Derivative - slope
\[\eqalign{\frac{dy}{dx} &=& \frac{d(4x - x^3)}{dx}\\
&=& 4-3x^2\\}\]
Key proof for derivative
\[\eqalign{ \frac{dy}{dx} &=& lim_{d\rightarrow 0} \left(\frac {4(x+d) - (x+d)^3 - (4x - x^3)}{d}\right)\\
&=&lim_{d\rightarrow 0} \left(\frac{4x + 4d -x^3 -3dx^2 - 3d^2x - d^3 - 4x + x^3}{d}\right)\\
&=&lim_{d\rightarrow 0} \left(\frac{ 4d -3dx^2 - 3d^2x - d^3}{d}\right)\\
&=&lim_{d\rightarrow 0} \left(4 -3x^2 - 3dx - d^2 \right) = 4 -3x^2 \\
}\]
Integration - area
\[\eqalign{
area_{rthnd} &=& h \left(f(x_1) + f(x_2) + ... + f(x_4)\right)\\
area_{midpt} &=& h \left({\Large f}\left(\frac{x_0 +x_1}{2}\right) + {\Large f}\left(\frac{x_1 +x_2}{2}\right) + ...+{\Large f}\left(\frac{x_3 +x_4}{2}\right)\right)\\
area_{trap} &=& \frac{h}{2}\left(f\left(x_0\right) + 2 f\left(x_1\right) + 2 f\left(x_2\right)+
+ 2 f\left(x_3\right) + + 2 f\left(x_1\right)\right)\\
area_{simp}&=& \frac{h}{3}\left(f\left(x_0\right) + 4 f\left(x_1\right) + 2 f\left(x_2\right)
+ 4 f\left(x_3\right) + f\left(x_4\right)\right)\\
area_{int} &=& \int_{x_0}^{x_4} f(x)dx\\
}\]
Numerical Integration
| Right hand rule |
3.50 |
| Midpoint rule |
4.125 |
| Trapezoidal rule |
3.75 |
| Simpson rule |
4 |
| Integration by antiderivative |
4 |
Application
- Injection of dye in a vein to measure cardiac blood flow
- Requires integrating dye concentration in arterial blood over time.
- Also useful for measuring the flow of rivers and glacier flows.
\[\eqalign{\left(\frac{mg}{area}\right) &=&\frac{mg_{total}}{\left(\frac{mg}{Litre_{blood}}\right)\cdot sec}\\
&=& \left(\frac{mg}{sec}\right)\cdot\left(\frac{Litre_{blood}}{mg}\right)\\
&=& \frac{Litre_{blood}}{sec}}\]
Dye concentration
$
\[\begin{matrix}
time & 5 & 7 & 9 & 11& 13\\
dye & 0 & 3.8 & 8.0 & 6.1 & 3.6\\
\\
time & 15 & 17 & 19 & 21 & 23\\
dye & 2.3 & 1.45 & 0.91 & 0.57 & 0.36 \\
\\
time & 25 & 27 & 29 & 31 & 33\\
dye & 0.23 & 0.14 & 0.09 & 0.05 & 0\\
\end{matrix}\]
$
Area of a rectangle
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Integration of a line
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Area of Parabola
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4th power
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Smaller intervals
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Even smaller intervals
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Yet smaller intervals
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Matrix multiplicartion (Trapezoid Rule)
\[\eqalign{area &=& y \times wt \times correction\\
&=& [0\quad 2.716049\quad 3.160494\quad 4\quad 5.530864\quad 5.679012\quad 0] \left[\begin{matrix}1\\ 2\\ 2\\ 2\\ 2\\ 2\\1\\\end{matrix}\right] \times \left(\frac{\frac{2}{3}}{2}\right)\\
