Primitives, Proof based on theorems for differentiation

[c.teixeira]

Hi there!

If one would want to prove that the indefined integral :

[itex]\int[f(x)+g(x)]dx[/itex] = [itex]\int f(x)dx[/itex] + [itex]\int g(x)dx[/itex].

Would this be apropriate:

A(x) = [itex]\int[f(x)+g(x)]dx[/itex];
B(x) = [itex]\int f(x)dx[/itex];
C(x) = [itex]\int g(x)dx[/itex].

And since the primitive of a fuction is another fuction whose derivative is the original fuction:

A'(x) = f(x) + g(x);
B'(x) + C'(x) = (B + C)'(x) = f(x) + g(x).

What would imply bt the Mean Value Theorem: A(x) = B(x) + C(x) + K.

Is this a approiate proof?

If so, who do we know that k = 0? Since there is no K in " [itex]\int[f(x)+g(x)]dx[/itex] = [itex]\int f(x)dx[/itex] + [itex]\int g(x)dx[/itex]."

Regards,

 

[pwsnafu]

If so, who do we know that k = 0?

k is not zero in general. The correct form is to define ##A(x) + k_A = \int [ f(x)+g(x) ] dx## for some constant ##k_A##. Similarly ##k_B## and ##k_C##, when you differentiate the constants vanish. You want to show ##k_A = k_B + k_C##.

 

[c.teixeira]

k is not zero in general. The correct form is to define ##A(x) + k_A = \int [ f(x)+g(x) ] dx## for some constant ##k_A##. Similarly ##k_B## and ##k_C##, when you differentiate the constants vanish. You want to show ##k_A = k_B + k_C##.

So, how can I prove that? And why would I want to show that? (##k_A = k_B + k_C##)

I don't understand. If my explnation is correct, and K is generally not 0, as you said, wouldn't that make [itex]\int[f(x) + g(x) dx[/itex] = [itex]\int f(x)dx[/itex] + [itex]\int g(x)dx[/itex] + K instead?

What I undersandt from the above expression, is that, that is valid for any constant K, being it 0, or not. So basically, the k doen'st do much diference, wether it is 0 or otherwise. Am I right or wrong?

Thank you,

 

[c.teixeira]

Also,

I believe that making A(x) + K[itex]_{A}[/itex] = [itex]\int[f(x)+g(x)]dx[/itex], and then showing that K[itex]_{A}[/itex] = K[itex]_{B}[/itex] + K[itex]_{C}[/itex], is the same thing as defining A(x) = [itex]\int[f(x)+g(x)]dx[/itex], and then showing that K = 0.

Spivak says that concer for this Constants is merely an annoyance, but not knowing exactly why [itex]\int[f(x)+g(x)]dx[/itex] = [itex]\int f(x)dx[/itex] + [itex]\int g(x)dx[/itex], without the C for constant is bothering me.

Regards,

 

[pwsnafu]

Spivak says that concer for this Constants is merely an annoyance,

Introductory texts trivialise this because students are not interested in technicalities. But its not a mere annoyance. Consider two variables then ##\int f(x,y) \, dx = F(x,y) + g(y)## for some non-constant function g. So its not trivial.

Here's the run down: the derivative is a linear operator ##D : C^1(U) \rightarrow \mathbb{R}^U## for some open set ##U \subset \mathbb{R}##. However it is not injective because if two functions differ by a constant they have the same derivative. It is not surjective because there are some functions which can never be integrated.

Let ##I(U)## be the set of integrable functions on ##U##. I'm not going to explain what this space looks like, just accept it exists. The indefinite integral is then an operator ##\int \, \cdot \, dx : I(U) \rightarrow C^1(U)/\sim## where the equivalence relation is given by ##f = g \iff f - g## is constant over ##U##.

In this space ##A = A + k## for any constant k, so notice that equality in this space is not the same equality you were working with. If you have done modulo arithmetic, this is easy: you are working with modulo constants. Then the proof is trivial, because any constant is equivalent to zero:
##k_A = k_B + k_C = 0 \mod \text{const}##

 

[HallsofIvy]

Hi there!

If one would want to prove that the indefined integral :

[itex]\int[f(x)+g(x)]dx[/itex] = [itex]\int f(x)dx[/itex] + [itex]\int g(x)dx[/itex].

Would this be apropriate:

A(x) = [itex]\int[f(x)+g(x)]dx[/itex];
B(x) = [itex]\int f(x)dx[/itex];
C(x) = [itex]\int g(x)dx[/itex].

And since the primitive of a fuction is another fuction whose derivative is the original fuction:

A'(x) = f(x) + g(x);
B'(x) + C'(x) = (B + C)'(x) = f(x) + g(x).

What would imply bt the Mean Value Theorem: A(x) = B(x) + C(x) + K.

Is this a approiate proof?

If so, who do we know that k = 0? Since there is no K in " [itex]\int[f(x)+g(x)]dx[/itex] = [itex]\int f(x)dx[/itex] + [itex]\int g(x)dx[/itex]."

Regards,

Yes, there is a "k" in [itex]\int[f(x)+g(x)]dx[/itex] = [itex]\int f(x)dx[/itex] + [itex]\int g(x)dx[/itex]. In fact, there is a "constant of integration" in each of the three integrals in that equation.

 

[c.teixeira]

Yes, there is a "k" in [itex]\int[f(x)+g(x)]dx[/itex] = [itex]\int f(x)dx[/itex] + [itex]\int g(x)dx[/itex]. In fact, there is a "constant of integration" in each of the three integrals in that equation.

So, my little "proof", is correct, and "rigorously" we should have [itex]\int[f(x)+g(x)]dx[/itex] = [itex]\int f(x)dx[/itex] + [itex]\int g(x)dx[/itex] + C?
Altough, the Constant C doesn't do any difference as a consequece to definite integrals.

So, I can think that Spivak didn't write the C, because of this non-consequece to definite integrals, and because of him stating 1 page earlier in the book that Constants where merely an annoyance?

 

[HallsofIvy]

No, you don't need the "C". The constant of integration is already in each of the integrals. When you write [itex]\int 3x^2 dx= x^3+ C[/itex] you do not need "C" on the left because it is part of the integral. You do need it on the right because there is no longer an integral.

 

[c.teixeira]

No, you don't need the "C". The constant of integration is already in each of the integrals. When you write [itex]\int 3x^2 dx= x^3+ C[/itex] you do not need "C" on the left because it is part of the integral. You do need it on the right because there is no longer an integral.

Let:

f(x) = 3x[itex]^{2}[/itex];
g(x) = 2x;

[itex]\int[3x^{2} + 2x] dx[/itex] = x[itex]^{3}[/itex] +x[itex]^{2}[/itex] + C[itex]_{1}[/itex],
[itex]\int3x^{2}dx[/itex] = x[itex]^{3}[/itex] + C[itex]_{2}[/itex],
[itex]\int2xdx[/itex] = x[itex]^{2}[/itex] + C[itex]_{3}[/itex]

Then, [itex]\int[3x^{2} + 2x] dx[/itex] = [itex]\int3x^{2}dx[/itex] + [itex]\int2xdx[/itex] only if C[itex]_{1}[/itex] = C[itex]_{2}[/itex] + C [itex]_{3}[/itex], right?

So how exacltly do you prove that? Using pwsnafu advice?