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My professor was trying to prove the linearity of the integral. That is,

$\int hf(x)+kg(x) dx=h\int f(x)dx + k \int g(x) dx$, where $f,g$ are continuous functions, and $h,k\in \mathbb{R}$.

The following is the proof he wrote on the board:

Proof:

Let $\frac{d}{dx}(F(x))=f(x)$ and $\frac{d}{dx}(G(x))=g(x).$

Since differentiation is linear, we have

$\frac{d}{dx}(hF(x)+kG(x))=h\frac{d}{dx}(F(x))+k\frac{d}{dx}(G(x))=hf(x)+kg(x)=\int hf(x)+kg(x) dx=hF(x)+kG(x)=h\int f(x)dx+k\int g(x)dx=h(F(x)+C_1)+k(G(x)+C_2)=hF(x)+hC_1+kG(x)+kC_2=hF(x)+kG(x)+hC_1+kC_2=hF(x)+kG(x)+C$, where $C$ is the constant of integration. QED

I understand most of the proof, however there is one step that makes no sense to me.

It is the following step:

$hf(x)+kg(x)=\int hf(x)+kg(x) dx$

If anyone could explain how this makes sense, I would be grateful.

By the way, it seems like it doesn't make sense to me, as if I tried a concrete example:

If $f(x)=2x$ and $g(x)=3x$, and $h=2$ and $k=3$ then that step is saying that

$2(2x)+3(3x)=\int 2(2x)+3(3x)\implies 4x+9x=\int 4x+9x \implies 13x=\frac{13x^2}{2}$, which doesn't make sense, right?

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Right, it doesn't make sense. I'm trying to figure out what the intention was. But don't hold your breath. –  Daniel Fischer Jun 26 '13 at 22:03
    
I think it's just a handwritten(?) equivalent of a typo, and the incriminated $=$ should have been an implication, $\Rightarrow$. It's still - in my opinion - a terrible and sloppy proof. –  Daniel Fischer Jun 26 '13 at 22:08
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1 Answer 1

up vote 0 down vote accepted

I'm guessing it was meant to be something like:

$\frac{d}{dx}(hF(x)+kG(x))=h\frac{d}{dx}(F(x))+k\frac{d}{dx}(G(x))=hf(x)+kg(x)$

therefore, by integrating both sides it follows that

$\int hf(x)+kg(x) dx=hF(x)+kG(x) +c$

etc.

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Hmm, that's what I was thinking as well. I even asked the professor if what he put on the board was right, and he said yes. –  Sujaan Kunalan Jun 26 '13 at 22:19
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