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

edits

Browse Source
This commit is contained in:
jverzani
2025-07-23 08:05:43 -04:00
parent 31ce21c8ad
commit c3a94878f3
50 changed files with 3711 additions and 1385 deletions
Download Patch File Download Diff File Expand all files Collapse all files

502
quarto/integrals/area.qmd
View File

@@ -16,6 +16,8 @@ using Roots
---
![A jigsaw puzzle needs a certain amount of area to complete. For a traditional rectangular puzzle, this area is comprised of the sum of the areas for each piece. Decomposing a total area into the sum of smaller, known, ones--even if only approximate--is the basis of definite integration.](figures/jigsaw.png)
The question of area has long fascinated human culture. As children, we learn early on the formulas for the areas of some geometric figures: a square is $b^2$, a rectangle $b\cdot h$, a triangle $1/2 \cdot b \cdot h$ and for a circle, $\pi r^2$. The area of a rectangle is often the intuitive basis for illustrating multiplication. The area of a triangle has been known for ages. Even complicated expressions, such as [Heron's](http://tinyurl.com/mqm9z) formula which relates the area of a triangle with measurements from its perimeter have been around for 2000 years. The formula for the area of a circle is also quite old. Wikipedia dates it as far back as the [Rhind](http://en.wikipedia.org/wiki/Rhind_Mathematical_Papyrus) papyrus for 1700 BC, with the approximation of $256/81$ for $\pi$.
@@ -81,39 +83,46 @@ gr()
f(x) = x^2
colors = [:black, :blue, :orange, :red, :green, :orange, :purple]
## Area of parabola
## Area of parabola
function make_triangle_graph(n)
title = "Area of parabolic cup ..."
n==1 && (title = "Area = 1/2")
n==2 && (title = "Area = previous + 1/8")
n==3 && (title = "Area = previous + 2*(1/8)^2")
n==4 && (title = "Area = previous + 4*(1/8)^3")
n==5 && (title = "Area = previous + 8*(1/8)^4")
n==6 && (title = "Area = previous + 16*(1/8)^5")
n==7 && (title = "Area = previous + 32*(1/8)^6")
n==1 && (title = L"Area $= 1/2$")
n==2 && (title = L"Area $=$ previous $+\; \frac{1}{8}$")
n==3 && (title = L"Area $=$ previous $+\; 2\cdot\frac{1}{8^2}$")
n==4 && (title = L"Area $=$ previous $+\; 4\cdot\frac{1}{8^3}$")
n==5 && (title = L"Area $=$ previous $+\; 8\cdot\frac{1}{8^4}$")
n==6 && (title = L"Area $=$ previous $+\; 16\cdot\frac{1}{8^5}$")
n==7 && (title = L"Area $=$ previous $+\; 32\cdot\frac{1}{8^6}$")
plt = plot(f, 0, 1, legend=false, size = fig_size, linewidth=2)
annotate!(plt, [(0.05, 0.9, text(title,:left))]) # if in title, it grows funny with gr
n >= 1 && plot!(plt, [1,0,0,1, 0], [1,1,0,1,1], color=colors[1], linetype=:polygon, fill=colors[1], alpha=.2)
n == 1 && plot!(plt, [1,0,0,1, 0], [1,1,0,1,1], color=colors[1], linewidth=2)
plt = plot(f, 0, 1;
legend=false,
size = fig_size,
linewidth=2)
annotate!(plt, [
(0.05, 0.9, text(title,:left))
]) # if in title, it grows funny with gr
n >= 1 && plot!(plt, [1,0,0,1, 0], [1,1,0,1,1];
color=colors[1], linetype=:polygon,
fill=colors[1], alpha=.2)
n == 1 && plot!(plt, [1,0,0,1, 0], [1,1,0,1,1];
color=colors[1], linewidth=2)
for k in 2:n
xs = range(0, stop=1, length=1+2^(k-1))
ys = map(f, xs)
k < n && plot!(plt, xs, ys, linetype=:polygon, fill=:black, alpha=.2)
ys = f.(xs)
k < n && plot!(plt, xs, ys;
linetype=:polygon, fill=:black, alpha=.2)
if k == n
plot!(plt, xs, ys, color=colors[k], linetype=:polygon, fill=:black, alpha=.2)
plot!(plt, xs, ys, color=:black, linewidth=2)
plot!(plt, xs, ys;
color=colors[k], linetype=:polygon, fill=:black, alpha=.2)
plot!(plt, xs, ys;
color=:black, linewidth=2)
end
end
plt
end
n = 7
anim = @animate for i=1:n
make_triangle_graph(i)
@@ -183,7 +192,7 @@ $$
S_n = f(c_1) \cdot (x_1 - x_0) + f(c_2) \cdot (x_2 - x_1) + \cdots + f(c_n) \cdot (x_n - x_{n-1}).
