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What is indefinite integral? This is the question that always perplexes me. First my book wrote that

Indefinite integral of $f(x)$ is $F(x)$ if on differentiation, it gives $f(x)$. In fact it is the family of functions that give rise to $f(x)$ on differentiation. It is represented by $\int f(x)dx$

Hmmm... In a word , the book is saying $F(x)$ is an equation.

But in Fundamental Theorem of Calculus(in the Differential and Integral Calculus by Richard Courant and ThomasCalculus), it written that

In $F(x) = \int_a^x f(t)dt$ , $F(x)$ is a function of the upper limit and is defined as an indefinite integral of $f(x)$ . $F(x)$ represents the area between $a$ and $x$ under the curve $y = f(t)$ .

So, here $F(x)$ is defining an area between $a$ and $x$!!

So, which is true? What is indefinite integral representing - function or area? Why are the two definitions different? What do they mean? Please help.

  • To me, both are correct! My book is telling about all the family of functions whose derivative is $f(x)$ and thus represent them as $\int f(x)dx$ ... –  Nov 19 '14 at 03:22
  • Does your book (which book?) really say it is a family of equations? I think it should say it is a family of functions. – miracle173 Nov 19 '14 at 03:29
  • Yes,yes, the book(by R.D. Sharma) wrote about family of equations.... What Richard Courant told about indefinite integral is more accurate. It is the function that represents the area between $a$ and $x$ and hence is a function of the upper limit. It is indefinite in the sense that it is the area between a fixed and a variable point,right?? –  Nov 19 '14 at 03:40
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    The Fundamental Theorem of Calculus tells us that both definitions are equivalent—that is, both definitions describe the exact same thing. – Akiva Weinberger Nov 19 '14 at 04:23
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    Often, in math, there are two possible definitions for something. For example, a scalene triangle can be defined either as "a triangle whose sides are all different" or as "a triangle whose angles are all different." It's not too difficult to prove that both of these definitions describe the same thing. Therefore, if you want to define a scalene triangle, you have the freedom to choose which definition you want. A similar thing happens here. – Akiva Weinberger Nov 19 '14 at 04:26
  • @columbus8myhw: Yes, sir, you are totally right. But can you tell how the two definitions of indefinite integral are same??? –  Nov 19 '14 at 05:22
  • @user36790 So, in other words, you want us to prove the Fundamental Theorem of Calculus for you? – Akiva Weinberger Nov 19 '14 at 11:56
  • @columbus8myhw: No, no, sir! I just requested you to tell me the relationship between the definitions like you wrote about the scalene triangle. If you help, I will be very grateful to you!!:) –  Nov 19 '14 at 12:11
  • In a word, the first book is not saying that the indefinite integral is an equation; it's saying that it's a set. – Toby Bartels Feb 21 '18 at 16:01

5 Answers5

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An indefinite integral is just an antiderivative of a function. For example, if $f(x)=2x$ then the indefinite integral $\int f(x)dx=x^2+C$ where C is some arbitraty constant, because $\frac d{dx} (x^2+C) = 2x$. The fundamental theorem of calculus states that the derivative of the area function shown is equal to the function itself, and thus it is an antiderivative of it, and that is why the two definitions are equal.

ASKASK
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  • You haven't probably read the second quote of my question and hence are not understanding, what my problem is:-C –  Nov 19 '14 at 04:10
  • If indefinite integration is only a function, why did Richard Courant tell that $F(x) = \int_a^x f(t)dt$ which is representing the function of area?? –  Nov 19 '14 at 04:12
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In short, the definite integral is just a number, which represents the area under the curve $f(x)$ from $x=a$ to $x=b$. The indefinite integral is function; think about this way: What function(s) when you differentiate yields you $f(x)$? That is why it is a family of functions.

To illustrate, consider $f(x) = x^2 +1 \implies f'(x) = 2x$, and $g(x) = x^2 +3 \implies g'(x) = 2x.$ Note that $f(x)$ and $g(x)$ differs only by some constant, but their derivatives are actually the same.

Joshua
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  • Have you read Richard Courant's definition of indefinite integral??? If $\int_a^x f(x)dx$ is indefinite integral, isn't it representing an area??? –  Nov 19 '14 at 03:58
  • @user36790 what you defined right there is a definite integral. – Joshua Nov 30 '14 at 02:47
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I don't understand what is meant by "a family of equations". I see two possible definitions:

  1. The indefinite integral of a function $f(x)$ is the set of all functions $F(x)$ such that $F'(x)=f(x)$. Such a function is also called an antiderivative of $f$.

  2. the indefinite integral is the set of function $F(x)=\int_a^xf(t)dt$ for arbitrary $a$

From the definition of the definite (Riemann) integral we know that the functions in 2. represent the signed area between $a$ and $x$ under the curve $f$.

