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education/calculus/precalculus/Composing Functions.md

@@ -1,3 +0,0 @@
-- To compose a function is to create a new function from multiple smaller functions.
-- They can be solved from the inside out
-- 

education/math/MATH1220 (calc II)/Integral Review.md

@@ -6,11 +6,20 @@ Divide $[a, b]$ into $n$ equal parts of width $\Delta x = \dfrac{b-a}{n}$.
 Let $x_0, x_1, x_2, \cdots, x_n$ be the endpoints of this subdivision. $x_0 = a$ and $x_n = b$. 
 
 Define $$\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^nf(x_i)\Delta x$$
-- $\Delta x$ refers to the width of each sub-interval
+- $\Delta x$ refers to the width of each sub-interval, or $\frac{b-a}{n}$.
 - $f(x_i)$ refers to the height of each subinterval, and can be found with the equation $x_i = \Delta xi + a$ 
+	- Or, the width of each interval times the interval index, plus the starting offset.
 
 Then let $f$ be a continous function on $[a, b]$ and let $F$ be the antiderivative of $f$ (i.e $F'(x) = f(x)$).
+
 Then $\int_a^b f(x) dx = F(b) - F(a)$.
 
+## Sum of an infinite series
+The sum of an infinite series can be defined as follows:
+$$ \sum_{i = 1}^n i = \frac{n(n+1)}{2}$$
+
 ## Examples
-$$ \int_0^1 x^2 dx = \frac{1}{3} x^3 \Big |_0^1 = 1/3(1^3)- 1/3 (0)^3 = 1/3$$
+1. Find the area under the curve between 0 and 1 of the function $f(x) = x^2$
+$$ \int_0^1 x^2 dx = \frac{1}{3} x^3 \Big |_0^1 = 1/3(1^3)- 1/3 (0)^3 = 1/3$$
+2. Find the Riemann Sum under the curve between -2 and 2 of the function $2x + 2$. 
+$$ \int_{-2}^2 (2x + 2)dx = \lim_{n \to \infty} \sum_{i = 1}^nf(x_i)\Delta x $$

education/math/MATH1220 (calc II)/Integration by Parts.md

@@ -1,11 +1,42 @@
 The integration by parts formula is:
 $$ \int udv = uv - \int vdu $$
-
+Broadly speaking, integration by parts is done by:
+1. Pick a part of integral to be $u$. 
+2. The rest of the integral will be $dv$, 
+3. Compute the derivative of $u$, $du$.
+4. Compute the antiderivative of $dv$
+5. Substitute those values in to the integration by parts formula.
 ## Deriving the Integration by Parts Formula
 $$ \frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x) $$
 1. Integrating both sides, we get:
 $$\int \frac{d}{dx} (f(x)g(x))dx = \int [f'(x)g(x) + f(x)]$$
 2. Through the distributive property of integrals,
 $$ = \int f'(x)g(x)dx + \int f(x)g'(x)dx $$
-3. Therefore:
-$$f(x)g(x) = \intf'(x)g(x)dx $$
+3. An integral cancels out an antiderivative, therefore:
+$$f(x)g(x) = \int f'(x)g(x)dx + \int f(x)g'(x)dx $$
+4. Moving terms around:
+$$ \int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx$$
+
+Now, let $u = f(x)$ and $v = g(x)$, then $dv = g'(x)dx$ and $du = f'(x)dx$. 
+
+# Examples
+>  Evaluate the below antiderivative using integration by parts.
+$$\int xe^{2x}dx$$
+1. Define $u$ to be a value you can take the derivative of easily, in this case $u = x$. The rest of the integral will be set to $dv$, in this case, $dv = e^{2x}dx$. 
+	- $u = x$
+	- $du = \frac{d}{dx}(x)= 1dx$ 
+	- $dv = e^{2x}dx$
+	- $v = \frac{1}{2}e^{2x}$ - The antiderivative of $dv$.
+2. Looking back at the integration by parts formula, we know that:
+	$$ \int udv = uv - \int v du $$
+	$$ \int xe^{2x}dx = (\frac{1}{2}e^{2x})(x)-\int (\frac{1}{2}e^{2x}) (1dx) $$
+3. The remaining integral can be solved with $u$ substitution, but we've already defined $u$, so we use $w$ as a replacement.
+	- $w = 2x$
+	- $dw = 2dx$
+	- $\frac{1}{2}dw=dx$ 
+		1. Substituting $w$ and $dw$ into the integral:
+		$$ \int \frac{1}{2}e^w \frac{1}{2}dw $$
+		2. This gives an integral that can be computed naively
+		$$ \int\frac{1}{2}e^{w}\frac{1}{2}dw = \frac{1}{4}\int e^w dw= \frac{1}{4}e^{2x} $$
+4. Combining everything together, we get:
+	$$ \int xe^{2x}dx = (\frac{1}{2}e^{2x})(x)- (\frac{1}{4}e^2x) + C$$

