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I have often read that the Fundamental Theorem of Calculus (FTC) tells us that integration is the opposite of differentiation. I have always found this summary confusing, so I will lay out what I think people mean when they make such a statement.

The First FTC implies the existence of antiderivatives for every function, $f$, that is continuous on a particular interval, say $[a,b]$. Generally, we denote this antiderivative as $F$. Differentiating $F$ gets back to our original function, $f$. So when people say that 'integration is the opposite of differentiation', what they mean is that an antiderivative of a function can be computed using a definite integral.

The Second FTC is more powerful than the First FTC, as it tells us that definite integrals can be computed using the antiderivative of a function (which is generally more useful than knowing that one possible antiderivative of $f$ can be computed using a definite integral, $F$). For the Second FTC, I don't understand how this is related to 'integration being the opposite of differentiation' at all. The Second FTC shows us the link between antiderivatives (indefinite integrals) and definite integrals. It is extremely useful for trying to find the area under a curve, but I'm not sure how this relates to integration and differentiation being 'opposites'.

Is there something about the First FTC or the Second FTC that has a bigger implication about integration being the opposite of differentiation, or is my understanding correct?

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    $\begingroup$ Let $\frac{d}{dx}$ be the derivative operator and let $I$ be the operator that takes a continuous function $f$ as input and returns the function $g(x) = \int_a^x f(t) \, dt$ as output. The first fundamental theorem of calculus tells us that $\frac{d}{dx} \circ I$ is the identity operator. Roughly speaking, integration followed by differentiation leaves us back where we started. $\endgroup$
    – littleO
    Jun 21, 2020 at 10:25
  • $\begingroup$ Also look at the related questions in the right column on the web page (desktop!) Some might be interesting for you. $\endgroup$
    – Henno Brandsma
    Jun 21, 2020 at 10:38
  • $\begingroup$ @littleO Thank you. That makes a lot of sense. One thing that has always bothered me (just a little bit) is that it seems we start off with a function $f(t)$ and end up with $f(x)$. I try to explain it by saying 'it doesn't matter what letter we are using, as we are still performing the same rule', but I wonder if you have a better explanation. $\endgroup$
    – Joe
    Jun 21, 2020 at 13:27
  • $\begingroup$ @Joe Maybe I should have named the derivative operator $D$ and stated that $D \circ I$ is the identity operator. In other words, if $f$ is a continuous function, then $D(I(f)) = f$. $\endgroup$
    – littleO
    Jun 21, 2020 at 22:46

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I think the first FTC:

If $f: [a,b] \to \Bbb R$ is continuous then $F: [a,b] \to \Bbb R$ defined by $F(x)=\int_a^x f(t)dt$ is differentiable and $F'(x)=f(x)$ for all $x \in [a,b]$.

is what people mean by saying the integration (which defines $F$) is the inverse of differentiation (as we have found a function with derivative $f$).

The second FTC

If $f: [a,b] \to \Bbb R$ is Riemann-integrable on $[a,b]$ and we have a function $F: [a,b] \to \Bbb R$ such that $F'(x)=f(x)$ on $[a,b]$, then $\int_a^b f(x)dx=F(b)-F(a)$.

is more of a "recipe" to find an integral: the target is to compute the definite integral and the tool we're given is to find an antiderivative. So not an inverse as such but a method. It's a bit of an iffy one, as an antiderivative $F$ need not exist at all (except when $f$ is continuous and the first FTC gives us one, but not explicitly, but at least we know some solution exists, but we don't have it in computable form yet). I think the first is closer to giving a direct "inverse" connection between integration and differentiation (and is often used in other contexts when we differentiate wrt boundaries of integrals, etc.). But that's just one view.

The first FTC can be summarised as $$\frac{d}{dx}\int_a^x f(t)dt = f(x)$$ so "Applying the integration operator to $f$, followed by the differentiation operator gives us back $f$ again".

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  • $\begingroup$ @Henro Brandsma Thanks, this was a very clear answer. I asked this in another comment, but I was wondering if you could shed some light on this: One thing that has always bothered me (just a little bit) is that it seems we start off with a function $f(t)$ and end up with $f(x)$. I try to explain it by saying 'it doesn't matter what letter we are using, as we are still performing the same rule', but I was wondering if you had a better explanation. $\endgroup$
    – Joe
    Jun 21, 2020 at 16:52
  • $\begingroup$ $x$ is the primary variable for differentiation (in analysis; in physics differentiation to a time variable $t$ (speed, acceleration) is common) in that functions on $\Bbb R$ are by default $f(x)$. But we need a variable too "to integrate along", i.e. the definite integral needs a $f(x)dx$ in that case. But FTC-1 we need the $x$ as a boundary to define a function, so we could have used $y dy$ but that suggests a second dimension and a two-dimensional integral so we use another "linear" variable so $tdt$ (as said, well known from applications, and curves are often so parametrised). $\endgroup$
    – Henno Brandsma
    Jun 21, 2020 at 17:01
  • $\begingroup$ @Henro Brandsma I see. And is there any difference between $f(x)$ and $f(t)$? I think there isn't because $x$ and $t$ are just variables, and it doesn't really matter. What matters is that you get back to your original function, $f$, that has the same domain, codomain, and range as before. $\endgroup$
    – Joe
    Jun 21, 2020 at 17:04
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    $\begingroup$ @Joe No, there is no difference. But it's a matter of clear presentation and what the reader expects. $\endgroup$
    – Henno Brandsma
    Jun 21, 2020 at 17:06
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    $\begingroup$ @Joe So to be very concise, that FTC1 identity can be written as $\displaystyle\frac{d}{dx}\int_a^x f = f(x),$ the understanding being that the variable of integration (bound variable) shall not be $x$ (free variable). $\endgroup$
    – ryang
    Jan 9, 2021 at 10:49

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