︠28343162-e9b1-4afc-a2e5-e5721cfdc7fesi︠
latex.add_package_to_preamble_if_available("color")
latex.add_package_to_preamble_if_available("tikz")
latex.add_package_to_preamble_if_available("pgfplots")
︡49d39df4-e980-4aad-8605-c136d624b93b︡{"done":true}︡
︠9ae6def7-22ac-46a3-8d49-b9d88edfbd62si︠
%latex
\textbf{What is the area?}
\begin{enumerate}
\item What does the area mean?
\item How to find the area?
\end{enumerate}

{\color{red} Please feel free to ignore the coding part.}

%{\color{red} The file will be post on BlackBoard.}
︡a3a63568-00ec-4c17-8532-8e5e32b71c36︡{"file":{"filename":"/tmp/tmp_9SJQF.png","show":true,"text":null,"uuid":"e2f85e63-23c8-4480-849c-fad4ab4ce952"},"once":false}︡{"done":true}︡
︠5e9ff6b5-e8f8-46bd-9014-5e7820f7e9ddsi︠
%latex
\begin{tikzpicture}
\draw[fill=orange] (0,0) rectangle (3,2);
\node at (1.5,-0.5) {$3$};
\node at (-0.5,1) {$2$};
\end{tikzpicture}

$\text{Rectangle Area}=\text{Base}\times\text{Height}=3\times 2=6 $
︡145b0f62-54a7-4f89-bb09-400f1adf320c︡{"file":{"filename":"/tmp/tmp0pMULr.png","show":true,"text":null,"uuid":"0db92e40-d4cc-45ce-a196-d5be2ce91103"},"once":false}︡{"done":true}︡
︠23bd992e-bdce-466c-a1e7-9d468c49e075si︠
%latex
\begin{tikzpicture}
\draw[fill=gray!20] (0,0) rectangle (3,2);
\draw[fill=orange] (0,0) -- (3,0) -- (0,2) -- cycle;
\node at (1.5,-0.5) {$3$};
\node at (-0.5,1) {$2$};
\end{tikzpicture}

$\text{Triangle Area}=\frac{1}{2}\times\text{Base}\times\text{Height}=\frac{1}{2}\times 3\times 2=3 $
︡6000e642-87c0-4f1c-8080-0de31d486d70︡{"file":{"filename":"/tmp/tmp4BJjsD.png","show":true,"text":null,"uuid":"3ba1f65c-7e89-400c-a136-46b28751c18e"},"once":false}︡{"done":true}︡
︠af098a34-2d26-4557-b9bd-15b7cfa6a8cesi︠
%latex
\begin{tikzpicture}
\draw[fill=orange] (3,0) arc (0:90:3) -- (0,0) -- cycle;
\node at (1.5,-0.5) {$3$};
\draw (0.5,0) -- (0.5,0.5) -- (0,0.5);
\end{tikzpicture}

$\text{Fan-shape Area}=\pi r^2\times \frac{\theta}{2\pi}=\frac{1}{2}r^2\theta=\frac{1}{2}\times 3^2\times \frac{\pi}{2}=\frac{9}{4}\pi\sim 7.068$

But WHY?
︡fdcb01db-afb1-4bbd-b38d-039e1e2e1719︡{"file":{"filename":"/tmp/tmpkZ_LKd.png","show":true,"text":null,"uuid":"93b2cc62-e496-4eec-8363-116b636914d7"},"once":false}︡{"done":true}︡
︠1b2e2669-4de5-4283-b064-d114abac86e2s︠
%latex
\begin{tikzpicture}
\draw[fill=orange] (3,0) arc (0:90:3) -- (0,0) -- cycle;
\node at (1.5,-0.5) {$3$};
\draw[step=1] (0,0) grid (3,3);
\end{tikzpicture}

Try changing step.
︡51c93457-2641-4e9d-b6c6-36481e402ba5︡{"file":{"filename":"/tmp/tmppcuYXB.png","show":true,"text":null,"uuid":"c92450e9-5bac-4006-8fe6-bb0288890c70"},"once":false}︡{"done":true}︡
︠8926773b-bc28-4814-bdd5-be765f634c5fs︠
gap=1;
unitarea=gap^2;
x,y=gap,gap
area,counter=0,0;
while y<=3:
    if x^2+y^2<=9:
        counter+=1;
        area+=unitarea;
        x+=gap;
    else:
        y+=gap;
        x=gap;
print counter,"*",unitarea,"=",area
print "Real answer:", N(9*pi/4);
print "Try changing gap"
︡01c6098b-7ce3-4bce-91bb-e4c174f2473b︡{"stdout":"4 * 1 = 4\n"}︡{"stdout":"Real answer: 7.06858347057703\n"}︡{"stdout":"Try changing gap\n"}︡{"done":true}︡
︠989e886b-a5b4-4495-b7c0-a6061d99f1aasi︠
%latex
\textbf{What does the area mean?}