$$
Clearly for a given partition and choice of $c_i$, the above can be computed. Each term $f(c_i)\cdot(x_i-x_{i-1})$ can be visualized as the area of a rectangle with base spanning from $x_{i-1}$ to $x_i$ and height given by the function value at $c_i$. The following visualizes left Riemann sums for different values of $n$ in a way that makes Beekman's intuition plausible – that as the number of rectangles gets larger, the approximate sum will get closer to the actual area.
Clearly for a given partition and choice of $c_i$, the above can be computed. Each term $f(c_i)\cdot(x_i-x_{i-1}) = f(c_i)\Delta_i$ can be visualized as the area of a rectangle with base spanning from $x_{i-1}$ to $x_i$ and height given by the function value at $c_i$. The following visualizes left Riemann sums for different values of $n$ in a way that makes Beekman's intuition plausible – that as the number of rectangles gets larger, the approximate sum will get closer to the actual area.
```{julia}
@@ -354,7 +363,7 @@ When the integral exists, it is written $V = \int_a^b f(x) dx$.
:::{.callout-note}
## History note
The expression $V = \int_a^b f(x) dx$ is known as the *definite integral* of $f$ over $[a,b]$. Much earlier than Riemann, Cauchy had defined the definite integral in terms of a sum of rectangular products beginning with $S=(x_1 - x_0) f(x_0) + (x_2 - x_1) f(x_1) + \cdots + (x_n - x_{n-1}) f(x_{n-1})$ (the left Riemann sum). He showed the limit was well defined for any continuous function. Riemann's formulation relaxes the choice of partition and the choice of the $c_i$ so that integrability can be better understood.
The expression $V = \int_a^b f(x) dx$ is known as the *definite integral* of $f$ over $[a,b]$. Much earlier than Riemann, Cauchy had defined the definite integral in terms of a sum of rectangular products beginning with $S=f(x_0) \cdot (x_1 - x_0) + f(x_1) \cdot (x_2 - x_1) + \cdots + f(x_{n-1}) \cdot (x_n - x_{n-1}) $ (the left Riemann sum). He showed the limit was well defined for any continuous function. Riemann's formulation relaxes the choice of partition and the choice of the $c_i$ so that integrability can be better understood.
:::
@@ -364,18 +373,6 @@ The expression $V = \int_a^b f(x) dx$ is known as the *definite integral* of $f$
The following formulas are consequences when $f(x)$ is integrable. These mostly follow through a judicious rearranging of the approximating sums.
The area is $0$ when there is no width to the interval to integrate over:
> $$
> \int_a^a f(x) dx = 0.
> $$
Even our definition of a partition doesn't really apply, as we assume $a < b$, but clearly if $a=x_0=x_n=b$ then our only"approximating" sum could be $f(a)(b-a) = 0$.
The area under a constant function is found from the area of rectangle, a special case being $c=0$ yielding $0$ area:
@@ -388,16 +385,65 @@ The area under a constant function is found from the area of rectangle, a specia
For any partition of $a < b$, we have $S_n = c(x_1 - x_0) + c(x_2 -x_1) + \cdots + c(x_n - x_{n-1})$. By factoring out the $c$, we have a *telescoping sum* which means the sum simplifies to $S_n = c(x_n-x_0) = c(b-a)$. Hence any limit must be this constant value.
