daviddoji/CalculusWithJuliaNotes.jl
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

layout fix

Browse Source
This commit is contained in:
jverzani
2023-04-23 12:40:35 -04:00
parent 8210c27844
commit a39e24323d
1 changed files with 6 additions and 5 deletions
Download Patch File Download Diff File Expand all files Collapse all files

11
quarto/integrals/area.qmd
View File

@@ -368,10 +368,11 @@ The area is invariant under shifts left or right.
Any partition $a =x_0 < x_1 < \cdots < x_n=b$ is related to a partition of $[a-c, b-c]$ through $a-c < x_0-c < x_1-c < \cdots < x_n - c = b-c$. Let $d_i=c_i-c$ denote this partition, then we have: Any partition $a =x_0 < x_1 < \cdots < x_n=b$ is related to a partition of $[a-c, b-c]$ through $a-c < x_0-c < x_1-c < \cdots < x_n - c = b-c$. Let $d_i=c_i-c$ denote this partition, then we have:
$$ \begin{align*}
f(c_1 -c) \cdot (x_1 - x_0) + f(c_2 -c) \cdot (x_2 - x_1) + \cdots + f(c_n -c) \cdot (x_n - x_{n-1}) = f(c_1 -c) \cdot (x_1 - x_0) &+ f(c_2 -c) \cdot (x_2 - x_1) + \cdots + f(c_n -c) \cdot (x_n - x_{n-1}) = \\
f(d_1) \cdot(x_1-c - (x_0-c)) + f(d_2) \cdot(x_2-c - (x_1-c)) + \cdots + f(d_n) \cdot(x_n-c - (x_{n-1}-c)). & f(d_1) \cdot(x_1-c - (x_0-c)) + f(d_2) \cdot(x_2-c - (x_1-c)) + \cdots + f(d_n) \cdot(x_n-c - (x_{n-1}-c)).
$$ \end{align*}
The left side will have a limit of $\int_a^b f(x-c) dx$ the right would have a "limit" of $\int_{a-c}^{b-c}f(x)dx$. The left side will have a limit of $\int_a^b f(x-c) dx$ the right would have a "limit" of $\int_{a-c}^{b-c}f(x)dx$.
@@ -498,7 +499,7 @@ Cauchy showed this using a *geometric series* for the partition, not the arithme
\begin{align*} \begin{align*}
S &= a^{k+1}\alpha u^0 + a^{k+1}\alpha u^1 + \cdots + a^{k+1}\alpha u^{n-1} S &= a^{k+1}\alpha u^0 + a^{k+1}\alpha u^1 + \cdots + a^{k+1}\alpha u^{n-1}\\
&= a^{k+1} \cdot \alpha \cdot (u^0 + u^1 + \cdot u^{n-1}) \\ &= a^{k+1} \cdot \alpha \cdot (u^0 + u^1 + \cdot u^{n-1}) \\
&= a^{k+1} \cdot \alpha \cdot \frac{u^n - 1}{u - 1}\\ &= a^{k+1} \cdot \alpha \cdot \frac{u^n - 1}{u - 1}\\
&= (b^{k+1} - a^{k+1}) \cdot \frac{\alpha}{(1+\alpha)^{k+1} - 1} \\ &= (b^{k+1} - a^{k+1}) \cdot \frac{\alpha}{(1+\alpha)^{k+1} - 1} \\