Module 2: Indefinite Integrals

2.1 Definition

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

2.2 Example

Consider $F_{1} (x) = x^{3}, F_{2} (x) = x^{3} + π$ $F_1(x)=x^3,F_2(x)=x^3+\pi$ and $F_{1} (x) = x^{3} - 47$ $F_1(x)=x^3-47$ . The derivative of the three of them is $f (x) = 3 x^{2}$ $f(x)=3x^2$

$F_{1}, F_{2}$ $F_1, F_2$ and $F_{3}$ $F_3$ are each an antiderivative of $f (x) = 3 x^{2}$ $f(x)=3x^2$ . Can you think of any other antiderivatives of $3 x^{2}$ $3x^2$ ?

2.3 Notation

We commonly use the following notation for antiderivatives:

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

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

2.4 Example

Compute

$\int \sin x \cos x d x$ $\int \sin x \cos x \; dx$

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

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

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

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

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

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

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

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

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

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

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

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

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

Similar argument with

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

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