2.1 Definition

A function $y = F (x)$ is called an antiderivative of another function $y = f (x)$ if $F^{'} (x) = f (x)$ for all $x .$ In other words, suppose you have a function $f (x)$ that is the derivative of some other function $F (x)$ . Then $F (x)$ is an antiderivative of $f (x)$ .

2.2 Example

Consider $F_{1} (x) = x^{3}, F_{2} (x) = x^{3} + π$ and $F_{1} (x) = x^{3} - 47$ . The derivative of the three of them is $f (x) = 3 x^{2}$

$F_{1}, F_{2}$ and $F_{3}$ are each an antiderivative of $f (x) = 3 x^{2}$ . Can you think of any other antiderivatives of $3 x^{2}$ ?

2.3 Notation

We commonly use the following notation for antiderivatives:

$F (x) = \int f (x) d x .$

This is called an indefinite integral. Note $\int f (x) d x$ is a collection (set) of functions.

2.4 Example

Compute

$\int \sin x \cos x d x$

Malcom recalls that $\sin (2 x) = 2 \sin x \cos x$ and that $\frac{d}{d x} \cos (2 x) = - 2 \sin (2 x) .$ So

$\frac{d}{d x} (\frac{- 1}{2} \cos (2 x)) = 2 \sin x \cos x, and therefore$

$\int \sin x \cos x d x = \frac{- 1}{4} \cos (2 x) + C$

Autumn recalls that $\frac{d}{d x} \cos^{2} x = - 2 \sin x \cos x .$ That is $\frac{d}{d x} (\frac{- 1}{2} \cos^{2} x) = \sin x \cos x$ . So she concludes that

$\int \sin x \cos x d x = \frac{- 1}{2} \cos^{2} x + C$

Xi proposes that since $\frac{d}{d x} \sin^{2} x = 2 \sin x \cos x$ , the answer is

$\int \sin x \cos x d x = \frac{1}{2} \sin^{2} x + C$

The three approaches were done correctly. So does this suggest that, for example

$\frac{- 1}{2} \cos^{2} x + C = \frac{- 1}{4} \cos (2 x) + C,$ that is $\frac{- 1}{2} \cos^{2} x = \frac{- 1}{4} \cos (2 x) ?!$

This does not seem right!
Barack steps in and reminds them that

$\cos (2 x) = 2 \cos^{2} x - 1.$

So $\frac{- 1}{4} \cos (2 x) + C = - \frac{1}{2} \cos^{2} x - \frac{1}{4} + C = - \frac{1}{2} \cos^{2} x + (some number)$

$LaTeX: \frac{-1}{4} \cos (2x) +\mbox{some number}=-\frac{1}{2}\cos^2 x + (\mbox{some other number}).$

Similar argument with

$\cos (2 x) = 1 - 2 \sin^{2} x .$

verifies the equality for case where the answer includes sin $x .$
This is a good reminder that $f (x) + C$ implies a collection of functions and not just one function. $C$ can take the value of any real number.