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jverzani
2025-07-23 08:05:43 -04:00
parent 31ce21c8ad
commit c3a94878f3
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97
quarto/integrals/integration_by_parts.qmd
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@@ -39,13 +39,13 @@ Now we turn our attention to the implications of the *product rule*: $[uv]' = u'
By the fundamental theorem of calculus:
$$
[u(x)\cdot v(x)]\big|_a^b = \int_a^b [u(x) v(x)]' dx = \int_a^b u'(x) \cdot v(x) dx + \int_a^b u(x) \cdot v'(x) dx.
[u(x)\cdot v(x)]\Big|_a^b = \int_a^b [u(x) v(x)]' dx = \int_a^b u'(x) \cdot v(x) dx + \int_a^b u(x) \cdot v'(x) dx.
$$
Or,
$$
\int_a^b u(x) v'(x) dx = [u(x)v(x)]\big|_a^b - \int_a^b v(x) u'(x)dx.
\int_a^b u(x) v'(x) dx = [u(x)v(x)]\Big|_a^b - \int_a^b v(x) u'(x)dx.
$$
:::
@@ -58,16 +58,16 @@ The following visually illustrates integration by parts:
#| label: fig-integration-by-parts
#| fig-cap: "Integration by parts figure ([original](http://en.wikipedia.org/wiki/Integration_by_parts#Visualization))"
let
## parts picture
## parts picture
gr()
u(x) = sin(x*pi/2)
v(x) = x
xs = range(0, stop=1, length=50)
a,b = 1/4, 3/4
p = plot(u, v, 0, 1, legend=false, axis=([], false))
plot!([0, u(1)], [0,0], line=(:black, 3))
plot!([0, 0], [0, v(1) ], line=(:black, 3))
plot!(p, zero, 0, 1)
p = plot(u, v, 0, 1; legend=false, axis=([], false), line=(:black,2))
plot!([0, u(1)], [0,0]; line=(:gray, 1), arrow=true, side=:head)
plot!([0, 0], [0, v(1) ]; line=(:gray, 1), arrow=true, side=:head)
xs′ = range(a, b, length=50)
plot!(Shape(vcat(u.(xs′), reverse(u.(xs′))),
@@ -81,21 +81,28 @@ plot!(p, [u(a),u(a),0, 0, u(b),u(b),u(a)],
[0, v(a), v(a), v(b), v(b), 0, 0],
linetype=:polygon, fill=(:brown3, 0.25))
annotate!(p, [(0.65, .25, "A"),
(0.4, .55, "B"),
(u(a),v(a) + .08, "(u(a),v(a))"),
(u(b),v(b)+.08, "(u(b),v(b))"),
(u(a),0, "u(a)",:top),
(u(b),0, "u(b)",:top),
(0, v(a), "v(a) ",:right),
(0, v(b), "v(b) ",:right)
annotate!(p, [(0.65, .25, text(L"A")),
(0.4, .55, text(L"B")),
(u(a),v(a), text(L"(u(a),v(a))", :bottom, :right)),
(u(b),v(b), text(L"(u(b),v(b))", :bottom, :right)),
(u(a),0, text(L"u(a)", :top)),
(u(b),0, text(L"u(b)", :top)),
(0, v(a), text(L"v(a)", :right)),
(0, v(b), text(L"v(b)", :right)),
(0,0, text(L"(0,0)", :top))
])
end
```
```{julia}
#| echo: false
plotly()
nothing
```
@fig-integration-by-parts shows a parametric plot of $(u(t),v(t))$ for $a \leq t \leq b$..
The total shaded area, a rectangle, is $u(b)v(b)$, the area of $A$ and $B$ combined is just $u(b)v(b) - u(a)v(a)$ or $[u(x)v(x)]\big|_a^b$. We will show that $A$ is $\int_a^b v(x)u'(x)dx$ and $B$ is $\int_a^b u(x)v'(x)dx$ giving the formula.
The total shaded area, a rectangle, is $u(b)v(b)$, the area of $A$ and $B$ combined is just $u(b)v(b) - u(a)v(a)$ or $[u(x)v(x)]\Big|_a^b$. We will show that $A$ is $\int_a^b v(x)u'(x)dx$ and $B$ is $\int_a^b u(x)v'(x)dx$ giving the formula.
