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I'm confused about the two parts of the Fundamental Theorem of Calculus, as I feel these two parts are somewhat contradictory?

The first part states that: $$ F(x)=\int_{a}^{x}f(t)dt $$

Whereas the second part states that: $$ F(b)-F(a)=\int_{a}^{b}f(t)dt $$

Well if the second part is true in evaluating the integral, then shouldn't the first part be: $$ F(x)-F(a)=\int_{a}^{x}f(t)dt $$

And this is how it works. Like, if I'm evaluating x=2 then I would calculate the area from a to 2, and use F(2)-F(a), for some value a.

So what exactly is the first part of the fundamental theorem of calculus telling us? Looking at the second part, I can't wrap my head on the purpose of the first part, and even to me it seems wrong, that it should be written as: $$ F(x)-F(a)=\int_{a}^{x}f(t)dt $$

LukeWu
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    The first part of the fundamental theorem how you defined it is the assumption in the fundamental theorem, not the conclusion. It says if $F(x)$ is defined how you define it, then $F$ is uniformly continuous on $[a,b]$, differentiable on $(a,b)$, and $F'(x) = f(x)$ for $x \in (a,b)$. – Ryan Shesler Apr 20 '20 at 23:32
  • The first part of what states the above? – copper.hat Apr 20 '20 at 23:33
  • I think the first equation is for indefinite integrals. –  Apr 20 '20 at 23:35
  • 2nd equation is for definite integrals when you want to find the area under the graph. –  Apr 20 '20 at 23:37
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    Those aren't contradictory. If $F(x)=\int_a^b f(x)dx$ then $F(b) -F(a) = \int_a^b f(x)dx - \int_a^af(x)dx = \int_a^b f(x)dx - 0 =\int_a^bf(x)dx =F(b)$. But that's not what the issue is. – fleablood Apr 20 '20 at 23:39
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    Please actually state the fundamental theorem of calculus that you are using. In "the first part" is $F(x)=\int_a^xf(t)dt$ a definition or a conclusion. If it's a conclusion then how was $F(x)$ defined in the first place. In the "second part" who are $F(a)$ and $F(b)$ being defined? – fleablood Apr 20 '20 at 23:42
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    The $F$ in both parts is not the same function and treating them as same is probably the source of your confusion. Both parts are actually independent theorems with different hypotheses. See here for details. – Paramanand Singh Apr 22 '20 at 12:27

4 Answers4

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What you call the “First part” is usually called the second part. And it doesn’t state what you say. What is states is that if $f(t)$ is continuous on $[a,b]$, and we define the function $F(x)$ by the formula $$F(x) = \int_a^x f(t)\,dt$$ then $F(x)$ is differentiable, and is in fact an antiderivative for $f$; that is, that $$\frac{d}{dx} F(x) = \frac{d}{dx}\int_a^x f(t)\,dt = f(x).$$ That’s what that part states.

What you call the second part, which is usually called the first part, also doesn’t quite say what you are writing. What it says is that if $f(x)$ is continuous and $G(x)$ is any antiderivative for $f(x)$ on $[a,b]$, then $$\int_a^b f(t)\,dt = G(b)-G(a).$$

Now, you could try to use the funtion $F(x)$ in your “first part”, but it doesn’t actually give you anything useful: it just says $$\begin{align*} \int_a^b f(t),dt &= F(b)-F(a)\\ &= \int_a^b f(t)\,dt - \int_a^a f(t)\,dt\\ &= \int_a^b f(t)\,dt \end{align*}$$ which doesn’t really give you a way to compute the integral.

The “Second part” (usually called the First Part) says:

If you have an antiderivative of the continuous function $f(x)$, then you can use the antiderivative to calculate the integral $\int_a^b f(t)\,dt$

(though implicit is that you need to be able to compute values of the antiderivative independently of the integral...)

The “First part” (usually called the Second Part) says:

If $f(x)$ is continuous, then it definitely has (at least one) antiderivative.

This tells you that the first part is not an empty promise: if you can find a (useful, independently calculable) antiderivative, then you are set, and antiderivatives exist to be found. However, the antiderivative it shows you is not a useful antiderivative as far as the first part is concerned.

