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# System prepended metadata

title: Math 5i Reference
tags: [Math 5i, Reference]

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# Math 5i Reference

View the book with "<i class="fa fa-book fa-fw"></i> Book Mode".

## Selected derivative formulas

Recall that for a function $f(x)$, we define its derivative as
$$f'(x) = \frac{df}{dx} = D_x f(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} = \lim_{u \to x} \frac{f(x) - f(u)}{x - u}.$$

1. $D_x (uv) = (D_x u) \cdot v + u \cdot (D_x v)$
1. $D_x \dfrac{u}{v} = \dfrac{(D_x u) \cdot v + u \cdot (D_x v)}{v^2}$
1. $D_x (u^n) = nu^{n - 1}\cdot D_x u$



## Selected antiderivative formulas

1. $\displaystyle\int u^n\,du = \frac{1}{n + 1}u^{n + 1} + C \quad (n \ne -1)$
1. $\displaystyle\int \frac{1}{u}\,du = \ln\left|u\right| + C$
1. $\displaystyle\int \tan(u)\,du = -\ln\left|\cos(u)\right| + C$
1. $\displaystyle\int \sec(u)\,du = \ln\left|\sec(u) + \tan(u)\right| + C = \frac{1}{2} \ln\left| \frac{1 + \sin(u)}{1 - \sin(u)} \right| + C$
1. Reduction formulas
a. $\displaystyle\int \sin^n(u)\,du = -\frac{1}{n}\sin^{n-1}(u)\cos(u) + \frac{n - 1}{n}\int \sin^{n -2}(u)\,du \quad (n > 1)$
b. $\displaystyle\int \cos^n(u)\,du = \frac{1}{n}\cos^{n-1}(u)\sin(u) + \frac{n - 1}{n}\int \cos^{n -2}(u)\,du \quad (n > 1)$




## Theorems

### Mean Value Theorem
**If:** $f$ is a function that is continuous on $[a,b]$ and differentiable on $(a,b)$,
**then:** there exists a number $c$ in $(a,b)$ such that $f(b) = f(a) + f'(c)(b - a)$ or equivalently $f'(c) = \dfrac{f(b) - f(a)}{b - a}$.
<iframe src="https://www.geogebra.org/calculator/wkud29gq?embed" width="800" height="400" allowfullscreen style="border: 10px solid #e4e4e4;border-radius: 20px;" caption="an illustration of the MVT" frameborder="2"></iframe>

**Don't fall into the trap where you use the mean value theorem when it isn't applicable (i.e. when your function isn't actually continuous or isn't actually differentiable on that interval)!**



### Extendend Mean Value Theorem
**If:** $f$ is a function that is $n+1$ times differentiable on $[a,b]$ (i.e. the $n$-th derivative of $f$ exists on $[a,b]$),
**then:** there exists a number $c$ in $(a,b)$ such that $f(b) = T_n (b) + \frac{f^{(n+1)}(c)}{(n+1)!} b^{n+1}$, there $T_n(x)$ is the $n$-th-degree Taylor polynomial of $f$ centered at $a$.<br>
**Recall:** The *Taylor polynomial* of $f$ (of degree $n$) centered at $a$ is $T_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \dots + \frac{f^{(n - 1)}(a)}{(n - 1)!}(x - a)^{n - 1} + \frac{f^{(n)}(a)}{n!}(x - a)^n$.


## Tests about infinite series

### Alternating series test
**If:** the sequence of postive numbers $a_0$, $a_1$, $a_2$, $\dots$ is decreasing and $\displaystyle\lim_{n\to\infty} a_n = 0$,
**Then:** the alternating series $\displaystyle\sum_{n=0}^{\infty} (-1)^n a_n$ converges.


### Direct comparison test
**Given:** $\sum a_n$ and $\sum b_n$ are two series such that $0 \le a_n \le b_n$ for all $n$ large enough,
**Then:**
(1) if $\sum b_n$ converges, then so does $\sum a_n$, or
(2) if $\sum a_n$ diverges, then so does $\sum b_n$.


### Divergence test
**If** $\displaystyle\lim_{n\to\infty} a_n \ne 0$, **then** $\sum a_n$ diveges.  Equivalently, we can state the contrapositive: **If** $\sum a_n$ converges, **then** it must be that $\displaystyle\lim_{n\to\infty} a_n = 0$.


### Integral test
**If:** $f(x)$ is continuous (thus integrable), positive and decreasing, and $f(n) = a_n$,
**Then:** the sum $\displaystyle\sum_{n=A}^{\infty} a_n$ and the integral $\displaystyle\int_A^{\infty} f(x)\,dx$ either both converge or both diverge.

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## A Few Things to Check for When You Have an Integral in Front of You

**Before anything else, consider plotting the function you're taking the integral of. This might save you a lot of work and at the very least it will help you check your work at the end.**
### Simplification 
* Do check to see if there are any basic algebra simplifications you can do. If there are, pause and ask yourself, "Is there a reason whoever wrote this problem left it unsimplified like this? Are they trying to hint at a way to do the problem?" If not, then simplify. 
* Do you have something in the denominator that you can use partial fractions on? If so, do that! Pay special attention to catch an $a^2-b^2$ in the denominator. 
* *fancy trick but optional:* Is this an even function? ([How can you tell?](#even)) If so, you can take advantage of that. Set the lower bound of integration to 0 and calculate twice the integral you now have. 