So if $f(x): = x^2$ the set defined by 1. contains the functions

$$ F(x)=\frac{x^3}{3}-\frac{8}{3}, F(x)=\frac{x^3}{3}+576,\ldots$$

and the set defined by 2. contains the functions $$F(x)=\int_2^x t^2dt,F(x)=\int_{-12}^xt^2dt,\ldots$$

But it is not trivial to see that both the sets defined in 2 are contained in 1. It is the "Fundamental Theorem of Calculus" that shows this. The difference of functions of these sets is always a constant but as @Micah pointed out in a comment not all functions of 1 must be members of 2. A counter example by Micah:

$$e^{2x} \neq\int_a^x \frac{1}{2}e^{2t} \, dt$$

For all $a \in \mathbb{R}$. This is because the right hand side will never become $0$ but the left hand side will become $0$ for $x=a$.

So I would prefer the first definition.

Especially we have $$\int_2^x t^2dt=\frac{x^3}{3}-\frac{8}{3}$$ and $$ \int_{-12}^xt^2dt=\frac{x^3}{3}+576$$

You can find more here:

http://mathworld.wolfram.com/IndefiniteIntegral.html http://mathworld.wolfram.com/FirstFundamentalTheoremofCalculus.html http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus http://en.wikipedia.org/wiki/Antiderivative

miracle173
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  • Sorry, sir , you are right they are family of functions; will edit it...and thnks as you only have understood my confusion between $\int f(x)dx$ and $\int_a^x f(t)dt$ . +1 –  Nov 19 '14 at 04:28
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    In general, those two sets aren't equal. For example, $e^{2x}$ is an antiderivative of $\frac{1}{2}e^{2x}$, but there is no $a$ such that $e^{2x}=\int_a^x \frac{1}{2}e^{2t} , dt$. What the Fundamental Theorem of Calculus says is that the first set contains the second. – Micah Nov 19 '14 at 04:53
  • you are right, thank you – miracle173 Nov 19 '14 at 05:00
  • @miracle173: So sir if both the interpretations are correct, then how are they connected to each other?? What is the relationship between them? Please help,sir. Thanks:) –  Nov 19 '14 at 06:08
  • @miracle173: Sir, if integral represents area(written in Wolfram), how can indefinite integral representing a function gives area?? If both the definitions are correct, how are they linked??? Please explain,sir. –  Nov 19 '14 at 13:01
  • For example select $a=0$ and $f(t)=t^2$ and therefore $F(x)=\int_0^x t^2 dt$. So for $x=3$ the function value $F(3)$ is the definite integral $\int_0^3 t^2 dt$, which is $9$, for $x=4$ the function value $F(4)$ is the definite integral $\int_0^4 t^2 dt$, which is $64/3$ – miracle173 Nov 19 '14 at 20:50
  • How is a function $F(x)=\int_0^xf(t)dt$ connected with the antiderivates? It is an antiderivate of $f$ if $f$ is continous, so $F'(x)=(\int_0^xf(t)dt)'=f(x)$. To see this one has to proof that $\lim_{h \to 0} \frac{F(x+h)-F(x)}{h}=\lim_{h \to 0} \frac{\int_0^{x+h}f(t)dt-\int_0^xf(t)dt}{h}=\lim_{h \to 0} \frac{\int_x^{x+h}f(t)dt}{h}$ is equal to $f(x)$ – miracle173 Nov 19 '14 at 21:03
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Fix a function $f$. An antiderivative or indefinite integral is a function $F$ with $F' = f$. If such a function exists, it's unique up to addition of a constant. The antiderivate $F$ is sometimes written as $F = \int f$ (without limits on the integral), sometimes throwing in an arbitrary constant $C$ because of the caveat above. Thus, for example, $\int x^2 = \frac{1}{3}x^3 + C$, since $F(x) = \frac{1}{3}x^3 + C_0$ has $F'(x) = x^2$ for any constant $C_0$.

The definite integral $\int_a^b f$ is defined to be a suitable Riemann (or etc.) sum, which carries the geometric meaning of being the area under the curve $f$ is $f$ is well-behaved. Assume $f$ is continuous. By the fundamental theorem of calculus, the antiderivative of $F$ satisfies $\int_a^b f = F(b) - F(a)$. Note that this difference is independent of the choice of constant $C$. In fact, we can set $F(x) = \int_b^x f$, since changing $b$ only changes the arbitrary constant applied to $F$.

anomaly
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Let $X=\mathbb{R}^{I}$ be the set of functions on an open interval $I\subset\mathbb{R}$.

For me, the anti-derivative is a partial function

$$\int\cdot\,dx:X\rightarrow X_{/\sim},$$

where $X_{/\sim}$ is the set of equivalence classes of $\sim$ where $\sim$ is the equivalence relation on $X$:

$$f\sim g\Leftrightarrow f(x)-g(x)=C$$ for some $C\in\mathbb{R}$ and all $x\in I$.

The notation $f(x)+C$ is the equivalence class of $f$: $$f(x)+C:=[f]=\{f+C:C\in\mathbb{R}\}.$$

We have $$\int f(x)\,dx=F(x)+C=[F]\Leftrightarrow \frac{d}{dx}F(x)=f(x).$$

Clearly the function is well-defined as $$\frac{d}{dx}(f(x)+C)=\frac{d}{dx}f(x).$$

JP McCarthy
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