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sorry if this has been asked before, but I can't seem to find my question in particular.

Anyway, in the Second FTM it says $$F(x)=\int_a^xf(x)dx$$ If I understand correctly is just the area under the curve. no problem there. Then it says that$$\int_a^bf(x)dx=F(b)-F(a)$$ if I'm thinking this correctly, it would make sense because in $F(a)=\int_a^af(x)dx$ the area is $0$ so I'm just left with $F(b)$ which is the integral from a to b, this is where I think I have to be wrong, because in every example I see, they take the value of $a$ and plug it in $F(x)$. For example in $$\int_2^5x^2dx$$ with $F(x)=x^3/3$ they take $F(5)$ and $F(2)$, in particular $F(2)=2^3/3$, but shouldn't $F(2)$ always be $0$ because it's basically just $\int_2^2x^2dx$?

p.s Sorry in advance for any mistake I made in formating or anything else.

Valus001
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    $\int_a^a f(x) dx = 0$ since $\int_a^a f(x) dx = F(a) - F(a) = 0$ not because $F(a)$ necessarily is equal to zero – eepperly16 Aug 22 '16 at 01:53
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    @eepperly16: The symbol $\int_{a}^{a}f(x),dx$ is defined to be $0$ for any function $f$. This definition has nothing to do with anti-derivatives. – Paramanand Singh Aug 23 '16 at 08:35

3 Answers3

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This kind of confusion is very prevalent and the primary reason behind the confusion is the wrong definition of definite integral as area under a curve.

It is important to first understand that the definition of symbol $\int_{a}^{b}f(x)\,dx$ has nothing do with area as such. The definition has to be based on numbers $a, b$ and the function $f$ defined on interval $[a, b]$. One such definition was provided by Bernhard Riemann and it assumes $f$ to be bounded on $[a, b]$. I will leave the definition of Riemann integral to the standard textbooks of analysis and focus next on the Fundamental Theorems of Calculus.

First Fundamental Theorem of Calculus: If $f$ is bounded on $[a, b]$ and the Riemann integral $\int_{a}^{b}f(x)\,dx$ exists then the function $F$ defined on $[a, b]$ by $$F(x) = \int_{a}^{x}f(t)\,dt$$ is continuous on $[a, b]$ and $$F'(c) = f(c)$$ for any point $c \in [a, b]$ where $f$ is continuous.

Second Fundamental Theorem of Calculus: If $F$ is differentiable on $[a, b]$ and the derivative $F' = f$ (say) is Riemann integrable on $[a, b]$ then $$F(b) - F(a) = \int_{a}^{b}F'(x)\,dx = \int_{a}^{b}f(x)\,dx$$

Note that the first FTC describes a function $F$ based on a given function $f$ which has some nice properties ($F$ is continuous on $[a, b]$ and differentiable at those points where $f$ is continuous). But this $F$ is not an anti-derivative of $f$ on $[a, b] $ because we are not guaranteed that $F'(x) = f(x)$ for all $x \in [a, b]$. We have $F'(x) = f(x)$ only at those points $x$ at which $f$ is continuous. At points where $f$ is discontinuous the function $F$ may or may not be differentiable.

The second FTC deals with anti-derivatives. It says that if $F$ is anti-derivative of $f$ on $[a, b] $ i.e. $F'(x) = f(x)$ for all $x \in [a, b]$ and further that if $f$ is Riemann integrable on $[a, b]$ then the definite integral $\int_{a}^{b}f(x)\,dx$ can be simply evaluated in terms of the difference between values of anti-derivative (i.e. as $F(b) - F(a)$).

The function $F$ used in first FTC has a different role than the function $F$ of the second FTC and it is useless to think of them as same. Things change drastically when the function $f$ (which is to be integrated) is guaranteed to be continuous on $[a, b]$. When $f$ is continuous then both the FTC get merged into one theorem which we can simply call FTC for continuous functions:

Fundamental Theorem of Calculus for Continuous Functions: If $f$ is continuous on $[a, b]$ then $\int_{a}^{b}f(x)\,dx$ exists and the function $F$ given by $$F(x) = \int_{a}^{x}f(t)\,dt$$ is an anti-derivative of $f$ on $[a, b] $. Moreover if $G$ is any anti-derivative of $f$ on $[a, b]$ then $$\int_{a}^{b}f(x)\,dx = G(b) - G(a)$$

And now you see the connection between the word "anti-derivative" and the integral $$\int_{a}^{x}f(t)\,dt = F(x)$$ The function $F$ is an anti-derivative of $f$ and not necessarily the anti-derivative of $f$. You also remember the fact that a function does not have a unique anti-derivative and two anti-derivatives of the same function differ by a constant and this is the reason for using the constant of integration while calculating indefinite integrals. This is also the reason that I have used a different letter $G$ for a generic anti-derivative in the theorem mentioned above and the letter $F$ denotes a very specific anti-derivative represented in the form of a definite integral.