### Actually Solving the Integral

When it comes right down to it, there are really only a few ways we know how to solve integrals. If we're given a problem, it will be solvable through one of those, we just have to be willing to think through which one will work in this place. 

1. Check for an item from the "simplification" section
2. Check for chain rules you can do in your head. Is there a function inside another function and the derivative of the inner function outside? If so, make a guess at the integral solution and then try it out by taking the derivative of that.
3. Are there two functions multiplied together? That means it's time for integration by parts!
4. Check if you have something with powers of sines and cosines. 
5. For many other integrals that involve combinations of sines and cosines, remember that the only formulas you need are $\sin(A + B) = \sin A \cos B + \cos A \sin B$,
$\sin(A − B) = \sin A \cos B − \cos A \sin B$,
$\cos(A + B) = \cos A \cos B − \sin A \sin B$, and
$\cos(A − B) = \cos A \cos B + \sin A \sin B$.
You can often rearange those to help you take the integral of other trig combinations. (For example: $\int \sin^2x \,dx$ or $\int \sin(3x) + \cos(4x) \,dx$)
5. If none of the above worked, that probably means a substitution is in order.

#### Substitutions
7. Check for something that can become $\sqrt{1-x^2}$, $\sqrt[]{1+x^2}$, or $\sqrt[]{x^2-1}$. If you have something like that, try trig subsitution (like making it $\sqrt[]{1-\sin^2(x)}$ ). **remember that trig substitution isn't always this obvious. Sometimes it might look like $\sqrt{9-x^2}^3$**. ([What do you do in cases like this (when there isn't a 1 under the square root?](#turn_to_one_trick))
7. Try a few $m$ substitutions to see what works. Make $m$ = whatever is inside a square root. Make $m$ equal the complicated argument of any function (like a ln or a tan). Make $m$ equal the denomonator of your function. 

### Bonus Section: What to do After Solving an Integral
1. Check to see if you forgot to add your constant "C"
2. Check to see if the integral you got makes sense with the graph you plotted at the begining
2. Take the derivative of your antiderivative to check if you get the original function back
4. Do a happy dance



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##### <a name="even"></a> By definition, an even function is one that will have the same value with $x$ and $-x$. Try plugging $-x$ in in your head and see what happens. <span style="color:blue"> *clicking "How can you tell" in the third bullet point of "ways to simplify" should bring you to this paragraph. If it doesn't, I am still figuring out how to do that. If you can figure it out, feel free to help by linking "How can you tell" to this paragraph. (:* </span>




##### <a name="turn_to_one_trick"></a> For situations like this, first factor out the number that isn't one (but you want to be one) inside the square root. For the example, $\sqrt[]{9-x^2}^3$, this will make the square root, $\sqrt[]{{9}(1-\frac{x^2}{9})}^3$. Take the coefficiant out of the square root: $3\sqrt[]{(1-\frac{x^2}{9})}^3$. Then put whatever you have into the form $a\sqrt[]{1-(\frac{x}{c})^2}^3$ Then do an $m$ substitution where $sin (m)$ = your term inside the square root with the x in it. Here, $sin(m)$ = $\frac{x}{c}$. From there, you have your usual trig substitution problem.  <span style="color:blue"> *and this paragraph should be linked to "(What do you do in cases like this (when there isn’t a 1 under the square root?)." I hope I figure out how to link things soon!*</span>

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## All About Lagrange Remainders and the Extended Mean Value Theorem

### Extended Mean Value Theorem (A repost from the "Theorem" section)
**If:** $f$ is a function that is $n+1$ times differentiable on $[a,b]$ (i.e. the $n$-th derivative of $f$ exists on $[a,b]$),
**then:** there exists a number $c$ in $(a,b)$ such that $f(b) = T_n (b) + \frac{f^{(n+1)}(c)}{(n+1)!} b^{n+1}$, there $T_n(x)$ is the $n$-th-degree Taylor polynomial of $f$ centered at $a$.<br>
#### A few potentially helpful points:
* **Recall:** The *Taylor polynomial* of $f$ (of degree $n$) centered at $a$ is $T_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \dots + \frac{f^{(n - 1)}(a)}{(n - 1)!}(x - a)^{n - 1} + \frac{f^{(n)}(a)}{n!}(x - a)^n$.
As n goes to infinity in the above Taylor polynomial, the series gets closer and closer to the infinite Taylor series. A common question is whether the Lagrange remainder goes to zero as n goes to infinity. All the Lagrange remainder is is the last term of a Taylor series. So we should picture this question as asking whether there is a point at which the Taylor series is "complete" and there is no extra term we could add to it to make it more like the function it's approximating.

* Not all functions equal their Taylor series (and not all functions that sometimes equal their Taylor series always equal their Taylor series). But some do. This isn't neccessarilly intuative....
* 





###### tags: `Reference` `Math 5i`
