Primitives - The undefinite integral.

Definition 8.1.1   Let $f$ and $F$ be two functions defined on the same interval $I$ . We say that $F$ is a primitive of $f$ on $I$ if $F$ is differentiable on $I$ and $F'=f$ .

Example 8.1.2   The sine function is a primitive a the cosine function.

Theorem 8.1.3   Let $f$ and $F$ be two functions defined on the same interval $I$ . If $F$ is a primitive of $f$ on $I$ , then the functions $F+C$ , with constant $C$ , are all the primitives of $f$ .

This is a consequence of Lagrange's theorem (v.s. 8.5), via the following proposition:

Proposition 8.1.4   Let $f$ be a function, differentiable on the interval $I$ . If for any $x \in I$ , $f'(x)=0$ , then $f$ is a constant function on $I$ .

We denote all the primitives of a given function $f$ by the undefinite integral:

$$\int \; f(x) \; dx$$

You can compare the following theorem with Thm 4.4.

Theorem 8.1.5 (Table of usual primitives)  

$f(x)$	$\int f(x) \; dx$	conditions
$k$	0
$x$	1
$x^{\alpha}$	$\frac {1}{\alpha +1} x^{\alpha +1} +C$	$\alpha \in \mathbb{Q}-{-1}$ , $x>0$ when $\alpha \not\in \mathbb{Z}$
$\frac {1}{2\sqrt{x}}$	$\sqrt{x} +C$	$x>0$
$\frac 1x$	$\ln x +C$	$x>0$
$e^x$	$e^x +C$
$\cos x$	$\sin x +C$
$\sin x$	$-\cos x +C$
$\cosh x$	$\sinh x +C$
$\sinh x$	$\cosh x +C$
$\frac {1}{1+x^2}$	$\arctan x +C$
$\frac {1}{\sqrt{1-x^2}}$	$\arcsin x + C$	$-1 < x < 1$
$\frac {1}{\sqrt{1-x^2}}$	$- \arccos x + C$	$-1 < x < 1$