Arturo Magidin
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  • Thanks, this cleared it up somewhat. The other problem I was having where you tried to use the function to compute the integral. This was in the textbook as a proof of the second part, and I just didn't get how F(a)=0. I think I'm mixing it up, but that to evaluate this integral, you would first find the antiderivative (which is F(x))and sub in F(b) and F(a) and subtract the two. But then F(a) wouldn't be 0. – LukeWu Apr 21 '20 at 00:07
  • @Hugetanks: use a single dollar sign for in-line formulas. As to why $F(a)=0$, again, because $F(x)$ is defined to be $\int_a^x f(t),dt$, then by definition $F(a)=\int_a^a f(t),dt$, and you should have already shown as part of the basic properties of definite integrals that the integral from $a$ to $a$ of a function is $0$, because there is no area in a line segment. – Arturo Magidin Apr 21 '20 at 00:09
  • @Hugetanks: To evaluate the integral, technically, you need to do a limit of Riemann sums. The “magic” of the Fundamental Theorem is that it tells you that if you can find an antiderivative of $f(x)$ that you can evaluate separately (without using the integral), then you can use the antiderivative to calculate the integral. The second part tells you that every continuous function has an antiderivative, though it does not give you a useful antiderivative (in the sense of being an antiderivative you can use to easily calculate the integral). – Arturo Magidin Apr 21 '20 at 00:11
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    @Hugetanks: Another way to think about the FTC is that it tells you that even though the derivative and the integral are defined to solve very different problems (the tangent problem and the area problem, respectively), they are actually intimately connected to one another and are kind of, but not exactly, inverse operations of each other, in that the derivative on a certain integral gives you the function (your “first part”), and the integral of the derivative can be calculated using the function: $\int_a^b (\frac{d}{dt}f(t)),dt = f(b)-f(a)$. – Arturo Magidin Apr 21 '20 at 00:12
  • Thanks for the tip, and I do understand that there is no area under this segment, but if $F(x)$ is the antiderivative of the function $f(x)$, then plugging a in $F(x)$ should yield $0$, which I'm not getting. Possibly the antiderivative and the $F(x)$ I'm alluding to are different or is there something I'm not getting? – LukeWu Apr 21 '20 at 00:12
  • @Hugetanks: First, there is not such thing as “the antiderivative” (singular definite article). Every function that has one antiderivative has infinitely many antiderivatives. And I have no idea what “plugging in $F(x)$” means. Plugging into what? What should yield $0$? – Arturo Magidin Apr 21 '20 at 00:15
  • @Hugetanks: And if you are still this confused, why did you accept the answer? Clearly it didn’t answer your question! – Arturo Magidin Apr 21 '20 at 00:15
  • This answer helped clear my initial question that I posted about, but I was then confused about the proof. I think writing an example will maybe show where I go wrong? Like if the integral is $\int_{-1}^{3}(x^3+1)dx$ then $F(x)$ would be $(x^4)/4+x$, and then I would plug $3$ as $x$ into $(x^4)/4+x$, and plug $-1$ into $(x^4)/4+x$, and then subtract the two. In this case plugging $-1$ into $(x^4)/4+x$, would be the $F(a)$. – LukeWu Apr 21 '20 at 00:21
  • @Hugetanks: Again: you are not understanding the theorem. For $\int_{-1}^3(x^3+1),dx$, what you call the second part says “If you can find a function $G(x)$ with $G’(x) = x^3+1$, then $\int_{-1}^3(x^3+1),dx = G(3)-G(-1)$. The function $G(x) = \frac{1}{4}x^4 +x$ is one such function, so you can use that $G$ to calculate $\int_{-1}^3 (x^3+1),dx$. it does not say that $F$ “is” that function. Doesn’t say that at all. (cont) – Arturo Magidin Apr 21 '20 at 00:24
  • @Hugetanks: What you call the first part says: If you define $F(x)$ as $F(x) = \int_{-1}^x (t^3+1),dt$ for $-1\leq x\leq 3$, then $F’(x) = x^3+1$. But $F(x)$ is defined by the integral. If you use the function $G$ to calculate $\int_{-1}^x (t^3+1),dt$, you get $F(x) = G(x)-G(-1) = (\frac{1}{4}x^4 + x) - (\frac{1}{4} - 1) = \frac{1}{4}x^4 + x +\frac{3}{4}$, and you can easily verify that given that $F(x) = \frac{1}{4}x^4+x+\frac{3}{4}$, it is in fact true that $F’(x) = x^3+1$, which is what the first part says. – Arturo Magidin Apr 21 '20 at 00:27
  • Alright thank you, I felt that this was where I was getting confused due to the lettering of $F(x)$ and this cleared it up. – LukeWu Apr 21 '20 at 00:28
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The $F$ or the "first part" is actually not the same $F$ as the $F$ of the second part.

...

The gyst of the FTC is that integration and derivation are inverses of each other.

The subtlety in putting this in plain words is distinguishing between "derivatives" as general functions vs. derivatives as specific values about the function at certain values of $x$, and distinguishing between "integrals" as a general indefinite function, vs. a definite integral evaluated between two specific points.

The first part of the theorem is trying to basically say if $F(x) = \int f(x)dx$ then $F'(x) = f(x)$. ... Except $F(x) = \int f(x)dx$ doesn't actually make any sense.

So if $f$ is an intergratable function, and we define a function as $F(x) = \int_a^x f(t)dt$, that is, $F(x)$ for any specific value of $x$ would be evaluated as the integral evaluated from some fixed point $a$ to $x$ were $x$ treated as a specific value.

