# Primitive functions/Relation to fundamental theorem/Section

## Definition

Let ${\displaystyle {}I\subseteq \mathbb {R} }$ denote an interval, and let

${\displaystyle f\colon I\longrightarrow \mathbb {R} }$

denote a function. A function

${\displaystyle F\colon I\longrightarrow \mathbb {R} }$

is called a primitive function for ${\displaystyle {}f}$, if ${\displaystyle {}F}$ is differentiable on ${\displaystyle {}I}$ and if ${\displaystyle {}F'(x)=f(x)}$ holds for all

${\displaystyle {}x\in I}$.

A primitive function is also called an antiderivative. The fundamental theorem of calculus might be rephrased, in connection with fact, as an existence theorem for primitive functions.

## Corollary

Let ${\displaystyle {}I}$ denote a real interval, and let

${\displaystyle f\colon I\longrightarrow \mathbb {R} }$

denote a continuous function. Then ${\displaystyle {}f}$ has a primitive function.

### Proof

Let ${\displaystyle {}a\in I}$ be an arbitrary point. Due to fact, there exists the function

${\displaystyle {}F(x)=\int _{a}^{x}f(t)\,dt\,,}$

and because of the Fundamental theorem, the identity ${\displaystyle {}F'(x)=f(x)}$ holds. This means that ${\displaystyle {}F}$ is a primitive function for ${\displaystyle {}f}$.

${\displaystyle \Box }$

## Lemma

Let ${\displaystyle {}I}$ denote a real interval, and let

${\displaystyle f\colon I\longrightarrow \mathbb {R} }$

denote a function. Suppose that ${\displaystyle {}F}$ and ${\displaystyle {}G}$ are primitive functions of ${\displaystyle {}f}$. Then ${\displaystyle {}F-G}$ is a constant function.

### Proof

We have

${\displaystyle {}(F-G)'=F'-G'=f-f=0\,.}$

Therefore, due to fact, the difference ${\displaystyle {}F-G}$ is constant.

${\displaystyle \Box }$

The following statement is also a version of the fundamental theorem, it is called the Newton-Leibniz-formula.

## Corollary

Let ${\displaystyle {}I}$ denote a real interval, and let

${\displaystyle f\colon I\longrightarrow \mathbb {R} }$

denote a continuous function. Suppose that ${\displaystyle {}F}$ is a primitive function for ${\displaystyle {}f}$. Then for ${\displaystyle {}a,b\in I}$, the identity

${\displaystyle {}\int _{a}^{b}f(t)\,dt=F(b)-F(a)\,}$

holds.

### Proof

Due to fact, the integral exists. With the integral function

${\displaystyle {}G(x):=\int _{a}^{x}f(t)\,dt\,,}$

we have the relation

${\displaystyle {}\int _{a}^{b}f(t)\,dt=G(b)=G(b)-G(a)\,.}$

Because of fact, the function ${\displaystyle {}G}$ is differentiable and

${\displaystyle {}G'(x)=f(x)\,}$

holds. Hence ${\displaystyle {}G}$ is a primitive function for ${\displaystyle {}f}$. Due to fact, we have ${\displaystyle {}F(x)=G(x)+c}$. Therefore,

${\displaystyle {}\int _{a}^{b}f(t)\,dt=G(b)-G(a)=F(b)-c-F(a)+c=F(b)-F(a)\,.}$
${\displaystyle \Box }$

Since a primitive function is only determined up to an additive constant, we sometimes write

${\displaystyle {}\int _{}^{}f(t)\,dt=F+c\,.}$

Here ${\displaystyle {}c}$ is called a constant of integration. In certain situations, in particular in relation with differential equations, this constant is determined by further conditions.