$\text{Area}=\text{number of building blocks}\times\text{area of building block}$
︡d73c3835-5a2a-40f3-86e9-430e1ec77359︡{"file":{"filename":"/tmp/tmp5zVeZR.png","show":true,"text":null,"uuid":"2e41061c-ef74-4055-bf06-93aea529b02f"},"once":false}︡{"done":true}︡
︠1734bb58-5379-4992-8714-6e9387f56c45si︠
%latex
\textbf{How to find the area?}
\begin{itemize}
\item Area is {\color{red} approximated} by total area of rectangles.
\item When {\color{red} taking the limit}, it becomes the area.
\end{itemize}
︡e66133dc-94b4-432f-9f3e-156a86457854︡{"file":{"filename":"/tmp/tmpSbh54K.png","show":true,"text":null,"uuid":"2752c421-255b-4a65-a47b-9ee83f6567d6"},"once":false}︡{"done":true}︡
︠f6ae4fb0-6559-46b4-86fb-a1b97772c71cs︠
n=5;
x=var("x")
f=(sin(x)).function(x);
area,pic=integration_approx(f,0,pi,n);
show(pic,title="sin(x), x in [0,pi]")
print " "
print "Divided into %s segments gives the area %s"%(n,N(area));
print "Real answer:", integral(f,(x,0,pi));
print "Try changing n"
︡294e17f2-bf74-46db-a394-a9ec80eff724︡{"file":{"filename":"/projects/5b57d5ed-ca32-426f-8aa4-df4c47b5d13d/.sage/temp/compute3-us/12328/tmp_asDgFp.svg","show":true,"text":null,"uuid":"5af61eec-e49c-45d8-9959-dc08901b0c84"},"once":false}︡{"stdout":" \n"}︡{"stdout":"Divided into 5 segments gives the area 1.93376559809280\n"}︡{"stdout":"Real answer: 2\n"}︡{"stdout":"Try changing n\n"}︡{"done":true}︡
︠bf9690bc-079c-47ff-a2e8-59efef05d9bbsi︠
%latex
\textbf{Mathematical formula}
\begin{enumerate}
\item $n:$ number of segments, so length of each segment is $\ell=\frac{b-a}{n}$;
\item $x_i=a+i\times \ell, i=0,1,\ldots ,n-1$;
\item each height is $f(x_i)$;
\item then total area approximation is 
\[\ell\cdot f(x_0)+\ell\cdot f(x_1)+\cdots +\ell\cdot f(x_{n-1})=\frac{b-a}{n}[f(x_0)+f(x_1)+\cdots +f(x_{n-1})].\]
\item The {\color{red} real area} is 
\[\lim_{n\rightarrow \infty} \frac{b-a}{n}[f(x_0)+f(x_1)+\cdots +f(x_{n-1})].\]
\end{enumerate}
︡901cd167-98b4-44db-af7b-d7e4f64f449b︡{"file":{"filename":"/tmp/tmp6bwYcc.png","show":true,"text":null,"uuid":"ff9037ff-8f23-423a-92b1-918c832a864c"},"once":false}︡{"done":true}︡
︠778b9f94-0443-4840-8f06-dea5b4386543si︠
%latex
\textbf{Definition of the definite integral}
\begin{align*}
\int_a^bf(x)dx & =\lim_{n\rightarrow \infty} \frac{b-a}{n}[f(x_0)+f(x_1)+\cdots +f(x_{n-1})]\\
 & =\lim_{x\rightarrow \infty} \frac{b-a}{n}\sum_{i=0}^{n-1}f(x_i)\\
 & =\lim_{x\rightarrow \infty} \frac{b-a}{n}\sum_{i=0}^{n-1}f(a+i\cdot \ell)\\
\end{align*}
︡e2739b9f-adcb-4cba-993f-f4596fd2c74e︡{"file":{"filename":"/tmp/tmp3FfQt1.png","show":true,"text":null,"uuid":"f6e31288-8fbd-4728-b17b-925fa8c4552c"},"once":false}︡{"done":true}︡
︠6f2168b8-3928-43b7-b430-c23a2e302951si︠
%latex
\textbf{Sigma notation:} $\displaystyle \sum_{i=1}^{k}a_i=a_1+a_2+\cdots +a_{k}$

