13.4 Anti-Differentiation

To compute an integral,

\[ \begin{align} &\Bigl\langle{r}\mathrel{\textstyle{\int}}{i}\Bigr\rangle \end{align} \]

it is our job to write a ruler \( r : \Ruler_{x} \) as a differential

\[ \begin{align} r &= \bar{\diff} z. \end{align} \]

We have lots of practice taking the differential, but for integrals, we need to reverse this process!

Definition.

Suppose \( r : \Ruler_{x} \) is a ruler. An anti-differential for \( r \) is a function

\[ \begin{align} z &\depends x \end{align} \]

whose differential is \( r \).

\[ \begin{align} r &= \bar{\diff} z \end{align} \]

Finding an anti-differential is often a difficult process. There is no one method that always delivers anti-differentials. Rather, we must be willing to make guesses, experiment with the differential laws, and learn from our mistakes.

Example.

Let's find an anti-differential for

\[ \begin{align} r &: \Ruler_{x} \\ r &= x^2 \mult \bar{\standard}_{x}. \end{align} \]

Taking the differential of \( x^3 \) will almost produce our ruler.

\[ \begin{align} \bar{\diff} \bigl(x^3\bigr) &= 3 x^2 \mult \bar{\standard}_{x} \end{align} \]

If we divide both sides of this equation by three, we'll have our ruler.

\[ \begin{align} &x^2 \mult \bar{\standard}_{x} = \frac{1}{3} \mult \bar{\diff} \bigl(x^3\bigr) \\ &x^2 \mult \bar{\standard}_{x} = \bar{\diff} \biggl(\frac{x^3}{3}\biggr) \end{align} \]

We've written our ruler as the differential of a function!

Let's calculate an integral by finding an anti-differential.

Example.

Consider the integral

\[ \begin{align} &\biggl\langle{\sqrt[2]{x} \mult \bar{\standard}_{x}}\mathrel{\displaystyle{\int}}{[1, 4]_{x}}\biggr\rangle. \end{align} \]

To compute this integral, we'll need to write the ruler as a differential. The differential law for base functions helps!

\[ \begin{align} \bar{\diff} \bigl(x^{3/2}\bigr) &= \frac{3}{2} \mult x^{1/2} \mult \bar{\standard}_{x} \end{align} \]

Let's solve for our ruler.

\[ \begin{align} &\frac{3}{2} \mult x^{1/2} \mult \bar{\standard}_{x} = \bar{\diff} \bigl(x^{3/2}\bigr) \\ &\sqrt[2]{x} \mult \bar{\standard}_{x} = \frac{2}{3} \mult \bar{\diff} \bigl(x^{3/2}\bigr) \\ &\sqrt[2]{x} \mult \bar{\standard}_{x} = \bar{\diff} \biggl(\frac{2}{3} \mult x^{3/2}\biggr) \end{align} \]

We've found an anti-differential. Let's take the integral!

\[ \begin{align} \biggl\langle{\sqrt[2]{x} \mult \bar{\standard}_{x}}\mathrel{\displaystyle{\int}}{[1, 4]_{x}}\biggr\rangle &= \biggl\langle{\bar{\diff} \biggl(\frac{2}{3} \mult x^{3/2}\biggr)}\mathrel{\displaystyle{\int}}{[1, 4]_{x}}\biggr\rangle \\ &= \biggl\langle{\frac{2}{3} \mult x^{3/2}}\mathrel{\displaystyle{\int}}{\boundary [1, 4]_{x}}\biggr\rangle \\ &= \biggl\langle{\frac{2}{3} \mult x^{3/2}}\mathrel{\displaystyle{\int}}{[4]_{x}}\biggr\rangle - \biggl\langle{\frac{2}{3} \mult x^{3/2}}\mathrel{\displaystyle{\int}}{[1]_{x}}\biggr\rangle \\ &= \frac{2}{3} \mult 8 - \frac{2}{3} \mult 1 \\ &= \frac{14}{3} \end{align} \]

Example.

Let's calculate the integral.

\[ \begin{align} &\biggl\langle{\log x \mult \bar{\standard}_{x}}\mathrel{\displaystyle{\int}}{[1, \naturalbase]_{x}}\biggr\rangle \end{align} \]

To do so, we'll need to write the ruler

\[ \begin{align} &\log x \mult \bar{\standard}_{x} \end{align} \]

as the differential of a function. This is easier said than done! Let's try taking the differential of the function \( x \mult \log x \).

\[ \begin{align} &\bar{\diff} (x \mult \log x) = \log x \mult \bar{\diff} x + x \mult \frac{1}{x} \mult \bar{\diff} x \\ &\bar{\diff} (x \mult \log x) = \log x \mult \bar{\diff} x + \bar{\diff} x \end{align} \]

Our ruler is a term in this equation. We'll solve for our ruler.

\[ \begin{align} &\log x \mult \bar{\diff} x = \bar{\diff} (x \mult \log x) - \bar{\diff} x \\ &\log x \mult \bar{\standard}_{x} = \bar{\diff} (x \mult \log x - x) \end{align} \]

We've found an anti-differential! Let's use it to compute the integral.

\[ \begin{align} \biggl\langle{\log x \mult \bar{\standard}_{x}}\mathrel{\displaystyle{\int}}{[1, \naturalbase]_{x}}\biggr\rangle &= \biggl\langle{\bar{\diff} (x \mult \log x - x)}\mathrel{\displaystyle{\int}}{[1, \naturalbase]_{x}}\biggr\rangle \\ &= \biggl\langle{x \mult \log x - x}\mathrel{\displaystyle{\int}}{\boundary [1, \naturalbase]_{x}}\biggr\rangle \\ &= \biggl\langle{x \mult \log x - x}\mathrel{\displaystyle{\int}}{[\naturalbase]_{x}}\biggr\rangle - \biggl\langle{x \mult \log x - x}\mathrel{\displaystyle{\int}}{[1]_{x}}\biggr\rangle \\ &= (\naturalbase \cdot 1 - \naturalbase) - (1 \mult 0 - 1) \\ &= 1 \end{align} \]