Integrate d/dx f(x) with respect to x

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Homework Statement



Suppose I have d/dx f(x) = 0, and I integrate both sides with respect to x. I have worked out two possibilities, and I am not sure which is correct:

1) ∫ d/dx f(x) dx = 0 → ∫ f'(x) dx = 0 → f(x) + C = 0

or

2) ∫ d/dx f(x) dx = 0 → d/dx ∫ f(x) dx = 0 → d/dx (F(x) + C) = 0 → f(x) = 0

where F(x) is the integral of f(x) and f'(x) is the derivative of f(x), both with respect to x.


Thanks.
 

Answers and Replies

  • #2
vela
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Neither looks right to me. When you start with
$$\frac{d}{dx} f(x) = 0,$$ integrating both sides gives you
$$\int \frac{d}{dx} f(x)\,dx = \int 0\,dx.$$ The righthand side isn't necessarily equal to 0.
 
  • #3
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Oh, so the indefinite integral of 0 gives you a constant. So I'm still confused with if the left side simplifies to f(x), or f(x) + a constant.

Besides the fact that I improperly integrated zero, I don't see any errors on the left side of my evaluations. Perhaps they are both correct, but method two gives you more information about the original function f?
 
  • #4
SammyS
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Oh, so the indefinite integral of 0 gives you a constant. So I'm still confused with if the left side simplifies to f(x), or f(x) + a constant.

Besides the fact that I improperly integrated zero, I don't see any errors on the left side of my evaluations. Perhaps they are both correct, but method two gives you more information about the original function f?
It doesn't matter whether the left side gives a constant or not. The right side does give a constant that is not generally the same as the one on the left side. However, it can be shown that only one constant is needed.

Can you show that?
 
  • #5
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Thank you. I can see that the results are equivalent since you can combine the two different constants into one, but wanted to verify my math was correct otherwise.
 
  • #6
WWGD
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Neither looks right to me. When you start with
$$\frac{d}{dx} f(x) = 0,$$ integrating both sides gives you
$$\int \frac{d}{dx} f(x)\,dx = \int 0\,dx.$$ The righthand side isn't necessarily equal to 0.
Doesn't this violate linearity of the integral? If we have ∫0dx , then use 0=0.0 , so

∫0dx =0[∫0dx] =0

Moreover: If we used the perspective of Riemann sums, then ∫0dx is the area under

a curve with height zero. And ∫0dx=∫(c-c)dx=∫cdx-∫cdx.
 
  • #7
SammyS
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Doesn't this violate linearity of the integral? If we have ∫0dx , then use 0=0.0 , so

∫0dx =0[∫0dx] =0

Moreover: If we used the perspective of Riemann sums, then ∫0dx is the area under

a curve with height zero. And ∫0dx=∫(c-c)dx=∫cdx-∫cdx.
This is an indefinite integral (i.e. it's an anti-derivative), it's not a definite integral, so the Riemann sum comment doesn't apply here.
 
  • #8
WWGD
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But how about the linearity part: isn't the indefinite integral linear? I understand

the idea: for any constant function f(x)=c , f'(x)=0 , and the converse, but I think

there are problems with the layout.
 

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