We can compute $A$ by a change of variables with $x=u^{-1}(t)$ (so $u'(x)dx = dt$):
@@ -109,6 +116,7 @@ $$
$B$ is similar with the roles of $u$ and $v$ reversed.
----
Informally, the integration by parts formula is sometimes seen as $\int udv = uv - \int v du$, as well can be somewhat confusingly written as:
@@ -131,10 +139,10 @@ Consider the integral $\int_0^\pi x\sin(x) dx$. If we let $u=x$ and $dv=\sin(x)
$$
\begin{align*}
\int_0^\pi x\sin(x) dx &= \int_0^\pi u dv\\
&= uv\big|_0^\pi - \int_0^\pi v du\\
&= x \cdot (-\cos(x)) \big|_0^\pi - \int_0^\pi (-\cos(x)) dx\\
&= uv\Big|_0^\pi - \int_0^\pi v du\\
&= x \cdot (-\cos(x)) \Big|_0^\pi - \int_0^\pi (-\cos(x)) dx\\
&= \pi (-\cos(\pi)) - 0(-\cos(0)) + \int_0^\pi \cos(x) dx\\
&= \pi + \sin(x)\big|_0^\pi\\
&= \pi + \sin(x)\Big|_0^\pi\\
&= \pi.
\end{align*}
$$
@@ -166,8 +174,8 @@ Putting together gives:
$$
\begin{align*}
\int_1^2 x \log(x) dx
&= (\log(x) \cdot \frac{x^2}{2}) \big|_1^2 - \int_1^2 \frac{x^2}{2} \frac{1}{x} dx\\
&= (2\log(2) - 0) - (\frac{x^2}{4})\big|_1^2\\
&= (\log(x) \cdot \frac{x^2}{2}) \Big|_1^2 - \int_1^2 \frac{x^2}{2} \frac{1}{x} dx\\
&= (2\log(2) - 0) - (\frac{x^2}{4})\Big|_1^2\\
&= 2\log(2) - (1 - \frac{1}{4}) \\
&= 2\log(2) - \frac{3}{4}.
\end{align*}
@@ -204,7 +212,7 @@ Were this a definite integral problem, we would have written:
$$
\int_a^b \log(x) dx = (x\log(x))\big|_a^b - \int_a^b dx = (x\log(x) - x)\big|_a^b.
\int_a^b \log(x) dx = (x\log(x))\Big|_a^b - \int_a^b dx = (x\log(x) - x)\Big|_a^b.
$$
##### Example
@@ -214,14 +222,14 @@ Sometimes integration by parts is used two or more times. Here we let $u=x^2$ an
$$
\int_a^b x^2 e^x dx = (x^2 \cdot e^x)\big|_a^b - \int_a^b 2x e^x dx.
\int_a^b x^2 e^x dx = (x^2 \cdot e^x)\Big|_a^b - \int_a^b 2x e^x dx.
$$
But we can do $\int_a^b x e^xdx$ the same way:
$$
\int_a^b x e^x = (x\cdot e^x)\big|_a^b - \int_a^b 1 \cdot e^xdx = (xe^x - e^x)\big|_a^b.
\int_a^b x e^x = (x\cdot e^x)\Big|_a^b - \int_a^b 1 \cdot e^xdx = (xe^x - e^x)\Big|_a^b.
$$
Combining gives the answer:
@@ -229,8 +237,8 @@ Combining gives the answer:
$$
\int_a^b x^2 e^x dx
= (x^2 \cdot e^x)\big|_a^b - 2( (xe^x - e^x)\big|_a^b ) =
e^x(x^2 - 2x + 2) \big|_a^b.
= (x^2 \cdot e^x)\Big|_a^b - 2( (xe^x - e^x)\Big|_a^b ) =
e^x(x^2 - 2x + 2) \Big|_a^b.
$$
In fact, it isn't hard to see that an integral of $x^m e^x$, $m$ a positive integer, can be handled in this manner. For example, when $m=10$, `SymPy` gives:
@@ -247,14 +255,29 @@ The general answer is $\int x^n e^xdx = p(x) e^x$, where $p(x)$ is a polynomial
##### Example
The same technique is attempted for this integral, but ends differently. First in the following we let $u=\sin(x)$ and $dv=e^x dx$:
The same technique is attempted for the integral of $e^x\sin(x)$, but ends differently.
