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Reversing differentiation: antiderivatives
From Mathematics Is A Science
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.