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This might sound very useless but I'd like to see what you think. Bear in mind that I'm just a novice student.

if $f$ is the original function, then it could be found this way

$C+\int f'(x)\, dx=f(x)$

I understand this is equivalent to saying $\int f '(x)\, dx=f(x)+c$, but this way gives rise to the wrong interpretation!

If $\int f '(x)\, dx$ means the sum of all infinitesimally small increments, it is impossible that if you take the sum of those increments you'll get the original function! You'll only get the original function MINUS some constant inherent to that function. I guess this is a trivial matter but what do you think? Interpreting the indefinite integral is really making my head hurt, how do you interpret it?

edit: The wrong interpretation is that the indefinite integral gives you the original function. Which is what my teachers have taught all along.

Thanks.

Adam Hughes
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DLV
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  • $\int{f'(x)dx}$ is that function so that if you differentiate you get $f'(x)$ which is $f(x)+C$ (or minus C if you wish) not just the original function. The constant is not inherent to the function, it is truly arbitrary. – Paul Sundheim Jul 26 '14 at 19:12
  • You constant can be positive or negative. You can just as well write $-C$ on the RHS if you wish. – John Machacek Jul 26 '14 at 19:12
  • Yeah I understand that. The wrong interpretation is "the indefinite integral will give you the original function". – DLV Jul 26 '14 at 19:16
  • http://math.stackexchange.com/questions/877907/is-this-proof-of-the-fundamental-theorem-of-calculus-correct –  Jul 26 '14 at 19:16
  • @David, "the indefinite integral will give you a set of functions, among which is the original function" or "the indefinite integral will give you the original function, if you pick the right constant". – Karolis Juodelė Jul 26 '14 at 19:23

2 Answers2

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First, $\int f'(x)dx$ does not mean "the sum of all infinitesimally small increments". That is what $\int_a^x f'(t)dt$ means. The former means "an anti-derivative of the derivative of f".

The source of the constant comes the $a$ in the following integral.

$$\int_a^x f'(t)dt = f(x)+C(a)$$ where $C = -f$

John Joy
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  • I see that this is the convention. But why use the symbol ∫ at all, if the idea was that it implied 'summa' or sum? Also, I think its contradictory that ∫f(x)dx has a constant added to it, because then ∫f(x)dx will not be equal in all cases. – DLV Jul 27 '14 at 01:27
  • I think that you make a good point, but I also believe that your question presupposes an intentional (and temporary) ignorance of the Fundamental Theorem of Calculus. – John Joy Jul 27 '14 at 12:05
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Nice question. The way I look at the indefinite integral's constant of integration (which seems to be the real crux of your question) is that the constant of integration is really a function that does not depend on the variable integrated with respect to. This "function" $C$ is the component of the original function $F(x)$ which does not depend on $x$ This really comes into play with the work of multivariable calculus.

Think about it this way. A multivariable function depends on more than one variable: $$z=x+y$$ http://www.wolframalpha.com/input/?i=z%3Dx%2By

The slope of this plane depends on both $x$ and $y$. As a result, if we differentiate $z$ with respect to $x$ - $y$ is considered totally independent of $x$ -, then the resultant slope - in the x direction is: $$\frac{\partial z}{\partial x}=1$$ If we integrate this function $\frac{\partial z}{\partial x}$ with respect to $x$, we will not get the original function - as you mentioned. $$\frac{\partial z}{\partial x}= 1 \rightarrow \partial z = \partial x$$ $$\int \partial z = \int \partial x \rightarrow z = x$$ Notice how we are completely missing the $y$ term! This is because we have a component of the original function that is not dependent on $x$. Thus, to note that there might be something missing, we insert an arbitrary function $C_1$ with is not dependent on the variable of integration.

In this 2D example, $C_1$ is a function of $y$. Thus, the true expression from the integrals above is: $$z = x + C_1(y)$$ In the case of 1D, which is what you are learning about, the only component of the original function $z$ that can be independent of $x$ is a constant number, so we remove the $(y)$ designation. I.E. $$C_1(y)=C$$ Thus, our simplified expression is: $$z=x+C$$

FundThmCalculus
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  • If an integral means antiderivative then I agree we should add '+C' because that was lost in the process of deriving. But if the integral means the sum of all infinitesimally small increments (f'(x)dx) then that sum gives you a change, not the original function itself, and no '+C' term should be added. Thanks for replying btw. Saying ∫dx=x+C is I think ridiculous, and the more honest answer is =(X-Xo) – DLV Jul 26 '14 at 20:13
  • When deriving the regular kinematics equations we usually say dv/dt=a ∫adt=∫dv=(v-vo) and thus we refer only to a change and proceed onwards. – DLV Jul 26 '14 at 20:38
  • I think the indefinite integral is the equivalent to writing a summation without any boundaries at all. Which is done to imply "you know what to insert as limits" . – DLV Jul 26 '14 at 20:47