What is indefinite integral actually - $\int f(x)dx$ or $\int_a^x f(t)dt$?

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

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 $F$.

Answer 5

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 $$\frac{d}{dx}(f(x)+C)=\frac{d}{dx}f(x).$$