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Question ;Let's suppose I have function f(x) and function u(x)

Now if $\frac{d}{dx}f(x) = q$ and $\frac{d}{dx} u(x)=q$ then this means $u(x)=f(x)$?"

But if I do integral of q would I get $f(x)$ or $u(x)$?

God Bless You.

New question; Let's suppose I have function f(x) and function u(x) with same constant (ex: f(x)=5x+c u(x)=5x+c)

Now if $\frac{d}{dx}f(x) = q$ and $\frac{d}{dx} u(x)=q$ then this means $u(x)=f(x)$?"

But if I do integral of q would I get $f(x)$ or $u(x)$?

God Bless You.

ALSO I WANT TO ACCEPT ANSWER OF AMYWHY BUT I CANT

ChristianPRO
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  • Welcome to MSE! For some basic information about writing math at this site see e.g. here, here, here and here. – user110822 Nov 25 '13 at 12:11
  • Please don't ask new questions, or change a question, once you've received answers. That makes answers look off target. If you want to clarify what you mean, do so below your original question, clearly marking it as an edit: e.g. "Question.... (then) Edit: ...." – amWhy Nov 25 '13 at 12:18
  • Christian, with respect to your most recent edit: to accept the answer, just click on the grey check mark $\large \checkmark$ to the left of the answer you want to accept: it turns green when you click on it. ;-) – amWhy Nov 25 '13 at 12:48

3 Answers3

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It is not true that if $\frac{d}{dx}f(x) = q(x)$ and $\frac{d}{dx} u(x) = q(x)$ then $f(x) = u(x)$.

For example, suppose $f(x) = x^2 + 3$, and $u(x) = x^2 - 7$. Then $f'(x) = u'(x) = 2x$, but $f(x)\neq u(x).\;$

What we can say is that if $f'(x) = u'(x),\,$ then $f(x)$ and $u(x)$ differ by at most a constant.

Note that if you find the indefinite integral of, say $2x$, you'll obtain a family of functions $$\int 2x \,dx = x^2 + \text{any constant}$$ which, again, differ by at most a constant of integration.

So to answer your updated (new) question: In your polynomial example, only if we know that the two functions have exactly the same constant term, then yes, of course, $$f'(x) = u'(x) = 5\;\;\implies f(x) = u(x)$$ In other words, in your example, we essentially already have that $f(x) = u(x) = 5x + c $, so clearly, we'd necessarily have that $f'(x) = u'(x) = 5$, and we'd necessarily have $f(x) = u(x)$ not because their derivatives are the same, but because we started with the exact same functions $f(x) = u(x)$.

With integration, you'd get neither (or both) $f(x)$ and $u(x)$: You'd integrate $\int 5\,dx$ to obtain $5x + \text{any constant}$, which is a specific form of a function to denote that the integral represents the family of all functions that differ only in their constant term, and so both $f(x)$ and $u(x)$ belong the the family of function given by $5x + \text{constant}$.

You might want to read/review what is highly relevant to your question:

The Fundamental Theorem of Calculus.

amWhy
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if $\frac{d}{dx} f(x) = \frac{d}{dx} u(x)$ does not mean $u(x) = f(x)$, it only means they have the same rate of change, e.g.: $f(x) = x$ and $u(x) = x+100$

also if you integrate $q$, you will get both, because you will get $x + c$, where $c = 0$ for $f(x)$ and $c = 100$ for $u(x)$ in my example

see here for more detail

Brett Schneider
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I think that the correct answer depends on the intersection between the domain of $f$ and the domain of $u$. If $E$ is such intersection, then you can deduce that, on every interval $I \subset E$, $f$ and $u$ differ by a constant possibly depending on $I$.

Finally, the indefinite integral of $q$ cosists, by definition, of all the primitives of $q$. In particular, $\int q(x)\, dx$ gives you *both $f$ and $u$. And it gives you any other function that differs from $f$ and $u$ by a constant.

Siminore
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