Thus when you wish to calculate $\int_{2}^{5}x^{2}\,dx$ by the use of anti-derivative $x^{3}/3$ you are just choosing one of the infinitely many anti-derivatives available and it will work as expected. You may wish to choose the specific anti-derivative $\int_{2}^{x}f(t)\,dt = (x^{3}/3) - 8/3$ and this will also work fine.

You can also think in this manner. The anti-derivative $F(x) = \int_{a}^{x}f(t)\,dt$ is a very specific one and has the property $F(a) = 0$. Other anti-derivatives of $f$ will not have this property of vanishing at $a$. And the fundamental theorem says that the choice of anti-derivatives does not matter in evaluating the definite integral.

Paramanand Singh
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  • A doubt here. Let us take $e^{x^2}$. The function is continuous and bounded in $[a, b]$. Here, the SFTC proves useless, does it not( the exponential function has no antiderivative )? – R004 Jun 24 '17 at 09:36
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    @R004: the SFTC ensures that $e^{x^{2}}$ has an anti-derivative. But it is another deep theorem of Liouville which says that this anti-derivative can not be expressed using a finite number of arithmetic and algebraic operations as well as composition of functions on algebraic functions, the trigonometric functions (direct as well as inverse) and logarithmic and exponential functions. You can thus not have a formula for it in terms of these functions. In technical language we say that anti-derivative of $e^{x^{2}}$ is a non-elementary function. – Paramanand Singh Jun 24 '17 at 09:42
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    @R004: a simple and familiar example of non-elementary function is $[x] $. You can guess by some trial and error that this function can not be expressed as a finite formula consisting of algebraic, trigonometric, exponential and logarithmic functions. – Paramanand Singh Jun 24 '17 at 09:46
  • With my level of understanding, I have made an attempt to dig deeper into the Fundamental Theorem of Calculus, Here. I hope this is acceptable. – R004 Jun 24 '17 at 09:47
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    @R004 : I have seen that question of yours. You are unnecessarily trying to complicate things by taking the limits of integration as functions $a(x), b(x) $. This is not needed at all. It is best to keep one limit as constant and other limit of integral as variable. Apart from that everything is fine in that question. – Paramanand Singh Jun 24 '17 at 09:50
  • Noted. This helps. – R004 Jun 24 '17 at 09:55
  • You stated, "It says that if $F$ is anti-derivative of $f$ i.e. $F'(x) = f(x)$ for all $x \in [a, b]$ and further that if $f$ is Riemann integrable on $[a, b]$ then the definite integral..." Doesn't $F'(x) = f(x)$ itself imply that $f$ is Riemann integrable? Was it necessary to further state if $f$ is Riemann integrable or not? – R004 Jun 26 '17 at 13:54
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    @R004: No! Not every derivative is Riemann integrable. In fact there are bounded derivatives which are not Riemann integrable. There are two broad popular classes of Riemann integrable functions (these are not exhaustive) : monotone functions and continuous functions. And a derivative is not necessarily monotone or continuous. So we may expect that there is some $F$ such that $F'$ is not Riemann integrable. An example of such an $F$ is difficult. – Paramanand Singh Jun 26 '17 at 13:59
  • @R004: I don't know if you can understand but an example of such an $F$ is given in this answer https://math.stackexchange.com/a/2295796/72031 but with less details. Let me know if you need more detail. – Paramanand Singh Jun 26 '17 at 14:02
  • @R004 : also see wiki reference https://en.wikipedia.org/wiki/Volterra%27s_function – Paramanand Singh Jun 26 '17 at 14:05
  • This probably is beyond what I can comprehend. I might get back to this once I touch upon advanced concepts of mathematical physics. – R004 Jun 26 '17 at 14:32
  • So, if $F'(x) =f(x) $ is not Riemann integrable, how do we compute $F(x) $ given $f$ and the bounds? Could you also talk about the Fundamental Theorem of Calculus here? – R004 Jun 27 '17 at 02:12
  • @R004 : in general evaluation of anti-derivative is worthwhile only if it is expressible in terms of elementary functions. In this case we are not dealing with elementary functions anyway. Moreover when $F'=f$ is not Riemann integrable then the equation of second FTC $\int_{a} ^{b} f(x) , dx=F(b) - F(a) $ does not make sense. There is more general theory of integration called Lebesgue integration where we have the result: If $F$ is differentiable on $[a, b] $ with a bounded derivative $f$ then $f$ is Lebesgue integrable on $[a, b] $ and $\int_{a} ^{b} f(x) , dx=F(b) - F(a) $. – Paramanand Singh Jun 27 '17 at 06:30
  • Let us continue this discussion in chat. – R004 Jun 27 '17 at 14:02
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There seems to have been a mix-up in what the letter $F$ refers to in each case.