&=& ( 0 + 5.432099 + 6.320988 + 8 + 11.061728 + 11.358025 + 0)\times \frac{1}{3}\\
&=& \frac{42.17284}{3} = 14.05761\\}\]
Matrix Multiplication (Simpson Rule)
\[\eqalign{area &=& y \times wt \times correction\\
&=& [0\quad 2.716049\quad 3.160494\quad 4\quad 5.530864\quad 5.679012\quad 0] \left[\begin{matrix}1\\ 4\\ 2\\ 4\\ 2\\ 4\\1\\\end{matrix}\right] \times \left(\frac{2}{3}\right)\left(\frac{1}{3}\right)\\
&=& (0.000000 + 10.864198 + 6.320988 + 16.000000 + 11.061728 + 22.716049 + 0)\times \frac{2}{9}\\
&=& \frac{66.96296}{4} = 14.88066\\}\]
Matrix multiplication (3/8 Rule)
\[\eqalign{area &=& y \times wt \times correction\\
&=& [0\quad 2.716049\quad 3.160494\quad 4\quad 5.530864\quad 5.679012\quad 0] \left[\begin{matrix}1\\ 3\\ 3\\ 2\\ 3\\ 3\\1\\\end{matrix}\right] \times \left(\frac{2-0}{3}\right)\left(\frac{3}{8}\right)\\
&=& (0 + 8.148147 + 9.481482 + 8 + 16.592592 + 17.037036 + 0)\times \frac{2}{8}\\
&=& \frac{59.25926}{4} = 14.81481\\}\]
Antideriative
\[\eqalign{\int_0^4 f\left(\frac{-x^4}{2}+\frac{7x^3}{2}-8x^2+8x\right) dx &=& \bigg|_0^4 \frac{-x^5}{10}+\frac{7x^4}{8}-\frac{8x^3}{3}+4x^2\\
&=&\frac{-\left(4^5\right)}{10}+\frac{7\left(4^4\right)}{8}-\frac{8\left(4^3\right)}{3}+4\left(4^2\right) - 0\\
&=&\frac{-1024}{10}+\frac{7\left(256\right)}{8}-\frac{8\left(64\right)}{3}+4\left(16\right)\\
&=&-102.4 + 224 - 170.6667 + 64 = 14.9333\\}\]
More rules
\[\eqalign{\int e^x dx &=& e^x +c\\
\int e^{kx} dx &=& \frac{e^{kx}}{k} +c,\quad k \ne 0\\
\int a^x dx &=& \frac{a^x}{\ln(a)} +c\\
\int a^{kx} dx &=& \frac{a^{kx}}{k\ln(a)} + C,\quad k \ne 0\\}\]
More samples
\[\eqalign{\int e^{9t} dt &=& ???\\
&=&\frac{e^{9t}}{9} dt\\
\int −x^2(2x \sin(2x+1) -3cos(2x+1) &=& \int 3x^2 cos(2x+1) -2x^3 sin(2x+1) &=& ???\\
&=& x^3 \cos(2x+1)\\
\int \frac{3}{3x+1} &=& \int \frac{1}{u} du = ???\quad u= 3x+1; du = 3dx\\
&=& \ln(u) = \ln(3x + 1) &
}\]
Landing a plane
\[\eqalign{y&=&\frac{\left(x-10\right)^{2}\left(x+5\right)}{50}\\
y&=&\frac{\left(x-10\right)^{2}\left(x+4\right)}{40.666}\\
y&=&\frac{\left(x-10\right)^{2}\left(x+3\right)}{32.6}\\
y&=&\frac{\left(x-10\right)^{2}\left(x+2\right)}{25.6}\\
y&=&\frac{\left(x-10\right)^{2}\left(x+1\right)}{19.6667}\\
y&=&\frac{\left(x-10\right)^{2}x}{14.8}\\
}\]
Landing a plane
\[\eqalign{y&=&\frac{\left(x-10\right)^{2}\left(x+5\right)}{50}\\
y&=&\frac{\left(x-10\right)^{2}\left(x+4\right)}{40.666}\\
y&=&\frac{\left(x-10\right)^{2}\left(x+3\right)}{32.6}\\
y&=&\frac{\left(x-10\right)^{2}\left(x+2\right)}{25.6}\\
y&=&\frac{\left(x-10\right)^{2}\left(x+1\right)}{19.6667}\\
y&=&\frac{\left(x-10\right)^{2}x}{14.8}\\
}\]
Landing a plane
\[\eqalign{y&=&\frac{\left(x-10\right)^{2}\left(x+5\right)}{50}\\