Examples: [p. 308 and 309]
\begin{itemize}
\item $\displaystyle \sum_{i=1}^k1=\underbrace{1+1+\cdots +1}_{k\text{ times}}=n$
\item $\displaystyle \sum_{i=1}^ki=1+2+\cdots +k=\frac{k(k+1)}{2}~~~[\text{Gau\ss}]$
\item $\displaystyle \sum_{i=1}^ki^2=1^2+2^2+\cdots +k^2=\frac{k(k+1)(2k+1)}{6}$
\item $\displaystyle \sum_{i=1}^ki^3=1^3+2^3+\cdots +k^3=\left(\frac{n(n+1)}{2}\right)^2$
\end{itemize}
︡9d33c7f7-5d9d-4948-ba07-7e42a1ec64ff︡{"file":{"filename":"/tmp/tmpBM_yuw.png","show":true,"text":null,"uuid":"bbe7d7cd-251d-4a20-afae-d71e32bda1ed"},"once":false}︡{"done":true}︡
︠54591125-7fba-4e89-a3be-f207226dbb05si︠
%latex
\textbf{Basic properties}
\begin{itemize}
\item $\displaystyle \sum_{i=1}^k (a_i\pm b_i)=\sum_{i=1}^k a_i+\sum_{i=1}^k b_k$
\item $\displaystyle \sum_{i=1}^kca_i=c\sum_{i=1}^ka_i$, since $(ca_1+ca_2+\cdots ca_k)=c(a_1+a_2+\cdots a_k)$
\end{itemize}

Note: $\displaystyle\sum_{i=1}^ki^2+b_i$ is a bad notation; write either $\displaystyle\sum_{i=1}^k(i^2+1)$ or $\displaystyle\left(\sum_{i=1}^ki^2\right)+1$.
︡97bfebd6-d6a9-4bae-9e79-60d9aebf2c18︡{"file":{"filename":"/tmp/tmpO8z0R2.png","show":true,"text":null,"uuid":"b74c12c2-36af-4a3c-80de-b5ec044f11b9"},"once":false}︡{"done":true}︡
︠3bbd45ee-0d21-4df4-a36f-091e353a0cf8si︠
%latex
\textbf{Finding area by definition}

[The essential concept of calculus, but computation is tedious.]


\textbf{e.g.} Find the area under $f(x)=x^2$ for $0\leq x\leq 2$.
\begin{align*}
\text{Area}& = \lim_{x\rightarrow \infty} \frac{b-a}{n}\sum_{i=0}^{n-1}f(a+i\cdot \ell)\\
 & =\lim_{x\rightarrow \infty}\frac{2}{n}\sum_{i=0}^{n-1}f(0+i\cdot \frac{2}{n}) \\
% & =\lim_{x\rightarrow \infty}\frac{2}{n}\sum_{i=0}^{n-1}(i\cdot \frac{2}{n})^2 \\
% & =\lim_{x\rightarrow \infty}\frac{2}{n}\sum_{i=0}^{n-1}\left(i^2\cdot \frac{4}{n^2}\right) \\
% & =\lim_{x\rightarrow \infty}\frac{8}{n^3}\sum_{i=0}^{n-1}i^2\\
% & =\lim_{x\rightarrow \infty}\frac{8(n-1)(n)(2n-1)}{6n^3}\\
% & =\frac{8}{3}\\
\end{align*}
\[\text{Note:} \sum_{i=1}^ki^2=\frac{k(k+1)(2k+1)}{6}.\]
︡63a0288b-5bb5-41e6-83a9-289ad16e9ab5︡{"file":{"filename":"/tmp/tmppAYjkf.png","show":true,"text":null,"uuid":"7169e93d-8653-4b02-b904-3a383123ffa6"},"once":false}︡{"done":true}︡
︠2387308e-9175-4616-a9a0-9713e5d20b21s︠
x=var("x")
f=(x^2).function(x);
pic=f.plot(xmin=0,xmax=2,fill=True,figsize=3);
show(pic)
︡9cb6e4ec-217a-4e27-be15-4bd55c0bfce4︡{"file":{"filename":"/projects/5b57d5ed-ca32-426f-8aa4-df4c47b5d13d/.sage/temp/compute3-us/12328/tmp_yUif0x.svg","show":true,"text":null,"uuid":"14ffd43f-08e7-4d6a-b1f8-60faf1e6d3b0"},"once":false}︡{"done":true}︡
︠81608952-a0cb-4683-8834-87654b6d264dsi︠
%latex
\textbf{Finding area by antiderivative} 

[Always work, so be familiar with finding the antiderivative!]

\textbf{e.g.} Find the area under $f(x)=x^2$ for $0\leq x\leq 2$.

Be definition, {\color{red} the area is $\int_0^2x^2dx$}.

To find this value, we need to know {\color{red} an antiderivative} $F(x)=\frac{1}{3}x^3$.

