#### Chapter 8 Integral Calculus

Section 8.1 Antiderivatives

# 8.1.2 Antiderivatives

In the context of this course we will discuss integral calculus for functions on "connected domains", which are of particular significance for many practical applications. In mathematical terms, the domains of the functions will be intervals. As the inverses of derivatives, antiderivatives will be also defined on intervals.
##### Antiderivative 8.1.1
Let an interval $D\subseteq ℝ$ and a function $f:D\to ℝ$ be given. If there exists a differentiable function $F:D\to ℝ$ that has $f$ as its derivative, i.e. $F\text{'}\left(x\right)=f\left(x\right)$ for all $x\in D$, then $F$ is called an antiderivative of $f$.

Let us first consider a few examples.
##### Example 8.1.2
The function $F$ with $F\left(x\right)=-\mathrm{cos}\left(x\right)$ has the derivative

$F\text{'}\left(x\right)=-\left(-\mathrm{sin}\left(x\right)\right)=\mathrm{sin}\left(x\right) .$

Thus, $F$ is an antiderivative of $f$ with $f\left(x\right)=\mathrm{sin}\left(x\right)$.

##### Example 8.1.3
The function $G$ with $G\left(x\right)=\frac{1}{3}e{}^{3x+7}$ has the derivative

$G\text{'}\left(x\right)=\frac{1}{3}·3·e{}^{3x+7} .$

Hence, $G$ is an antiderivative of $g$ with $g\left(x\right)=e{}^{3x+7}$.

Next we will consider another very simple example, which illustrates an important point to note when calculating antiderivatives.
##### Example 8.1.4
Let a constant function $H$ with the function value $H\left(x\right)=18$ be given on an interval. Then the function $H$ has the derivative

$H\text{'}\left(x\right)=0 .$

Hence, $H$ is an antiderivative of $h$ with $h\left(x\right)=0$.

The last example is a little surprising because the derivative of a constant function is the zero function. Thus, every constant function $F$ is an antiderivative of $f$ with $f\left(x\right)=0$ on an interval, i.e. $F\left(x\right)$ is equal to any number $C$ for every value of $x$. However, the antiderivative $F\left(x\right)$ cannot be any other function than a constant one if $f$ is defined on an interval.
##### All Derivatives of the Zero Function 8.1.5
The function $F$ is an antiderivative of $f$ with $f\left(x\right)=0$ on an interval if and only if $F$ is a constant function, i.e. if a real number $C$ exists such that $F\left(x\right)=C$ for all values of $x$ in the interval.

If the functions $F$ and $G$ have the same derivative, i.e. $f=F\text{'}=G\text{'}$, then we have $G\text{'}\left(x\right)-F\text{'}\left(x\right)=0$. Taking the antiderivative on an interval on both sides of the equation results in the relation $G\left(x\right)-F\left(x\right)=C$. Thus, we have $G\left(x\right)=F\left(x\right)+C$. Therefore, if $F$ is an antiderivative of $f$, then $G$ with $G\left(x\right)=F\left(x\right)+C$ is also an antiderivative of $f$.
##### Statement on Antiderivatives 8.1.6
If $F$ and $G$ are antiderivatives of $f:D\to ℝ$ on an interval $D$, then there exists a real number $C$ such that

This is also written as

$\int f\left(x\right) dx=F\left(x\right)+C,$

to express how all antiderivatives of $f$ look.

The set of all antiderivatives is also called the indefinite integral and is written according to the statement above as

$\int f\left(x\right) dx=F\left(x\right)+C ,$

where $F$ is any antiderivative of $f$.
This notation of the indefinite integral emphasises that it is a function $F$ with $F\text{'}=f$ that is calculated for a given function $f$. How this expression is used to calculate the (definite) integral of a continuous function $f$ is described by the fundamental theorem of calculus discussed in the next section in Info Box 8.2.3.
How do we know the value of this constant $C$? If we only look for an antiderivative of $f$ with $f\left(x\right)=0$ on an interval without knowing any other conditions, then the constant $C$ is indefinite. $C$ is only definite if an additional function value ${y}_{0}=F\left({x}_{0}\right)$ of $F$ at a point ${x}_{0}$ is given.
##### Example 8.1.7
For example, for $f$ with $f\left(x\right)=2x+5$, we have

$\int \left(2x+5\right)dx={x}^{2}+5x+C .$

If we look for the antiderivative $F$ of $f$ with $F\left(0\right)=6$, then we set $6=F\left(0\right)={0}^{2}+5·0+C=C$ and hence, $C=6$. Thus, the antiderivative is in this case $F\left(x\right)={x}^{2}+5x+6$.