Scaling the $y$ axis by a constant can be done before or after computing the area:
::: {#fig-consequence-rectangle-area}
```{julia}
#| echo: false
gr()
let
c = 1
a,b = 0.5, 1.5
f(x) = c
Δ = 0.1
plt = plot(;
xaxis=([], false),
yaxis=([], false),
legend=false,
)
plot!(f, a, b; line=(:black, 2))
plot!([a-Δ, b + Δ], [0,0]; line=(:gray, 1), arrow=true, side=:head)
plot!([a-Δ/2, a-Δ/2], [-Δ, c + Δ]; line=(:gray, 1), arrow=true, side=:head)
plot!([a,a],[0,f(a)]; line=(:black, 1, :dash))
plot!([b,b],[0,f(b)]; line=(:black, 1, :dash))
annotate!([
(a, 0, text(L"a", :top, :right)),
(b, 0, text(L"b", :top, :left)),
(a-Δ/2-0.01, c, text(L"c", :right))
])
current()
end
```
```{julia}
#| echo: false
plotly()
nothing
```
Illustration that the area under a constant function is that of a rectangle
:::
The area is $0$ when there is no width to the interval to integrate over:
> $$
> \int_a^b cf(x) dx = c \int_a^b f(x) dx.
> \int_a^a f(x) dx = 0.
> $$
Let $a=x_0 < x_1 < \cdots < x_n=b$ be any partition. Then we have $S_n= cf(c_1)(x_1-x_0) + \cdots + cf(c_n)(x_n-x_{n-1})$ $=$ $c\cdot\left[ f(c_1)(x_1 - x_0) + \cdots + f(c_n)(x_n - x_{n-1})\right]$. The "limit" of the left side is $\int_a^b c f(x) dx$. The "limit" of the right side is $c \cdot \int_a^b f(x)$. We call this a "sketch" as a formal proof would show that for any $\epsilon$ we could choose a $\delta$ so that any partition with norm $\delta$ will yield a sum less than $\epsilon$. Here, then our "any" partition would be one for which the $\delta$ on the left hand side applies. The computation shows that the same $\delta$ would apply for the right hand side when $\epsilon$ is the same.
Even our definition of a partition doesn't really apply, as we assume $a < b$, but clearly if $a=x_0=x_n=b$ then our only"approximating" sum could be $f(a)(b-a) = 0$.
#### Shifts
A jigsaw puzzle piece will have the same area if it is moved around on the table or flipped over. Similarly some shifts preserve area under a function.
The area is invariant under shifts left or right.
@@ -424,6 +470,59 @@ $$
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$.
::: {#fig-consequence-rectangle-area}
```{julia}
#| echo: false
gr()
let
f(x) = 2 + cospi(x^2/10)*sinpi(x)
plt = plot(;
xaxis=([], false),
yaxis=([], false),
legend=false,
)
a, b = 0,4
c = 5
plot!(f, a, b; line=(:black, 2))
plot!(x -> f(x-c), a+c, b+c; line=(:red, 2))
plot!([-1, b+c + 1], [0,0]; line=(:gray, 2), arrow=true, side=:head)
for x ∈ (a,b)
plot!([x,x],[0,f(x)]; line=(:black,1, :dash))
end
for x ∈ (a+c,b+c)
plot!([x,x],[0,f(x)]; line=(:red,1, :dash))
end
annotate!([
(a+c,0, text(L"a", :top)),
(b+c,0, text(L"b", :top)),
(a,0, text(L"a-c", :top)),
(b,0, text(L"b-c", :top)),
(1.0, 3, text(L"f(x)",:left)),
(1.0+c, 3, text(L"f(x-c)",:left)),
])
current()
end
```
```{julia}
#| echo: false
plotly()
nothing
```
Illustration that the area under shift remains the same
:::
Similarly, reflections don't effect the area under the curve, they just require a new parameterization:
@@ -433,7 +532,79 @@ Similarly, reflections don't effect the area under the curve, they just require
The scaling operation $g(x) = f(cx)$ has the following:
::: {#fig-consequence-reflect-area}
```{julia}
#| echo: false
gr()
let
f(x) = 2 + cospi(x^2/10)*sinpi(x)
g(x) = f(-x)
plt = plot(;
xaxis=([], false),
yaxis=([], false),
legend=false,
)
a, b = 1,4
plot!(f, a, b; line=(:black, 2))
plot!(g, -b, -a; line=(:red, 2))
plot!([-5, 5], [0,0]; line=(:gray,1), arrow=true, side=:head)
plot!([0,0], [-0.1,3.15]; line=(:gray,1), arrow=true, side=:head)
for x in (a,b)
plot!([x,x], [0,f(x)]; line=(:black,1,:dash))
plot!([-x,-x], [0,g(-x)]; line=(:red,1,:dash))
end
annotate!([
(a,0, text(L"a", :top)),
(b,0, text(L"b", :top)),
(-a,0, text(L"-a", :top)),
(-b,0, text(L"-b", :top)),
])
current()
end
```
```{julia}
#| echo: false
plotly()
nothing
```
Illustration that the area remains constant under reflection through $y$ axis.
:::
The "reversed" area is the same, only accounted for with a minus sign.
> $$
> \int_a^b f(x) dx = -\int_b^a f(x) dx.
> $$
#### Scaling
Scaling the $y$ axis by a constant can be done before or after computing the area:
> $$
> \int_a^b cf(x) dx = c \int_a^b f(x) dx.
> $$
Let $a=x_0 < x_1 < \cdots < x_n=b$ be any partition. Then we have $S_n= cf(c_1)(x_1-x_0) + \cdots + cf(c_n)(x_n-x_{n-1})$ $=$ $c\cdot\left[ f(c_1)(x_1 - x_0) + \cdots + f(c_n)(x_n - x_{n-1})\right]$. The "limit" of the left side is $\int_a^b c f(x) dx$. The "limit" of the right side is $c \cdot \int_a^b f(x)$. We call this a "sketch" as a formal proof would show that for any $\epsilon$ we could choose a $\delta$ so that any partition with norm $\delta$ will yield a sum less than $\epsilon$. Here, then our "any" partition would be one for which the $\delta$ on the left hand side applies. The computation shows that the same $\delta$ would apply for the right hand side when $\epsilon$ is the same.
The scaling operation on the $x$ axis, $g(x) = f(cx)$, has the following property:
> $$
@@ -448,8 +619,9 @@ The scaling operation shifts $a$ to $ca$ and $b$ to $cb$ so the limits of integr
Combining two operations above, the operation $g(x) = \frac{1}{h}f(\frac{x-c}{h})$ will leave the area between $a$ and $b$ under $g$ the same as the area under $f$ between $(a-c)/h$ and $(b-c)/h$.
---
#### Area is additive
When two jigsaw pieces interlock their combined area is that of each added. This also applies to areas under functions.
The area between $a$ and $b$ can be broken up into the sum of the area between $a$ and $c$ and that between $c$ and $b$.
@@ -463,35 +635,185 @@ The area between $a$ and $b$ can be broken up into the sum of the area between $
For this, suppose we have a partition for both the integrals on the right hand side for a given $\epsilon/2$ and $\delta$. Combining these into a partition of $[a,b]$ will mean $\delta$ is still the norm. The approximating sum will combine to be no more than $\epsilon/2 + \epsilon/2$, so for a given $\epsilon$, this $\delta$ applies.
This is due to the area on the left and right of $0$ being equivalent.