What the following equation means $$F(x) = \int_a^x f(x)\, dx$$ is that $F$ is a particular choice of antiderivative of $f$; that is, $F$ is a function such that $F'=f$, and moreover $F(a)=0$ for this choice of antiderivative.

However, the following equation $$\int_a^b f(x)\, dx = F(b)-F(a)$$ is true for any choice of antiderivative $F$; that is, any function $F$ for which $F'=f$ will suffice. Antiderivatives are only defined up to a constant of integration, but the constant of integration cancels out since you're subtracting one from another.

In the first of these equations $F$ is a specific choice of antiderivative; whereas in the second equation, the constant of integration can be anything.

To resolve your issue in the case where $a=2$, $b=5$ and $f(x)=x^2$, note that $$\int_2^x f(x)\, dx = \frac{x^3}{3} - \frac{8}{3}$$ so in this case, choosing $F(x) = \frac{x^3}{3} - \frac{8}{3}$, rather than simply $\frac{x^3}{3}$, gives you what you want.

Clive Newstead
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  • The confusion over F is so common on this site – operatorerror Aug 22 '16 at 01:56
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    @qbert: Indeed, and in textbooks, and schools, and ... :P – Clive Newstead Aug 22 '16 at 01:57
  • SO $F(x)$ is always $F(a)-F(b)$ but in the first case we choose the constant ($F(a)$) to be 0?

    Also, if $F(a)$ is the constant, is there a particular reason to choose it to be $F(a)$ or could it be anything else?

     – Valus001 Aug 22 '16 at 02:13
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    @Valus001: Absolutely not! The definite integral $\int_a^b f(x), dx$ is equal to $F(b)-F(a)$ for any choice of antiderivative $F$. But if you write $F(x) = \int_a^x f(x), dx$ then you're picking a particular choice of antiderivative, namely the one whose constant of integration corresponds to choosing the lower limit of the integral to be equal to $a$. – Clive Newstead Aug 22 '16 at 02:16
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    The confusion arises because the letter $F$ in each case refers to (potentially) different things. – Clive Newstead Aug 22 '16 at 02:16
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    ...so if you instead choose to write $G(x) = \int_a^x f(x), dx$, then you can safely say that $G(x) = F(x)-F(a)$ for any choice of antiderivative $F$. – Clive Newstead Aug 22 '16 at 02:17
  • So what we do is choose $F(x)$ to have $F(a)$ as it's constant? Is that just for simplicity or any other reason? – Valus001 Aug 22 '16 at 02:27
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    Not quite. Let's put it this way: when you write the indefinite integral $\int f(x), dx$, you typically write something with "$+C$" at the end. The point is that when you fix the lower limit at some constant $a$, this determines the constant of integration $C$. Half of the fundamental theorem of calculus says that $\int_a^x f(x), dx$ is an antiderivative of $f$; but so are all other functions equal to $\int_a^x f(x), dx + C$ for any constant $C$. The second part oft he theorem says that, no matter which value of $C$ you pick (zero or otherwise), we have $\int_a^b f(x), dx = F(b) - F(a)$. – Clive Newstead Aug 22 '16 at 02:44
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    Hmm I think I get it now, or at least much better than before, it seems I had the definitions of the $F$'s kind of mixed up. I'll keep rereading what you wrote to understand it better, but I think you cleared most of the mess I had, Thanks for your patience! – Valus001 Aug 22 '16 at 02:51
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Here are some aspects which might be helpful

  • Be careful about not mixing up in the first part of FTC the variable of integration with the variable $x$. \begin{align*} F(x)=\int_a^xf(t)\,dt\tag{1} \end{align*} The variable $t$ in (1) is a bound variable similar as the index $k$ in a sum $\sum_{k=0}^nf(k)$, while $x$ is a free variable.

  • The function $F$ is not the area under the curve. The area is a number while $F$ is not a number but a function in $x$.

    But, when we evaluate the function $F$ at a specific value $x=b$ we obtain the number \begin{align*} F(b)=\int_{a}^bf(t)\,dt+F(a) \end{align*} resp. \begin{align*} \int_{a}^bf(t)\,dt=F(b)-F(a)\tag{2} \end{align*} and $F(b)-F(a)$ in (2) can be interpreted as (signed) area under the curve in $[a,b]$.

Markus Scheuer
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