Then the theorem states $F'(x) = f(x)$.

In other words: the derivative of the integral is the function itself.

....

The "second" part is the other way around. The integral of the derivative is the function itself.

If we start with a function $\mathscr F$ and we know that $\mathscr F' = f$. Then if you integrate $f$ between two points $a,b$ we get $\int_a^b f(x)dx =\mathscr F(x)|_a^b = \mathscr F(b) -\mathscr F(a)$

....

If it were up to me, I'd define the FTC as this.

If $f$ is integratable and we define $G(x) = \int_a^x f(t)dt$ then $G'(x) = f(x)$. (The derivative of the integral is the function.)

ANd if $f$ is differentiable, then $\int_a^b f'(x)dx = f(x)|_a^b = f(b) - f(a)$. (The integral of the derivative is the function.)

fleablood
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One of the fundamental theorems of calculus states that the function $F$ defined by $$ \tag{1} F(x) = \int_a^x f(t) \, dt $$ is an antiderivative of $f$ (assuming that $f$ is continuous).

Since $F$ is an antiderivative of $f$, you are correct to note that the other fundamental theorem of calculus implies that $$ \tag{2} \int_a^x f(t) \, dt = F(x)- F(a). $$ But this does not contradict (1), of course, because $F(a) = 0$.

littleO
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    I disagree with your implication of the first part; for one thing, if you unpack it, what you are saying is that the first part of the fundamental theorem says that $\int_a^x f(t),dt = \int_a^xf(t),dt$, which seems a silly thing to have a theorem state.... – Arturo Magidin Apr 20 '20 at 23:55
  • @ArturoMagidin I don't understand your comment; which implication do you disagree with? – littleO Apr 21 '20 at 00:00
  • I’m saying that claiming that the implication of the (what is commonly called the First Part) FTC yield that $\int_a^x f(t),dt = F(x)-F(a)$ is remarkably silly, because given that $F$ is defined the way it is defined in (1), all you are saying is that $\int_a^x f(t),dt = \int_a^x f(t),dt$. – Arturo Magidin Apr 21 '20 at 00:00
  • @ArturoMagidin But OP's question was precisely that OP thought (2) contradicts (1), so I was stating that while (2) is indeed a consequence of the other fundamental theorem of calculus, as OP argued, it is not true that (2) contradicts (1). Anyway, there's an art to guessing what comments will clarify OP's confusion and it's possible I missed the mark this time. – littleO Apr 21 '20 at 00:03
  • Well, OP got the statements of both parts wrong, so granted, trying to guess what will solve the confusion is not trivial. – Arturo Magidin Apr 21 '20 at 00:07
  • @ArturoMagidin: I think this answer is trying to point out that there is a part of theorem which guarantees the anti-derivative$F$ of $f$ and that anti-derivative can also be used in second part of the theorem. For continuous integrands both parts of the theorem are equivalent as one can be used to prove another. That's why it's best to deal with general Riemann integrable integrands and then these two parts are really independent theorems and one can't be used to prove another. – Paramanand Singh Apr 22 '20 at 12:38
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You have the order of the theorems backwards.   Also, they are not referring to the exact same functions.

https://mathworld.wolfram.com/FundamentalTheoremsofCalculus.html

The first fundamental theorem of calculus states that, if $f$ is continuous on the closed interval $[a..b]$ and $F$ is the indefinite integral of $f$ on $[a..b]$, then $$\int_a^b f(x)~\mathrm d x=F(b)-F(a)\tag 1$$

This refers to an $F$ which is an indefinite integral of a given function $f$.

The second fundamental theorem of calculus holds for $f$, a continuous function on an open interval $I$, and $a$ any point in $I$, and states that if $F$ is defined by $$F(x):=\int_a^x f(t)\mathrm d t\tag 2$$

then $$F'(x)=f(x)\tag 3$$

at each point $x$ in $I$.

This refers to an $F$ which is a definite integral of function $f$ over $[a..x]$.

I usually recall it as simply:

$$f(x)=\dfrac{\mathrm d ~~}{\mathrm d x}\int_a^x f(t)~\mathrm d t$$

Community
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Graham Kemp
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    Ugh; “the indefinite integral” is not a function, it is a family of functions, so talking about it as if it were a single function, and a singularly determined one at that, is... a disturbance in the force. – Arturo Magidin Apr 20 '20 at 23:52
  • From what I understand, the first fundamental theorem of calculus tells you how to evaluate the integral, and the second theorem states F(x) as an actual function describing this integral? – LukeWu Apr 20 '20 at 23:52
  • @Hugetanks: No. The first part tells you that you can use any antiderivative to evaluate the definite integral (under certain conditions), and the second part tells you that $F(x)$ is an antiderivative; the fact that $F(x)$ is a function follows just from the fact that the integral exists and is a number. – Arturo Magidin Apr 20 '20 at 23:54