If the relation between the derivative $f=F\text{'}$ and the antiderivative $F$ is written in the way discussed above for the types of functions considered so far, then one obtains the following table:
##### A Small Table of Antiderivatives 8.1.8
The functions $f$ are considered on an interval. The antiderivatives of these functions are given as an indefinite integral:

$\begin{array}{cc}\text{Function}\mathrm{ }f\hfill & \text{Antiderivatives}\mathrm{ }F\hfill \\ f\left(x\right)=0\hfill & F\left(x\right)=\int 0 dx=C\hfill \\ f\left(x\right)={x}^{n}\hfill & F\left(x\right)=\int {x}^{n} dx=\frac{1}{n+1}·{x}^{n+1}+C\hfill \\ f\left(x\right)=\mathrm{sin}\left(x\right)\hfill & F\left(x\right)=\int \mathrm{sin}\left(x\right) dx=-\mathrm{cos}\left(x\right)+C\hfill \\ f\left(x\right)=\mathrm{sin}\left(kx\right)\hfill & F\left(x\right)=\int \mathrm{sin}\left(kx\right) dx=-\frac{1}{k} \mathrm{cos}\left(kx\right)+C\hfill \\ f\left(x\right)=\mathrm{cos}\left(x\right)\hfill & F\left(x\right)=\int \mathrm{cos}\left(x\right) dx=\mathrm{sin}\left(x\right)+C\hfill \\ f\left(x\right)=\mathrm{cos}\left(kx\right)\hfill & F\left(x\right)=\int \mathrm{cos}\left(kx\right) dx=\frac{1}{k} \mathrm{sin}\left(kx\right)+C\hfill \\ f\left(x\right)=e{}^{x}\hfill & F\left(x\right)=\int e{}^{x} dx=e{}^{x}+C\hfill \\ f\left(x\right)=e{}^{kx}\hfill & F\left(x\right)=\int e{}^{kx} dx=\frac{1}{k} e{}^{kx}+C\hfill \\ f\left(x\right)={x}^{-1}=\frac{1}{x}\hfill & F\left(x\right)=\int \frac{1}{x} dx=\mathrm{ln}|x|+C\mathrm{ }\text{for}\mathrm{ }x>0\mathrm{ }\text{or}\mathrm{ }x<0\hfill \end{array}$

Here, $k$ and $C$ denote arbitrary real numbers with $k\ne 0$, and $n$ is an integer with $n\ne -1$.

The next example shows how the table is used.
##### Example 8.1.9
Find the indefinite integral of the function $f$ with $f\left(x\right)=10{x}^{2}-6=10{x}^{2}-6{x}^{0}$.
From the table above we read off the antiderivatives of $g$ with $g\left(x\right)=x$ and $h$ with $h\left(x\right)={x}^{0}=1$: The function $G$ with $G\left(x\right)=\frac{1}{1+1}·{x}^{1+1}=\frac{1}{2}·{x}^{2}$ is an antiderivative of $g$, and the function $H$ with $H\left(x\right)=\frac{1}{0+1}·{x}^{0+1}=x$ is an antiderivative of $h$. Thus, the function $F:ℝ\to ℝ$ with

$F\left(x\right)=10·\frac{1}{2}{x}^{2}-6·x=5{x}^{2}-6x$

is an antiderivative of $f$. We see that

$\int \left(10x-6\right)dx=5{x}^{2}-6x+C$

describes the set of antiderivatives of $f:ℝ\to ℝ$ with $f\left(x\right)=10x-6$, where $C$ is an arbitrary real number.
The notation using the constant $C$ expresses that, for example, $G:ℝ\to ℝ$ with $G\left(x\right):=5{x}^{2}-6x-7$ is also an antiderivative of $f$, where $C=-7$, since $G\text{'}\left(x\right)=5·2x-6=f\left(x\right)$ for all $x\in ℝ$.

In table books the antiderivatives are generally listed neglecting the constants. However, for calculations it is necessary to state that several functions differing by a constant can exist. In solving problems of applied mathematics, the constant $C$ is often determined by additional conditions, such as a given function value of the antiderivative.
##### Practical Note 8.1.10
It is very easy to check whether the antiderivative of a given function $f$ was found correctly. Take the derivative of the found antiderivative and compare it to the initially given function $f$. If the functions coincide, then the calculation was correct. If the result does not coincide with the function $f$, then the antiderivative has to be checked again.