Primitives, Proof based on theorems for differentiation

[c.teixeira]

Hi there!

If one would want to prove that the indefined integral :

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

Would this be apropriate:

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

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

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

Spivak says that concer for this Constants is merely an annoyance, but not knowing exactly why \int[f(x)+g(x)]dx = \int f(x)dx + \int g(x)dx, 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 :

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

Would this be apropriate:

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

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

Regards,

Yes, there is a "k" in \int[f(x)+g(x)]dx = \int f(x)dx + \int g(x)dx. 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 \int[f(x)+g(x)]dx = \int f(x)dx + \int g(x)dx. 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 \int[f(x)+g(x)]dx = \int f(x)dx + \int g(x)dx + 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 \int 3x^2 dx= x^3+ C 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 \int 3x^2 dx= x^3+ C 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^{2};
g(x) = 2x;

\int[3x^{2} + 2x] dx = x^{3} +x^{2} + C_{1},
\int3x^{2}dx = x^{3} + C_{2},
\int2xdx = x^{2} + C_{3}

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

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