Definition

Concept	Notation	Meaning
Antiderivative of $f$	$F$ with F'(x)=f(x)	A function whose derivative is $f$
Indefinite integral	${\displaystyle \int f(x)dx}$	$F(x)+C$, the family of all antiderivatives
Integral sign	$\int$	Indicates the operation of integration
Integrand	$f(x)$	The function being integrated
Variable of integration	$x$ in $dx$	The variable with respect to which we integrate
Constant of integration	$C$	Arbitrary constant added to any antiderivative

Antiderivatives

For example, because ${\displaystyle \frac{d}{dx}}{x}^3=3x^2$, the function $F(x)=x^3$ is an antiderivative of $f(x)=3x^2$. Because the derivative of a constant is zero, the derivatives of the following functions

$$x^3+1,x^3-\pi ,x^3+\sqrt{5}$$

and in general $x^3+C$, are the same.

• In general, if $F(x)$ is an antiderivative of $f(x)$, so that F'(x) = f(x), then $F(x)+C$ is also an antiderivative of $f(x)$, where $C$ is an arbitrary number, since (F(x)+C)' = F'(x) + 0 = f(x). This shows that if $f$ has an antiderivative, it has infinitely many antiderivatives.
• If $F(x)$ is a function whose derivative is $f(x)$ on an interval, then every function having the same derivative $f(x)$ on that interval can be written as $F(x)+C$ for some constant $C$. This means that if we have one antiderivative of $f$, then we have all antiderivatives of $f$.

• The above theorem depends on the domain of definition being an interval. For example, let F(x) = \begin{cases} x^2 & \text{if } -1 < x < 1 \\ x^2 - 4 & \text{if } 2 < x < 3 \end{cases}

  and

  G(x) = \begin{cases} x^2 + 5 & \text{if } -1 < x < 1 \\ x^2 - 3 & \text{if } 2 < x < 3 \end{cases}.

  Then F'(x) = G'(x) = 2x for all $x\in (-1,1)\cup (2,3)$, but $F(x)$ is not of the form $G(x)+C$.

  

Indefinite Integrals

It follows from the above theorem that if $F(x)$ is an antiderivative of $f(x)$ in an interval $I$, the general form of the function which has the derivative $f(x)$ or the differential $f(x)dx$ in $I$ is $F(x)+C$, where $C$ is an arbitrary constant.² This expression is called the indefinite integral of $f(x)$.

²Any constant value can be assigned to $C$. It is therefore called "arbitrary constant."

It follows from the above definition that:

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\begin{aligned} \int f(x)\,dx &= F(x)+C \quad \Leftrightarrow \\ \frac{d}{dx}F(x) = f(x) \quad &\text{or equivalently} \quad dF(x) = f(x)\,dx. \end{aligned}}

For example, because ${\displaystyle \frac{d}{dx}}\sin x=\cos x$ or $d(\sin x)=\cos xdx$, we have

$$\int \cos xdx=\sin x+C.$$

Also because $d(x^3)=3x^{2}dx$, we have

$$\int 3x^{2}dx=x^3+C.$$

• An indefinite integral is a family of functions $y=F(x)+C$. From a geometric perspective, an indefinite integral is a collection of curves. Each curve is formed by moving one of the curves parallel to itself either upward or downward along the $y$-axis.
• In writing $\int f(x)dx=F(x)+C$, it is assumed, although not explicitly stated, that this formula holds true for all values of $x$ within a certain interval $I$ where both $f(x)$ and $F(x)$ are defined.

The notation ${\displaystyle \int f(x)dx}$ can be interpreted in two ways:

1. We can regard $\int \dots dx$ as a single symbol, which means "integrate with respect to $x$." It operates on a function $f(x)$ and produces its general antiderivative, or integral $F(x)+C$. In this interpretation the integral sign $\int$ and the $dx$ go together; $\int$ indicates the operation of integration and $dx$ specifies that the variable of integration is $x$. In this interpretation $\int \dots dx$ is viewed as the inverse of differentiation with respect to $x$, i.e., ${\displaystyle \frac{d}{dx}}\cdots$.
2. We may rewrite ${\displaystyle \frac{d}{dx}}F(x)=f(x)$ using differentials as $dF(x)=f(x)dx$, so we can say $f(x)dx$ is the differential of $F(x)$. Writing the integral sign in front of each side of this equation, we get $\int dF(x)=\int f(x)dx$. But ${\displaystyle \int f(x)dx=F(x)+C}$. Therefore, \int dF(x) = \int F'(x)\,dx = F(x)+C. In this viewpoint, $\int$ is an operation that acts on the differential of a function and returns the original function (plus a constant). Thus, the symbol $\int$ for integration (without thinking of the $dx$ as part of the symbol) can be viewed as the inverse of the operation denoted by the symbol $d$.

Both interpretations are valid, and we will use both of them throughout this text. The second one is particularly useful, not only for evaluating integrals, but also for solving certain differential equations (a differential equation is an equation that contains the derivatives of an unknown function, and we try to discover that unknown function).

• Notice that \bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\frac{d}{dx}\left[\int f(x)\,dx\right] = f(x),}

  or equivalently $d(\int f(x)dx)=f(x)dx$, and

  \bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\begin{aligned} \int dF(x) &= \int\frac{d(F(x))}{dx}\,dx \\ &= \int F'(x)\,dx = F(x)+C. \end{aligned}}

  We observe that the signs ${\displaystyle \frac{d}{dx}}$ and ${\displaystyle \int \cdots dx}$, or $d$ and ${\displaystyle \int }$, annul each other in either order. However, when ${\displaystyle \int }$ follows $d$; that is, when we integrate after differentiating, we must add an arbitrary constant $C$, since differentiation discards any constant that was present in the original function. When ${\displaystyle \frac{d}{dx}}$ (or $d$) follows ${\displaystyle \int }$, no constant arises: differentiating $F(x)+C$ returns $f(x)$ exactly.

Does Every Function Have an Antiderivative?

A natural question arises: do antiderivatives, and thus indefinite integrals, exist for every function $f(x)$? The answer is no. Nevertheless, it can be shown that every continuous function has an antiderivative (or hence indefinite integral); that is, if $f$ is a continuous function, then there exists a function $F$ such that F'(x) = f(x) and $\int f(x)dx=F(x)+C$. However, this does not mean that we can necessarily find $\int f(x)dx$ or that we can express it in terms of familiar functions, because there are many non-elementary functions whose derivatives are elementary functions. For example, there is no combination of the functions that we have been dealing with so far whose derivative is $\sin (x^2)$, $\sqrt{1+x^4}$, $e^x/x$, or $e^{-x^2}$. In other words,

$$\int \sqrt{1+x^4}dx,\int \sin \left(x^2\right)dx,\int \frac{e^x}{x}dx,\int e^{-x^2}dx$$

cannot be expressed in terms of familiar functions.