symbol to be a unique function.
For example If suppose we say that [itex] \int x dx [/itex] represents unique function [itex] x^2/2 + 1 [/itex].
and the rule you quoted says [itex] \int 2x [/itex] is the unique function [itex] 2( x^2/2 + 1) = x^2 + 2[/itex]. This would contradict any calculation that said [itex] \int 2 x = x^2 [/itex] or [itex] \int 2 x = x^2 + 5 [/itex] etc.
It doesn't specify a unique function unless you interpret [itex] C [/itex] as symbolizing a unique number.MohammedRady97 said:it would imply that the indefinite integral is a single unique function?
There are areas of mathematics where terminology is imprecise. I think most peopl treat "indefinite integral" as a synonym for "antiderivative". Many people say "the" indefinite integral or "the" antiderivative even though they know the functions are only unique "up to a constant". If we want to study the symbolic calculations of calculus in a rigorous way as a manipulation of strings (the way a symbolic algebra computer program woud do them) then we'd have to use more precise terminology. As I recall, a mathematician named Ritt did such work. We can investigate what terminology he used.Should we draw a line between the often interchangeable terms "Indefinite Integral" and "Antiderivative" by expressing Indefinite Integrals as sets of all possible Antiderivatives? Or are they synonyms?
MohammedRady97 said:So what you're saying is I should treat the indefinite integral as a set of functions ## {F(x) + C | C ∈ ℝ} ##, but not as a single function?
Nick O said:You can safely think of an antiderivative as exactly what the name says: the inverse of a derivative.
d/dx 5 = 0.
So, the antiderivative of 0 dx has to be the same shape as 5, namely, a constant.
I don't think there is any reason to restrict C to the reals. C could just as easily be 5+3i, because d/dx (5+3i)=0
pwsnafu said:The derivative is not injective. It does not have an inverse. It has a right inverses.
You restrict to the co-domain of the function you are integrating. If the OP chose R, you restrict the constant to R.
Stephen Tashi said:It doesn't specify a unique function unless you interpret [itex] C [/itex] as symbolizing a unique number.
There are areas of mathematics where terminology is imprecise. I think most peopl treat "indefinite integral" as a synonym for "antiderivative". Many people say "the" indefinite integral or "the" antiderivative even though they know the functions are only unique "up to a constant". If we want to study the symbolic calculations of calculus in a rigorous way as a manipulation of strings (the way a symbolic algebra computer program woud do them) then we'd have to use more precise terminology. As I recall, a mathematician named Ritt did such work. We can investigate what terminology he used.
?MohammedRady97 said:So should I interpret the indefinite integral of a function as a "set", or as a function with a "variable constant of integration"?
MohammedRady97 said:So should I interpret the indefinite integral of a function as a "set", or as a function with a "variable constant of integration"?
Stephen Tashi said:As PeroK mentioned, such rules are not about symbols representing unique functions. The symbol [itex] \int f(x) [/itex] does not represent a unique function. Such rules don't work if you interpret the anti-derivative symbol to be a unique function.
For example If suppose we say that [itex] \int x dx [/itex] represents unique function [itex] x^2/2 + 1 [/itex].
and the rule you quoted says [itex] \int 2x [/itex] is the unique function [itex] 2( x^2/2 + 1) = x^2 + 2[/itex]. This would contradict any calculation that said [itex] \int 2 x = x^2 [/itex] or [itex] \int 2 x = x^2 + 5 [/itex] etc.
I, for one, don't agree! If the function is a constant function, f(x)= a, where a is fixed number, then [tex]\int a dx= a \int dx= ax+ CTim77 said:I am not quite at the education level to give a rigorous answer based on Riemann integrals. But I do see that case B is invalid because you are not using linearity to pull a constant out of the integral. In this case 0 is not being treated as a constant but as a function. I'm sure we agree you can't pull the function that is being integrated over out of the integral.
Another viewpoint that might explain why the indefinite integral of 0 = C is that you can think of the output of an indefinite integral as sum of the rates of change of an unknown family of functions,which in this case is all of the constant valued functions that all differ by a constant:). A definite integral is the difference in that function evaluated at two points which in turn is the difference in total rates of change. Where C(b) -C(a) =C-C= 0 .Technically the constant of integration is always there we just omit it in the definite case because it is subtracted out.
Thinking in terms of functions being defined by sums of their rates of change (derivatives or total differentials) as opposed to the traditional Riemann picture and area underneath curves helped me to understand what is happening a little better.And it gives better intuition of why Taylor series work.The function must be analytic though:)
Sorry for a long winded post and I apologize also for not using Latec I will get there though.
HallsofIvy said:I, for one, don't agree! If the function is a constant function, f(x)= a, where a is fixed number, then [tex]\int a dx= a \int dx= ax+ C