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Reversing differentiation: antiderivatives

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Antidifferentiation is the "reverses" the effect of differentiation. In that sense it's similar to inverse function.

Let's review how inverse functions appear in algebra: $$\begin{aligned} x^{2} & = 4 \to x = 2 \quad \text{ to find it we need } \sqrt{} \\ x^{3} & = 27 \to x = 3 \quad \text{ to find it we need } \sqrt[3]{} \\ 2^{x} & = 8 \to x = 3 \quad \text{ to find it we need } \log_{2} \\ \sin x & = 0 \to x = 0 \quad \text{ to find it we need } \sin^{-1} \end{aligned}$$ etc.

Similarly in calculus, what if we know the result of differentiation and want to know where it came from? $$\begin{aligned} f^{\prime}(x) &= 2x \to f(x) = x^{2} \\ f^{\prime}(x) &= \cos x \to f(x) = \sin x \\ f^{\prime}(x) &= \frac{1}{x} \to f(x) = \ln x \end{aligned}$$ Observe that is either case what's on the left is simpler than what's on the right.

To illustrate graphically:

Find $x$, number

Why is this important?

Let's go back to motion.

Until now:

  • Given position (as a function of time),
  • find velocity.

Now:

  • Given velocity,
  • find position.

These two problems can be seen in these simple situations:

  1. Speedometer is broken. You watch your odometer and your watch to find the speed.
  2. Odometer is broken. You watch your speedometer and your watch to find the location.

Is this a real problem?

Yes, unlike cars an airplane has no odometer. It replies on sensors on its surface to determine its speed and then computes its location.

Airplane has sensors on surface to read wind speed.

To summarize the idea:

 "The antiderivative of the velocity is the location." 

Generally, if $F^{\prime}(x) = f(x)$ then $F$ is called an antiderivative of $f$.

Why "an"? Is the definition ambiguous?

Antiderivative of $2x$ is $x^{2}$. Are there any others? Not $(3x^{2})^{\prime} = 6x$ but $(x^{2} + C)^{\prime} = 2x $, where $C$ is any constant because $$ (x^{2} + C)^{\prime} = (x^{2})^{\prime} +(C)^{\prime} = (x^{2})^{\prime} = 2x. $$

Theorem. If $F$ is an antiderivative of $f$ then so is $F+ C$. where $C$ is any constant.

The proof comes from an an old theorem:

if $F^{\prime} = G^{\prime}$ then $F-G$ is a constant.

It's the theorem that follows from the Mean Value Theorem: two runners with equal (possibly variable) speed keep the same distance.

Now, to find the position $F$ from the velocity $F^{\prime}$, need the initial position $F(0)$.

Shift by $C$ of the graph of $F$.

How many antiderivatives are there? Infinite, one for each $C$.

Fortunately, you need just one to find all of them. Why?

Now about the algebra and computations.

Recall this table of derivatives: $$\begin{alignat}{3} &\text{function} & \quad & \rightarrow & \quad & \text{derivative} \\ & x^{r} & \quad & & \quad & rx^{r-1} \\ & e^{x} & \quad & & \quad & e^{x} \\ & \ln x & \quad & & \quad & \frac{1}{x} \\ & \sin x & \quad & & \quad & \cos x \\ & \cos x & \quad & & \quad & -\sin x\\ &\text{antiderivative} & \quad & \leftarrow & \quad & \text{function} \end{alignat}$$ To find antiderivatives, reverse the order, read from right to left!

Example. Find antiderivatives of $x^{3}$.

Use the Power Formula for Differentiation: $$(x^{r})^{\prime} = rx^{r-1}.$$

How do I find an antiderivative of $x^{r}$? Divide PF by $r$: $$\frac{1}{r}(x^{r})^{\prime} = x^{r-1}$$ then apply the Constant Multiple Rule: $$ \quad \left(\frac{1}{r} x^{r}\right)^{\prime} = x^{r-1} $$ Convert to $x^{s}$. How? Set $r-1 = s$. Then $$\left( \frac{1}{s + 1}x^{s + 1} \right)^{\prime} = x^{s} $$ Then we have the Power Formula for Integration:

  • an antiderivative of $x^{s}$ is $\frac{1}{s + 1}x^{s + 1}$, provided $s \neq -1$.

What if $s = -1$? Then we read the answer from elsewhere in the table:

  • an antiderivative of $x^{-1}$ is $\ln |x|$.

More:

  • an antiderivative of $e^{x}$ is $e^{x}$.
  • an antiderivative of $\cos x$ is $\sin x$.
  • an antiderivative of $\sin x$ is $-\cos x$.


So, we have a table of antiderivatives:

Function Antiderivative
$x^{s}$ $\frac{1}{s + 1} x^{s + 1}, \quad s \neq -1$
$\frac{1}{x}$ $\ln x, \quad x \neq 0$
$e^{x}$ $e^{x}$
$\sin x$ $\cos x$
$\cos x$ $-\sin x$

Now, just like with differentiation we need the "rules" of antidifferentiation.