Then the answer is $F(2)-F(0)=\frac{1}{3}\cdot (2)^3-\frac{1}{3}\cdot (0)^3=\frac{8}{3}$.  Done!
︡c594cb60-5621-423b-957a-b45bd322408d︡{"file":{"filename":"/tmp/tmplt9E07.png","show":true,"text":null,"uuid":"bd612b24-d21a-44b0-b0ea-bb3ba23bf05e"},"once":false}︡{"done":true}︡
︠6f71d3a5-3f26-4f14-abff-4738cbdf8665si︠
%latex
\textbf{Finding area by basic geometry}

[The easist way! But only work for few cases.]

\textbf{e.g.} Let $f(x)$ be a function shown as the graph.  

\begin{tikzpicture}
\draw[->] (0,0) -- (4.5,0);
\draw[->] (0,0) -- (0,4.5);
\draw[fill=orange] (0,2) arc (180:0:1) -- (3,0) -- (0,0) -- cycle;
\draw[fill=green] (3,0) -- (4,-1) -- (4,0) -- cycle;
\foreach \i in {1,2,3,4}{
\draw (\i,0.2) -- (\i,-0.2);
\draw (0.2,\i) -- (-0.2,\i);
}
\end{tikzpicture}

\begin{enumerate}
\item Find $\int_0^2f(x)dx=$ 
\tikz{\draw (0,0) rectangle (2,2);} 
$+$ \tikz{\draw (2,0) arc (0:180:1) -- cycle;} 
$=2\cdot 2+\frac{1}{2}\cdot \pi \cdot 1^2=4+\frac{\pi}{2}$.
\item Find $\int_1^3f(x)dx=$ 
\tikz{\draw (0,0) rectangle (2,2);} 
$+$ \tikz{\draw (2,0) arc (0:180:1) -- cycle;}
$+$ \tikz{\draw (0,0) -- (1,0) -- (0,2) -- cycle;}
$=2\cdot 2+\frac{1}{2}\cdot \pi\cdot 1^2+\frac{1}{2}\cdot 1\cdot 2=5+\frac{\pi}{2}$.
\item Find $\int_2^4f(x)dx=$.
\tikz{\draw (0,0) -- (1,0) -- (0,2) -- cycle;}
{\color{red} $-$} \tikz{\draw (0,0) -- (1,0) -- (1,-1) -- cycle;}
$= \frac{1}{2}\cdot 1\cdot 2{\color{red}-}\frac{1}{2}\cdot 1\cdot 1=\frac{1}{2}$
\end{enumerate}
︡20e0c850-7622-4339-b81b-15b415f4c08a︡{"file":{"filename":"/tmp/tmp0JqWQ3.png","show":true,"text":null,"uuid":"5fa8c990-9e8e-4dd8-a2e6-6e443f644880"},"once":false}︡{"done":true}︡
︠be26c9fc-4ead-4a12-b332-a2c3d528ec18si︠
%latex
\textbf{Summary}

\begin{itemize}
\item Finding the area by definition is the concept to keep in mind.
\item Findging the area by the antiderivative is the most efficient way.
    \begin{itemize}
    \item But you need to know the antiderivative!
    \end{itemize}
\item Finding the area by geometry is easy and fast, but does not always work!
\end{itemize}
︡d42faeda-0960-4a94-bbb4-128505c09e7f︡{"file":{"filename":"/tmp/tmpdCL4x4.png","show":true,"text":null,"uuid":"1631c795-df47-4fa5-847f-8f9535701d67"},"once":false}︡{"done":true}︡
︠18bd70ba-1462-409b-ada4-f92bda7643a8s︠
#Run this cell before all cells

def integration_approx(f,a,b,pieces):
    """
    Input:
        f: function defined on [a,b];
        pieces: number of pieces;
    Output:
        By dividing [a,b] into "pieces" pieces, find the area under f.
        Return area, picture;
    """
    gap=(b-a)/pieces;
    base=[(a+i*gap,a+i*gap+gap) for i in range(pieces)];
    height=[f(a+i*gap) for i in range(pieces)];
    area=gap*sum(height);
    pic=f.plot(xmin=a,xmax=b,fill=True,color="orange");
    recs=sum([polygon([(base[i][0],0),(base[i][1],0),(base[i][1],height[i]),(base[i][0],height[i])],edgecolor="black",fill=False) for i in range(pieces)])
    xlabel=sum([text("x%s"%i,(base[i][0],-0.2)) for i in range(pieces)])
    ylabel=sum([text("f(x%s)"%i,(base[i][0],height[i]/2)) for i in range(pieces)])
    return area,pic+recs+xlabel+ylabel;
︡6a19de1b-d35a-42a7-b20a-fc5b35b8de91︡{"done":true}︡
︠f7aafecc-3a09-4eb3-a9bb-a62a3cac920cs︠
︡a0f07bb4-8ace-42cc-abd9-c9fd6605c4c3︡{"done":true}︡
︠96c1b95a-17a3-46e4-8722-cd5dcc455666︠












Collaborative Calculation and Data Science

colby

sagemath

cornellcollege

facebook