education/math/MATH1220 (calc II)/Integration with Trig Identities.md

@@ -0,0 +1,77 @@
+The below integration makes use of the following trig identities:
+1. The Pythagorean identity: $\sin^2(x) + \cos^2(x) = 1$
+2. The derivative of sine: $\frac{d}{dx}sin(x) = cos(x)$
+3. The derivative of cosine: $\dfrac{d}{dx} \cos(x) = -\sin(x)$
+4. Half angle cosine identity: $\cos^2(x) = \frac{1}{2}(1 + \cos(2x))$
+5. Half angle sine identity: $\sin^2(x) = \frac{1}{2}(1 - \cos(2x))$
+6. $tan^2(x) + 1 = sec^2(x)$
+7. $\dfrac{d}{dx}(\tan(x)) = \sec^2(x) \Rightarrow \int \sec^2(x)dx = \tan(x) + C$ 
+8. $\dfrac{d}{dx}(\sec x) = \sec(x)\tan(x) \Rightarrow \int\sec(x)\tan(x) dx = \sec(x) + C$
+# Examples
+> Evaluate the integral $\int\sin^5(x)dx$
+1. With trig identities, it's common to work *backwards* with u-sub. In the above example, we can convert the equation into simpler cosine functions by setting $du$ to $-\sin(x)dx$. This means that $u$ is equal to $cos(x)$. 
+$$ \int\sin^4(x)\sin(x)dx$$
+2. Rewrite $sin^4(x)$ to be $(\sin^2(x))^2$ to take advantage of the trig identity $1 - \cos^2(x) = \sin^2(x)$ 
+$$ \int(\sin^2x)^2 \sin(x)dx$$
+3. Apply the above trig identity and substitute $u$:
+$$ \int(1 - u^2)^2 (-du) $$
+4. Foil out and move negative out of integral:
+$$ -\int(1 - 2u^2 + u^4)du $$
+5. Take advantage of the distributive property of integrals:
+ $$ - (u - \frac{2}{3}u^3 + \frac{1}{5}u^5) + C $$
+ 6. Substituting $\cos(x)$ back in for $u$, we get the evaluated (but not entirely simplified) integral:
+ $$-(\cos(x)- \frac{2}{3}\cos^3x + \frac{1}{5}\cos^5x) $$
+# Trigonometric Substitutions
+Trigonometric substitution is useful for equations containing  $\sqrt{a^2 + x^2}$ or $a^2 + x^2$, where $a$ is any constant. It removes any addition or subtraction.
+
+The general process involves the use of a trig identity and multiplying everything in that identity by a constant.
+
+Consider the identity:
+$$ 1 + \tan^2(\theta) = \sec^2(\theta)$$
+Multiplying both sides of the identity by $a^2$, we get:
+$$a^2 + a^2\tan^2(\theta) = a^2\sec^2(\theta)$$
+This enables us to make use of **substitution** to simplify many integrals.
+- $x = a\tan \theta$
+- $dx = a \sec^2\theta d\theta$
+- for $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$
+# Examples
+> Evaluate the integral $\int\frac{3}{4+x^2}dx$ 
+1. Move the constant coefficient out of the integral:
+$$ \int \frac{3}{4+x^2}dx = 3\int \frac{1}{4 + x^2}dx$$
+2. Let $x = 2tan\theta$ and $dx = (2sec^2\theta d\theta)$, substitute accordingly
+$$ = 3\int\frac{1}{4 + 4\tan^2\theta}(2\sec^2\theta)d\theta$$
+3. Factor $4$ out in the denominator
+$$ = 3\int\frac{1}{4(1 + \tan^2\theta)}(2\sec^2\theta)d\theta$$
+4. Considering the identity $1 + \tan^2 \theta = \sec^2 \theta$
+$$ = 3\int\frac{1}{4(\sec^2\theta)}(2\sec^2\theta)d\theta$$
+5. $\sec^2\theta$ is present in the numerator and the denominator, so we can cancel those out. This means that:
+$$ 3\int\frac{2}{4}d\theta = \frac{3}{2} \theta + C$$
+6. At this point, we want to determine what $\theta$ is equal to relative to $x$. 
+	1. Look back to step 2 we defined $x = 2\tan\theta$
+	2. Moving $2$ to the other side, we get $\frac{x}{2} = \tan\theta$ 
+	3. Because we defined bounds for our definition of $\theta$, we can take advantage of $\arctan$, therefore:
+$$ \theta = \arctan(\frac{x}{2}) $$
+7. Rewriting the equation with $\theta$ in terms of x, we get:
+$$ \frac{3}{2}\arctan(\frac{x}{2}) + C$$
+This means that:
+$$ \int\frac{3}{4+x^2}dx = \frac{3}{2}\arctan(\frac{x}{2}) + C $$
+
+# A VERY LARGE LIST OF TRIG IDENTITIES FOR CALCULUS
+
+| Reciprocal           |
+| ------------------------------------ |
+| $\csc(x) = \dfrac{1}{sin(x)}$        |
+| $\sec(x) = \dfrac{1}{\cos(x)}$       |
+| $\cot(x) = \dfrac{1}{\tan(x)}$       |
+
+| Quotient |
+| -- |
+| $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$  |
+| $\cot(x) = \dfrac{\cos(x)}{\sin(x)}$  |
+
+| Pythagorean |
+| -- |
+| $\sin^2(x) + \cos^2(x) = 1$          |
+| $1 + \tan^2(x) = \sec^2(x)$          |
+| $1 + \cot^2(x) = \csc^2(x)$          |
+| $1 