Recall the rules of differentiation: $$ \text{SR: } \qquad (f + g)^{\prime} = f^{\prime} + g^{\prime} $$ Let's read from right to left.

Sum Rule 
If $F$ is an antiderivative of $f$ and
if $G$ is an antiderivative of $g$,
then
$F+G$ is an antiderivative of $f+g$.

Similarly:

Constant Multiple Rule 
If $F$ is and antiderivative of $f$ and $c$ is a constant, then
$cF$ is an antiderivative of $cf$.

Example. Find an antiderivative of $$3x^{2} + 5e^{2} + \cos x.$$ Simply find antiderivatives of each term: $$ 3\cdot \frac{x^{3}}{3} + 5e^{x} + \sin x. $$ How do I know this is correct? Differentiate the antiderivative: $$\begin{aligned} (x^{3} + 5e^{x} + \sin x)^{\prime} &= (x^{3})^{\prime} + 5(e^{x})^{\prime} + (\sin x)^{\prime} \\ &= 3x^{2} + 5e^{x} + \cos x \end{aligned}$$ This is the original function! The answer checks out.

Check for antiderivative.

So far, easy. But bad news, for antidifferentiation there is no

  • Product Rule,
  • Quotient Rule,
  • Chain Rule,

not in the same sense.

Question: What is the antiderivative of $xe^{x}$?

Question: How do you antidifferentiate $2x\cdot e^{x^{2}}$?

Answer: Recognize where the function comes from. Like here: $$ (e^{x^{2}})^{\prime} = 2x\cdot e^{x^{2}} $$

Example: Ball Thrown

  1. Ball thrown up at 47 ft/sec off a
  2. Cliff of height 432 ft.
Ball thrown off cliff.

Find the hight of the ball as a function of time.

We define: $t$ is time, starts at $t=0$, $h$ is the height above ground, $v$ is velocity and $a$ is acceleration. Here $h$, $v$ and $a$ are functions of time.

We also need some physics : $a$ is constant, equal to 32 ft/sec2. Hence $$ a(t) = -32 $$ $$\begin{aligned} a(t) &= v^{\prime}(t) \\ v(t) &= h^{\prime}(t) \end{aligned}$$

Now we use this set-up to get some data, translate (1) and (2) above. $$\begin{aligned} (1) \to v(0) &= 47 (3)\\ (2) \to h(0) &= 432 (4) \end{aligned}$$ We call these initial conditions.

Step 1. Find $v$.

$$v^{\prime} = -32 $$ So, antidifferentiate: $$ v = -32 t + C ,$$ but $C$ is an unknown constant.

Find $C$ from (3). $$\begin{aligned} v(0) &= 47 \\ v(0) &= -32 \cdot 0 + C = C \\ \therefore C &= 47 \end{aligned}$$ So, we've found $4 v(t) = -32t + 47$.

Step 2. Find $h$.

$$h^{\prime} = v = -32t + 47 $$ So, antidifferentiate: $$ h = -32 \frac{t^{2}}{2} + 47 t + K. $$ Find $K$ from (4): $$\begin{aligned} h(0) &= 432 \\ &= -32 \cdot \frac{0^{2}}{2} + 47\cdot 0 + K \\ \therefore K &= 432 \end{aligned}$$

Answer: $$h(t) = -\underbrace{16}_{\frac{1}{2} \text{ acceleration}} t^{2} + \underbrace{47t}_{\text{initial velocity}} + \underbrace{432}_{\text{initial position}}. $$

Q: When does it reach the maximum height?

When the velocity is zero! $$\begin{aligned} -32t + 47 &= 0 \\ t &= \frac{47}{32} \end{aligned}$$

Graph of Antiderivative function

How do we graph an antiderivative based on the graph of the function?

Answer: Reverse the procedure of graphing a derivative based on the graph of the function.

Plot $y=f(x)$. Look for special points: the local max and min points. These correspond to the zero ($x$-intercepts) of $f^{\prime}$.

Plotting $f^{\prime}(x)$ from a plot of $f(x)$.

These three are the points on the graph of $y=f^{\prime}(x)$. How do we get from one to the next? Look at $\nearrow\searrow$ of $f$.

Inverse Problem: How do we get $f$ from $f^{\prime}$? Reverse the procedure.

Plot the graph of $y=f^{\prime}(x)$ and look at important points - $x$-intercepts. These correspond to local maximum and minimum points of $f$.

Plotting $f(x)$ from a plot of $f^{\prime}(x)$.

How do we deal with the multiple antiderivatives?

Plot any of them. Pick a point and start graph there.

Note: The +- of $f^{\prime}$ may come from algebra done previously when sketching. This time the data about $f^{\prime}$ comes from its graph, not the formula.