y&=&\frac{\left(x-10\right)^{2}\left(x+4\right)}{40.666}\\
y&=&\frac{\left(x-10\right)^{2}\left(x+3\right)}{32.6}\\
y&=&\frac{\left(x-10\right)^{2}\left(x+2\right)}{25.6}\\
y&=&\frac{\left(x-10\right)^{2}\left(x+1\right)}{19.6667}\\
y&=&\frac{\left(x-10\right)^{2}x}{14.8}\\
}\]
Area between curves
\[\eqalign{\int_0^1 \left(\sqrt{x} - x^3\right) dx &=& \bigg|_0^1 \frac{2x^{3/2}}{3}-\frac{x^4}{4}\\
&=& \frac{2}{3}-\frac{1}{4}= \frac{8-3}{12}\\
&=& \frac{5}{12}=0.4166667\\}\]
Area between curves
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Area between curves
\[\eqalign{Area&=&\int_{-2}^{2} \left(f(x)-g(x)\right) dx\\
&=&\int_{-2}^{2} \left(2x+6\right) - \left(4-x^2\right) dx\\
&=&\bigg|_{-2}^{2} \frac{x^3}{3} + x^2 +2x\\
&=&\left(\frac{8}{3} +4 +4\right) - \left(\frac{-8}{3} + 4 -4\right)\\
&=&\frac{16}{3} + 8\\
&=&13.33333}\]
\[\eqalign{Area&=&\int_{-2}^{2} f(x) dx - \int_{-2}^{2} g(x) dx \\
&=&\int_{-2}^{2} \left(2x +6\right) dx - \int_{-2}^{2} \left(4-x^2\right) dx \\
&=&\bigg|_{-2}^{2} \left(x^2 +6x\right) - \bigg|_{-2}^{2} \left(4x - \frac{x^3}{3}\right)\\
&=&\left((4+12) - (4 -12)\right) - \left(\left(8 - \frac{8}{3}\right) - \left(-8 +\frac{8}{3]}\right)\right)\\
&=&24-16+ \frac{16}{3}\\
&=&13.33333\\
}\]
Integration by substitution
\[\eqalign{\int f(x) dx &=& \int \frac{f(g'(u))}{d(g'(u))}du \\
\\
\int (2x+1)^{16} dx &=& \int u^{16} \frac{1}{2}du = \frac{1}{2(17)}u^{17} +C = \frac{1}{34}(2x+1)^{17} + C,\quad u = 2x + 1\\
\int \frac{2x}{\sqrt{x^2 + 1}}dx &=& \int \frac{1}{u^{\frac{1}{2}}} du = 2u^{\frac{1}{2}}
+ C= 2\sqrt{x^2 + 1} + C,\quad u= x^2 +1\\
}\]
Basic Calculus
Area in 2D
\[\int_0^1 4x\ dx = \bigg|_0^1 2x^2 = 2 (1^2 - 0^2) = 2\]
Volume in 3D
\[\int_0^1 \pi(4x)^2 dx = \bigg|_0^1 \frac{16\pi x^3}{3} = \frac{16\pi}{3} - 0 = \frac{16\pi}{3}\]
Cylinder
\[\int_0^1 \pi(4x) dx = \bigg|_0^1 \frac{16\pi x^3}{3} = \frac{16\pi}{33} - 0 = \frac{16\pi}{3}\]
Surface area of a sphere
\[\eqalign{\int_{-\pi/2}^{\pi/2} 2\pi r \cos(A) r dA&=&4\pi r^2\bigg|_{-\pi/2}^{\pi/2} sin(A)\\
&=&4\pi r^2 (1-0)\\
&=&4\pi r^2}\]
Volume of a sphere
\[\eqalign{Area &=& 4\pi r^2\\
Volume &=& 4\pi r^2 dr\\
&=& \int_{0}^{R} 4\pi r^2 dr\\
&=& \frac{4\pi}{3} \bigg|_0^R r^3 \\
&=& \frac{4\pi r^3}{3} (R^3-0)\\
&=& \frac{4\pi R^3}{3}}\]
Integration by parts
\[\eqalign{
\frac{d(uv)}{dx} &=& u\frac{dv}{dx} + v\frac{du}{dx}\\
d(uv) &=& u\ dv + v\ du\\
\int d(uv) &=& \int u\ dv + \int v\ du\\
uv &=& \int u\ dv + \int v\ du\\
\int u\ dv &=& uv -\int v du\\}
\]
An example
\[\eqalign{
\int x\ e^{5x} &:&\quad (Set-up)\\
dv&=& e^{5x}\\
u &=& x\\
du &=& dx\\
v &=& \int dv = \int e^{5x} dx
&=& \frac{xe^{5x}}{5} - \frac{e^5x}{25}\\
