## Antidifferentiation (Part 1)

1. Derivatives Properties Review. Let ${F(x)}$ be given. We recall the derivative
(1)$\displaystyle \frac{\mathrm{d} F(x)}{\mathrm{d} x}=f(x)$
is defined by
(2)$\displaystyle f(x) = F'(x) = \lim_{h\rightarrow 0}\frac{F(x+h)-F(x)}{h}.$
We have several rules for differentiation:
Power Rule: If ${F(x)=x^{n}}$, then ${F'(x)=nx^{n-1}}$;
Product Rule: We have ${[F(x)G(x)]'=F'(x)G(x)+F(x)G'(x)}$;
Quotient Rule: We have ${[F(x)/G(x)]'=[F'(x)G(x)-G'(x)F(x)]/G(x)^{2}}$;
Composition Rule: $\displaystyle{\frac{\mathrm{d}}{\mathrm{d} x}F(G(x))=F'(G(x))G'(x)}$.

We want to introduce an “inverse” of differentiation. We will use the notation
(3)$\displaystyle \frac{\mathrm{d} f(x)}{\mathrm{d} x}=f'(x)$
interchangeably.

2. Antidifferentiation. A function ${F(x)}$ is an “Antiderivative” of ${f(x)}$ on an interval ${I}$ (usually it’s an open interval) iff
(4)$\displaystyle F'(x)=f(x).$

Example 1. Let ${f(x)=3x}$. What is its antiderivative?

Well, we see its of the form
(5)$\displaystyle F(x)=ax^{2}$
We take its derivative
(6)$\displaystyle F'(x)=2ax$
Set ${F'(x)}$ to be equal to ${f(x)}$
(7)$\displaystyle 2ax=3x$
and solve for ${a}$
(8)$\displaystyle a=\frac{3}{2}.$
Thus
(9)$\displaystyle F(x)=\frac{3x^{2}}{2}$
is the antiderivative for ${f(x)}$.

Note that ${G(x)=F(x)+C}$ is also an antiderivative for ${f(x)}$, where ${C}$ is any constant.

Example 2. Consider
(10)$\displaystyle g(x)=\cos(x).$
What is its antiderivative?

We recall
(11)$\displaystyle \frac{\mathrm{d}}{\mathrm{d} x}\sin(x)=\cos(x)$
which implies
(12)$\displaystyle G(x)=\sin(x)+C,$
for some constant ${C}$, is the antiderivative.

Example 3. Take
(13)$\displaystyle h(x)=\frac{1}{x}+2\mathrm{e}^{2x}$
and find its antiderivative.

We see that
(14)$\displaystyle \frac{\mathrm{d}}{\mathrm{d} x}\ln|x|=\frac{1}{x}.$
We also remember that
(15)$\displaystyle \frac{\mathrm{d}}{\mathrm{d} x}\mathrm{e}^{2x}=2\mathrm{e}^{2x}$
which means
(16)$\displaystyle \frac{\mathrm{d}}{\mathrm{d} x}\left(\ln|x|+\mathrm{e}^{2x}\right)=\frac{1}{x}+2\mathrm{e}^{2x}.$
This gives us the antiderivative
(17)$\displaystyle H(x)=\ln|x|+\mathrm{e}^{2x}+C$
where ${C}$ is some constant.