education/physics/PHYS2210/Unit 1.md

@@ -0,0 +1,67 @@
+Motion in a straight line is one dimensional.
+
+Kinematics or the physics of motion has 4 noteworthy parameters: time ($t$), position ($x$), velocity ($v$), and acceleration ($a$). 
+
+Kinematic problems have a start and end of motion.
+
+# Displacement
+
+Displacement is calculated with the formula:
+$$\Delta x = \text{x-value of final position} - \text{x-value of initial posiion}$$
+
+# Velocity
+Average velocity over a time interval $\Delta t$ is defined to be: **the displacement** (net change in position), **divided by** **the time taken**.
+
+$$ \bar{v} = \dfrac{\text{final position-initial position}}{\text{final time - initial time}} = \dfrac{x_2 - x_1}{t_2 - t_1} = \dfrac{\Delta x}{\Delta t}$$
+**Speed (m/s)** is defined to be the total distance traveled divided by the time taken. Speed and velocity are *not the same*. 
+
+
+$$ v_{\text{instant}} = v = \lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}$$
+- $x(t)$ -> **position** as a function of time
+- $v(t)$ -> **slope** of position-vs-time (derivative of $x(t)$)
+- $a(t)$ -> **slope** of velocity-vs-time (derivative of $v(t)$)
+# Acceleration
+To find the instantaneous acceleration, we can apply the formula:
+
+$$a_{\text{instant}} = a = \frac{dv}{dt} = \frac{d}{dt} \frac{dx}{dt} = \frac{d^2x}{dt^2}$$
+## Equations of Motion for Constant Acceleration
+1. $v = v_0 + at$  - Use when missing $x$
+2. $x = x_0 + \frac{1}{2}(v_0 + v)t$  - Use when missing $a$
+3. $x = x_0 + v_0 t + \frac{1}{2} a t^2$  - Use when missing $v$
+4. $v^2 = v_0^2 + 2a(x - x_0)$  - Use when missing $t$ 
+
+(Actually useful equations)
+$$ t = \frac{v_f - v_i}{a} $$
+$$ a = \frac{v - v_0}{t}
+$$s = v_it + \frac{1}{2}at^2 $$
+| -------------- | ----- |
+| $t_0$          | $t$   |
+| $v_0$          | $v$   |
+| $x_0$          | $x$   |
+| $a$ (constant) | $a$   |
+## Examples
+
+ > Sally aggressively drives her Alfa Romeo from rest to 50 m/s in 6s. What is her acceleration?
+1. Start:
+	- $v_0 = 0 m/s$
+	- $t_0 = 0s$
+	- $x_0 = 0m$
+2. Final:
+	- $v = 50 m/s$
+	- $t = 6s$
+	- $x = ?$ 
+3. Using the equation $v = v_0 + at \to a = \frac{(v - v_0)}{t}$, $a$ = $(50 m/s - 0 m/s)/ 6s = 8.3m/s^2$ 
+4. Assess: do the units make sense? Is the answer reasonable?
+
+> An object slides until coming to a rest. It traveled 15 meters in 3 seconds. What was the object's acceleration?
+1. The problem is a kinematics problem, refer to the above formula.
+2. Acceleration is a constant
+3. Problem has two unknowns, $v_0$ and $a$, let's solve for $v_0$ first.
+4. $x = x_0 + \frac{1}{2}(v_0 + v) t \to v_0 = 2*(x - x_0)/ t \to 2*(15)/3 = 10$
+5. $v = v_0 + at \to a = (v - v_0) /t \to a = (-10 m/s) / 3s = -3.33m/s^2$ 
+6. Asses: Units? Answer make sense? Significant figures? Acceleration speeding up or slowing down?
+# Gravity
+- Is the problem 1 dimensional?
+- Could draw axis pointing upwards, call it y axis
+- If not specified, assume acceleration is $g = 9.8 m/s^2$.
+- Direction is important, $g$ is down towards the earth, so it's often negative. The sign depends on your choice of axis.