\int u \ dv &=& uv - \int v du\\
}\]
\[\eqalign{
\int x\ e^{5x} &=&x\left(\frac{e^{5x}}{5}\right) - \int \frac{e^{5x}}{5} dx\\
&=&\frac{e^{5x}}{5} - \frac{e^{5x}}{25} + C\\
&=&\frac{e^{5x}}{25}(5x - 1) + C\\
}\]
\[\pmatrix{D & &I\\
x & + & e^{5x}\\
1 & - & \frac{e^{5x}}{5}\\
0 & + & \frac{e^{5x}}{25}\\
}\longrightarrow \pmatrix{x\frac{e^{5x}}{5}\\
-\frac{e^{5x}}{25}\\}\longrightarrow \\
x\frac{e^{5x}}{5}
-\frac{e^{5x}}{25} \longrightarrow\\ \frac{e^{5x}}{25}(5x-1)\]
Another example
\[\eqalign{\int \ln(x) dx &:& (Setup)\\
dv &=& dx\\
u &=& \ln(x)\\
v &=& x\\
du &=& \frac{1}{x}dx\\}\]
\[\eqalign{\int \ln(x) dx &=& x\ln(x) - \int x\left(\frac{1}{x} dx\right)\\
&=& x\ln(x) - \int dx\\
&=& x\ln(x) - x + C\\}\]
Practice
\[\eqalign{\int (3x^2+4) e^{2x} dx &=& (6x^2 − 6x + 11)\frac{e^{2x}}{4} +C\\
\int \frac{\ln(x)}{x^2} dx&=& −\frac{\ln(x)+1}{x}+C\\
\int x^2 \ln(x) dx&=& \frac{x^3(3\ln(x)-1)}{9}+ C\\
\int \frac{1}{x\sqrt{4-x^2}} dx&=& −\frac{\ln\left(\frac{\sqrt{4-x^2}}{x}+\frac{2}{x}\right)}{2}+C\\
\int \frac{1}{4-x^2} dx &=& \frac{\ln\left(|x + 2|\right)
−\ln\left(|x-2|\right)}{4} + C\\}\]
Practice solutions
\[\eqalign{
\int (3x^2+4) e^{2x} dx &=& ???\\
\int \frac{\ln(x)}{x^2} dx&=& ???\\
\int x^2 \ln(x) dx&=& ???\\
\int \frac{1}{x\sqrt{4-x^2}} dx&=& ???\\
\int \frac{1}{4-x^2} dx &=& ???\\}\]
Landing a drone: Rate of descent
\[y=\ ax^{2}\ +\ bx\]
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Drone landing: height profile
\[y=\frac{ax^{3}}{3}\ +\frac{bx^2}{2} + C\]
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Population growth
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Geometric Example
Quinine is administered 50mg/day and 77% is metabolized by the end of the day.
| BCF+ Int |
0 |
11.5 |
14.15 |
14.75 |
14.89 |
14.93 |
| Deposit |
50 |
50 |
50 |
50 |
50 |
50 |
| Balance |
50 |
61.5 |
64.50 |
64.75 |
64.89 |
64.93 |
|
|
|
|
|
|
|
| Calculated |
50 |
61.50 |
64.15 |
64.75 |
64.89 |
64.93 |
\[S_n = \frac{a(1-r^n)}{1-r} = \frac{50(1-0.23^n)}{1-0.23}\]
Equilbrium
\[\eqalign{S_{eq} &=& S_{eq} r + a\\
S_{eq} - S_{eq} r &=& a \\
S_{eq} &=& \frac{a}{1-r} \\
&=& \frac{50}{0.77} = 64.93506\\}\]
Recursive definition
Two pennies are added every day
\[\eqalign{\frac{dy}{dx} &=& 2\\
dy &=& 2dx\\
\int dy &=& \int 2 dx\\
y &=& 2x + C\\}\] | Day | 1 | 2| 3| 4 | 5 | 6 | 7 | |:——:|:——:|:——:|:——:|:——:|:——:|:——:|:——:| |Incr | 2 | 2| 2 | 2 | 2 | 2| 2| |Balance | 2 | 4 | 6 | 8 | 10 | 12 | 14 | | | | | | | | | |Calculated | 2 | 4 | 6 | 8 | 10 | 12 | 14 |
Recursive definition to direct equation
\[\eqalign{\frac{dy}{dx} &=& -\frac{x}{y}\\
y\ dy &=& -x\ dx\\
\int y\ dy &=& \int -x\ dx\\
\frac{y^2}{2} &=& -\frac{x^2}{2} + k\\
x^2 + y^2 &=& C\\}\]
