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I am struggling with assessing the validity of this statement.

$$\int ^{x}_{a}f'\left( t\right) dt \neq f\left( x\right) $$

I can understand that the left side yields a class of functions $F(x)$ whose derivative is $f(x)$, but doesn't that mean that the left side evaluates to $f(x) + C$ and that constant pairs with whatever constant exists in the original $f(x)$? For example, if $f(x)=2x+6$ then the antiderivative is $2x + C$ but $C$ here is just $6$, right? So doesn't the equality hold?

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3 Answers 3

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Assuming $f$ is differentiable, then the fundamental theorem of calculus says

$$\int_3^xf'(s) \, ds=f(x)-f(3)$$

Hence, unless $f(3)=0$, the integral expression is not $f(x)$.

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  • $\begingroup$ Thank you! This makes a lot of sense. I didn't think to be looking at FTC part 2. $\endgroup$
    – rb612
    Commented Dec 21, 2016 at 4:53
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    $\begingroup$ To be pedantic, it's not enough just to assume that $f'$ exists. (Counterexample: Volterra's function, mentioned already in another comment.) One must also assume that $f'$ is integrable. $\endgroup$ Commented Dec 21, 2016 at 8:08
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    $\begingroup$ You might want to change the integration variable from $x$ to anything else. $\endgroup$
    – gented
    Commented Dec 21, 2016 at 10:12
  • $\begingroup$ @HansLundmark - to be more pendantic, you are assuming the question is about Riemann integrability, which is not stated. If $f'$ exists, then it is automatically Henstock-Kurzweil integrable. $\endgroup$ Commented Dec 21, 2016 at 17:37
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    $\begingroup$ @TheGreatDuck - "Integration" is not a technique for finding anti-derivatives. It is a concept in and of itself. The FTC proves that you can use anti-derivatives to find the values of integrals and use integrals to find anti-derivatives, but many functions are integrable that do not meet the demands of the FTC and so cannot be determined from anti-derivatives. The theory of integration is far larger than anti-derivatives. The statement in the post is not about anti-derivatives, it is about integration, and so the method of integration determines whether it is true. $\endgroup$ Commented Dec 22, 2016 at 3:51
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Actually, I disagree with your statement 'the left side yields...'

You are talking about indefinite integrals, but here you have a definite integral. In particular, you have $$ \int_3^x f'(x)\,dx=g(x)-g(3), $$ where $g$ is any antiderivative of $f'(x)$. In particular, we know that all antiderivatives of $f'(x)$ are of the form $f(x)+C$ for some constant $C$, so that $$ \int_3^x f'(x)\,dx=[f(x)+C]-[f(3)+C]=f(x)-f(3). $$ So, your question boils down to this: is $f(x)=f(x)-f(3)$ true for all $x$? The answer will depend on the particular value that your function $f$ assigns to the input $3$.

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  • $\begingroup$ Ah this makes a lot of sense! What if the left side was instead the indefinite integral? $\endgroup$
    – rb612
    Commented Dec 21, 2016 at 4:53
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    $\begingroup$ Then the question cannot be answered, for much the reason you suggested: the indefinite integral gives you an entire class of functions, not a single function. The best you can say is that $f(x)$ is a member of that class, when you take the constant $C=0$. $\endgroup$ Commented Dec 21, 2016 at 4:54
  • $\begingroup$ thank you so much! Final question - how come you need to have the constant be 0 in order for it to be a member of the class of antiderivatives? $\endgroup$
    – rb612
    Commented Dec 21, 2016 at 4:55
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    $\begingroup$ That's not what I meant -- I meant that it IS a member of the class, witnessed by taking $C=0$. $\endgroup$ Commented Dec 21, 2016 at 4:55
  • $\begingroup$ okay I think I'm understanding, so if I have the function that I used in my example $f(x)=2x+6$ and now I use the indefinite integral on the left side, I get $2x+C$ so choosing a $C$ here of $6$ will show at f(x) is in the set of functions which differ by a constant? $\endgroup$
    – rb612
    Commented Dec 21, 2016 at 4:59
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In addition to the excelent answers already given, there are a few subleties one should explicitly point out.

There are two main concepts for integration, the first being indefinite integration, that is finding the antiderrivative of a function, and the second is definite integration, finding the measure of the (signed) area enclosed by the graph of a function and the x-axis. There are various ways to codify these concepts in a rigourous mathematical language.

For exaple, if we know that $f$ is a real function over $[a, b]$, $a,b \in \mathbb{R}$, we use Riemann's definition of $$\int_{a}^b f(x) \, \operatorname{d}\!x \ ,$$ which can be found in any elementary calculus textbook. Provited that $f$ meets some specific conditions, we say that $f$ is Riemann-integrable over $[a,b]$ and we assign the above symbolic expression a unique real value. (There is also a theorem according to which if $f$ is Riemann-integrale over $[a,b]$, it's also Riemann-integrable over any closed subinterval of $[a,b]$).

But what about indefinite integration? Remember that, given a real function $h$ over an interval $I$, function $H:I \to \mathbb{R}$ is called an antiderivative of $h$ just in case $H'=h$. You shall note that whenever $H$ is an antiderivative of $h$, $H+c$ is likewise an antiderivative of $h$, and you can prove that any antiderivative of $h$ is of the form $H+c$ for some constant c.

Now, get ready for the hard truth...

While acknowledging that there is an infinite number of antiderivatives for $h$ (all functions $H+c$), mathematicians go mad, break their own rules, and proudly state: $$\int h(x) \, \operatorname{d}\!x = H(x),$$ without any clue of shame! Sometimes, we even go as far as claiming that ‘the indefinite integral of $h(x)$ is H(x) for all $x \in I$’! There is not really such a thing as the indefinite integral of a function, for there is not such a thing as the antiderivative of a function; there are uncountably many of them. This is a clear abuse of notation/terminology, but due to historical reasons, it is accepted as the standard to this day.

To illustrate the above mentioned facts, some books prefer to write $$\int h(x) \, \operatorname{d}\!x = H(x)+c,$$ and say that $c$ is an arbitrary constant, named constant of integration. This is a good way to remind us that if we have an integral equation like this: $$\int h(x) \, \operatorname{d}\!x + x^2 = 6\ln x + 42,$$ we get the equivalent statement $$\exists c \in \mathbb{R} \quad H(x) + c + x^2 = 6\ln x +42.$$

$$\star \ \star \ \star$$

Now you may understand what exactly that peculiar constant is, and why it's not supposed to much a specific function, as you say in your question. There are not constants ‘existing’ in integrable functions!

I'll finish this long (too long?) post by noticing that the integral $$\int_{a}^x f'(t) \, \operatorname{d}\!t$$ in question is not just a definite integral (definite integrals equal a real number) but a function $$F(x) = \int_{a}^x f'(t) \, \operatorname{d}\!t, \ x \in D,$$ where $f'$ should be Riemann-integrable over $D$ (recall also that $a \in D$, and $D$ should be a closed and bounted interval). After all this, you can use the Fundamental Theorem of Calculus and find that $$F(x) = f(x) - f(a), \ x \in D,$$ as already mentioned by others.

I know I wrote a lot, but I think this helps the OP figure out some usual misconceptions that underlie her/his question. Any comments/corrections welcome!

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