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:

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

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$

Noah Dana-Picard 2007-12-28