::: {#fig-consequence-additive-area}
```{julia}
#| echo: false
gr()
let
f(x) = 2 + cospi(x^2/7)*sinpi(x)
a,b,c = 0.1, 8, 3
xs = range(a,c,100)
A1 = Shape(vcat(xs,c,a), vcat(f.(xs), 0, 0))
xs = range(c,b,100)
A2 = Shape(vcat(xs,b,c), vcat(f.(xs), 0, 0))
The "reversed" area is the same, only accounted for with a minus sign.
plt = plot(;
xaxis=([], false),
yaxis=([], false),
legend=false,
)
plot!([0,0] .- 0.1,[-.1,3]; line=(:gray, 1), arrow=true, side=:head)
plot!([0-.2, b+0.5],[0,0]; line=(:gray, 1), arrow=true, side=:head)
plot!(A1; fill=(:gray60,1), line=(nothing,))
plot!(A2; fill=(:gray90,1), line=(nothing,))
plot!(f, a, b; line=(:black, 2))
for x in (a,b,c)
plot!([x,x], [0, f(x)]; line=(:black, 1, :dash))
end
> $$
> \int_a^b f(x) dx = -\int_b^a f(x) dx.
> $$
annotate!([(x,0,text(latexstring("$y"),:top)) for (x,y) in zip((a,b,c),("a","b","c"))])
end
```
```{julia}
#| echo: false
plotly()
nothing
```
Illustration that the area between $a$ and $b$ can be computed as area between $a$ and $c$ and then $c$ and $b$.
:::
A consequence of the last few statements is:
> If $f(x)$ is an even function, then $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$. If $f(x)$ is an odd function, then $\int_{-a}^a f(x) dx = 0$.
> If $f(x)$ is an even function, then $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$.
> If $f(x)$ is an odd function, then $\int_{-a}^a f(x) dx = 0$.
Additivity works in the $y$ direction as well.
If $f(x)$ and $g(x)$ are two functions then
> $$
> \int_a^b (f(x) + g(x)) dx = \int_a^b f(x) dx + \int_a^b g(x) dx
> $$
For any partitioning with $x_i, x_{i-1}$ and $c_i$ this holds:
$$
(f(c_i) + g(c_i)) \cdot (x_i - x_{i-1}) =
f(c_i) \cdot (x_i - x_{i-1}) + g(c_i) \cdot (x_i - x_{i-1})
$$
This leads to the same statement for the areas under the curves.
The *linearity* of the integration operation refers to this combination of the above:
> $$
> \int_a^b (cf(x) + dg(x)) dx = c\int_a^b f(x) dx + d \int_a^b g(x)dx
> $$
The integral of a shifted function satisfies:
> $$
> \int_a^b \left(D + C\cdot f(\frac{x - B}{A})\right) dx = D\cdot(b-a) + C \cdot A \int_{\frac{a-B}{A}}^{\frac{b-B}{A}} f(x) dx
> $$
This follows from a few of the statements above:
$$
\begin{align*}
\int_a^b \left(D + C\cdot f(\frac{x - B}{A})\right) dx &=
\int_a^b D dx + C \int_a^b f(\frac{x-B}{A}) dx \\
&= D\cdot(b-a) + C\cdot A \int_{\frac{a-B}{A}}^{\frac{b-B}{A}} f(x) dx
\end{align*}
$$
#### Inequalities
Area under a non-negative function is non-negative
> $$
> \int_a^b f(x) dx \geq 0,\quad\text{when } a < b, \text{ and } f(x) \geq 0
> $$
Under this assumption, for any partitioning with $x_i, x_{i-1}$ and $c_i$ it holds the $f(c_i)\cdot(x_i - x_{i-1}) \geq 0$. So any sum of non-negative values can only be non-negative, even in the limit.
If $g$ bounds $f$ then the area under $g$ will bound the area under $f$, in particular if $f(x)$ is non negative, so will the area under $f$ be non negative for any $a < b$. (This assumes that $g$ and $f$ are integrable.)
> If $0 \leq f(x) \leq g(x)$ then $\int_a^b f(x) dx \leq \int_a^b g(x) dx.$
If $g$ bounds $f$ then the area under $g$ will bound the area under $f$.
