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

brushman

## 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.

Staff Emeritus
Homework Helper
$$\frac{d}{dx} f(x) = 0,$$ integrating both sides gives you
$$\int \frac{d}{dx} f(x)\,dx = \int 0\,dx.$$ The righthand side isn't necessarily equal to 0.

brushman
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?

Staff Emeritus
Homework Helper
Gold Member
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?

brushman
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.

Gold Member
$$\frac{d}{dx} f(x) = 0,$$ 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.

Staff Emeritus
Homework Helper
Gold Member
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.