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# Reversing differentiation: antiderivatives

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:

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:

*Speedometer is broken.*You watch your odometer and your watch to find the speed.*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.

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 $F(0)$.

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.

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**

- Ball thrown up at 47 ft/sec off a
- Cliff of height 432 ft.

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/sec^{2}. 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 $f^{\prime}$.

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 $f$.

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.