> $$
> $\int_a^b f(x) dx \leq \int_a^b g(x) dx \quad\text{when } a < b\text{ and } 0 \leq f(x) \leq g(x)
> $$
For any partition of $[a,b]$ and choice of $c_i$, we have the term-by-term bound $f(c_i)(x_i-x_{i-1}) \leq g(c_i)(x_i-x_{i-1})$ So any sequence of partitions that converges to the limits will have this inequality maintained for the sum.
::: {#fig-consequence-0-area}
```{julia}
#| echo: false
gr()
let
f(x) = 1/6+x^3*(2-x)/2
g(x) = 1/6+exp(x/3)+(1-x/1.7)^6-0.6
a, b = 0, 2
plot(; empty_style...)
plot!([a-.5, b+.25], [0,0]; line=(:gray, 1), arrow=true, side=:head)
plot!([0,0] .- 0.25, [-0.25, 1.8]; line=(:gray, 1), arrow=true, side=:head)
xs = range(a,b,250)
S = Shape(vcat(xs, reverse(xs)), vcat(f.(xs), g.(reverse(xs))))
plot!(S; fill=(:gray70, 0.3), line=(nothing,))
S = Shape(vcat(xs, reverse(xs)), vcat(zero.(xs), f.(reverse(xs))))
plot!(S; fill=(:gray90, 0.3), line=(nothing,))
plot!(f, a, b; line=(:black, 4))
plot!(g, a, b; line=(:black, 2))
for x in (a,b)
plot!([x,x], [0, g(x)]; line=(:black,1,:dash))
end
annotate!([(x,0,text(t, :top)) for (x,t) in zip((a,b),(L"a", L"b"))])
current()
end
```
```{julia}
#| echo: false
plotly()
nothing
```
Illustration that if $f(x) \le g(x)$ on $[a,b]$ then the integrals share the same property. The excess area is clearly positive.
:::
(This also follows by considering $h(x) = g(x) - f(x) \geq 0$ by assumption, so $\int_a^b h(x) dx \geq 0$.)
For non-negative functions, integrals over larger domains are bigger
> $$
> \int_a^c f(x) dx \le \int_a^b f(x) dx,\quad\text{when } c < b \text{ and } f(x) \ge 0
> $$
This follows as $\int_c^b f(x) dx$ is non-negative under these assumptions.
### Some known integrals
@@ -606,7 +928,7 @@ The main idea behind this is that the difference between the maximum and minimum
For example, the function $f(x) = 1$ for $x$ in $[0,1]$ and $0$ otherwise will be integrable, as it is continuous at all but two points, $0$ and $1$, where it jumps.
* Some functions can have infinitely many points of discontinuity and still be integrable. The example of $f(x) = 1/q$ when $x=p/q$ is rational, and $0$ otherwise is often used as an example.
* Some functions can have infinitely many points of discontinuity and still be integrable. The example of $f(x) = 1/q$ when $x=p/q$ is rational, and $0$ otherwise is often used to illustrate this.
## Numeric integration
@@ -636,11 +958,11 @@ deltas = diff(xs) # forms x2-x1, x3-x2, ..., xn-xn-1
cs = xs[1:end-1] # finds left-hand end points. xs[2:end] would be right-hand ones.
```
Now to multiply the values. We want to sum the product `f(cs[i]) * deltas[i]`, here is one way to do so:
We want to sum the products $f(c_i)\Delta_i$. Here is one way to do so using `zip` to iterate over the paired off values in `cs` and `deltas`.
```{julia}
sum(f(cs[i]) * deltas[i] for i in 1:length(deltas))
sum(f(ci)*Δi for (ci, Δi) in zip(cs, deltas))
```
Our answer is not so close to the value of $1/3$, but what did we expect - we only used $n=5$ intervals. Trying again with $50,000$ gives us:
@@ -652,7 +974,7 @@ n = 50_000
xs = a:(b-a)/n:b
deltas = diff(xs)
cs = xs[1:end-1]
sum(f(cs[i]) * deltas[i] for i in 1:length(deltas))
sum(f(ci)*Δi for (ci, Δi) in zip(cs, deltas))
```
This value is about $10^{-5}$ off from the actual answer of $1/3$.
@@ -745,7 +1067,7 @@ plot!(zero)
We could add the signed area over $[0,1]$ to the above, but instead see a square of area $1$, a triangle with area $1/2$ and a triangle with signed area $-1$. The total is then $1/2$.
This figure may make the above decomposition more clear:
This figure--using equal sized axes--may make the above decomposition more clear:
```{julia}
#| echo: false
@@ -758,6 +1080,7 @@ let
plot!(Shape([-1,0,1], [0,-1,0]); fill=(:gray90,))
plot!(Shape([1,2,2,1], [0,1,0,0]); fill=(:gray10,))
plot!(Shape([2,3,3,2], [0,0,1,1]); fill=(:gray,))
plot!([0,0], [0, g(0)]; line=(:black,1,:dash))
end
```
@@ -844,7 +1167,9 @@ We have the well-known triangle [inequality](http://en.wikipedia.org/wiki/Triang
This suggests that the following inequality holds for integrals:
> $\lvert \int_a^b f(x) dx \rvert \leq \int_a^b \lvert f(x) \rvert dx$.
> $$
> \lvert \int_a^b f(x) dx \rvert \leq \int_a^b \lvert f(x) \rvert dx$.
> $$
@@ -922,7 +1247,8 @@ The formulas for an approximation to the integral $\int_{-1}^1 f(x) dx$ discusse
$$
\begin{align*}
S &= f(x_1) \Delta_1 + f(x_2) \Delta_2 + \cdots + f(x_n) \Delta_n\\
&= w_1 f(x_1) + w_2 f(x_2) + \cdots + w_n f(x_n).
&= w_1 f(x_1) + w_2 f(x_2) + \cdots + w_n f(x_n)\\
&= \sum_{i=1}^n w_i f(x_i).
\end{align*}
$$
@@ -957,7 +1283,7 @@ f(x) = x^5 - x + 1
quadgk(f, -2, 2)
```
The error term is $0$, answer is $4$ up to the last unit of precision (1 ulp), so any error is only in floating point approximations.
The error term is $0$, the answer is $4$ up to the last unit of precision (1 ulp), so any error is only in floating point approximations.
For the numeric computation of definite integrals, the `quadgk` function should be used over the Riemann sums or even Simpson's rule.
@@ -1476,6 +1802,70 @@ val, _ = quadgk(f, a, b)
numericq(val)
```
###### Question
Let $A=1.98$ and $B=1.135$ and
$$
f(x) = \frac{1 - e^{-Ax}}{B\sqrt{\pi}x} e^{-x^2}.
$$
Find $\int_0^1 f(x) dx$
```{julia}
#| echo: false
let
A,B = 1.98, 1.135
f(x) = (1 - exp(-A*x))*exp(-x^2)/(B*sqrt(pi)*x)
val,_ = quadgk(f, 0, 1)
numericq(val)
end
```
###### Question
A bound for the complementary error function ( positive function) is
$$
\text{erfc}(x) \leq \frac{1}{2}e^{-2x^2} + \frac{1}{2}e^{-x^2} \leq e^{-x^2}
\quad x \geq 0.
$$
Let $f(x)$ be the first bound, $g(x)$ the second.
Assuming this is true, confirm numerically using `quadgk` that
$$
\int_0^3 f(x) dx \leq \int_0^3 g(x) dx
$$
The value of $\int_0^3 f(x) dx$ is
```{julia}
#| echo: false
let
f(x) = 1/2 * exp(-2x^2) + 1/2 * exp(-x^2)
val,_ = quadgk(f, 0, 3)
numericq(val)
end
```
The value of $\int_0^3 g(x) dx$ is
```{julia}
#| echo: false
let
g(x) = exp(-x^2)
val,_ = quadgk(g, 0, 3)
numericq(val)
end
```
###### Question