First we let $u=\sin(x)$ and $dv=e^x dx$, then
$$
du = \cos(x)dx \quad \text{and}\quad v = e^x.
$$
So:
$$
\int e^x \sin(x)dx = \sin(x) e^x - \int \cos(x) e^x dx.
$$
Now we let $u = \cos(x)$ and again $dv=e^x dx$:
Now we let $u = \cos(x)$ and again $dv=e^x dx$, then
$$
du = -\sin(x)dx \quad \text{and}\quad v = e^x.
$$
So:
$$
@@ -301,7 +324,7 @@ $$
This is called a reduction formula as it reduces the problem from an integral with a power of $n$ to one with a power of $n - 2$, so could be repeated until the remaining indefinite integral required knowing either $\int \cos(x) dx$ (which is $-\sin(x)$) or $\int \cos(x)^2 dx$, which by a double angle formula application, is $x/2 + \sin(2x)/4$.
`SymPy` is quite able to do this repeated bookkeeping. For example with $n=10$:
`SymPy` is able and willing to do this repeated bookkeeping. For example with $n=10$:
```{julia}
@@ -350,7 +373,7 @@ Using right triangles to simplify, the last value $\cos(\sin^{-1}(x))$ can other
The [trapezoid](http://en.wikipedia.org/wiki/Trapezoidal_rule) rule is an approximation to the definite integral like a Riemann sum, only instead of approximating the area above $[x_i, x_i + h]$ by a rectangle with height $f(c_i)$ (for some $c_i$), it uses a trapezoid formed by the left and right endpoints. That is, this area is used in the estimation: $(1/2)\cdot (f(x_i) + f(x_i+h)) \cdot h$.
Even though we suggest just using `quadgk` for numeric integration, estimating the error in this approximation is still of some theoretical interest.
Even though we suggest just using `quadgk` for numeric integration, estimating the error in this approximation is of theoretical interest.
Recall, just using *either* $x_i$ or $x_{i-1}$ for $c_i$ gives an error that is "like" $1/n$, as $n$ gets large, though the exact rate depends on the function and the length of the interval.
@@ -359,18 +382,18 @@ Recall, just using *either* $x_i$ or $x_{i-1}$ for $c_i$ gives an error that is
This [proof](http://www.math.ucsd.edu/~ebender/20B/77_Trap.pdf) for the error estimate is involved, but is reproduced here, as it nicely integrates many of the theoretical concepts of integration discussed so far.
First, for convenience, we consider the interval $x_i$ to $x_i+h$. The actual answer over this is just $\int_{x_i}^{x_i+h}f(x) dx$. By a $u$-substitution with $u=x-x_i$ this becomes $\int_0^h f(t + x_i) dt$. For analyzing this we integrate once by parts using $u=f(t+x_i)$ and $dv=dt$. But instead of letting $v=t$, we choose to add - as is our prerogative - a constant of integration $A$, so $v=t+A$:
First, for convenience, we consider the interval $x_i$ to $x_i+h$. The actual answer over this is just $\int_{x_i}^{x_i+h}f(x) dx$. By a $u$-substitution with $u=x-x_i$ this becomes $\int_0^h f(t + x_i) dt$. For analyzing this we integrate once by parts using $u=f(t+x_i)$ and $dv=dt$. But instead of letting $v=t$, we choose to add--as is our prerogative--a constant of integration $A$, so $v=t+A$:
$$
\begin{align*}
\int_0^h f(t + x_i) dt &= uv \big|_0^h - \int_0^h v du\\
&= f(t+x_i)(t+A)\big|_0^h - \int_0^h (t + A) f'(t + x_i) dt.
\int_0^h f(t + x_i) dt &= uv \Big|_0^h - \int_0^h v du\\
&= f(t+x_i)(t+A)\Big|_0^h - \int_0^h (t + A) f'(t + x_i) dt.
\end{align*}
$$
We choose $A$ to be $-h/2$, any constant is possible, for then the term $f(t+x_i)(t+A)\big|_0^h$ becomes $(1/2)(f(x_i+h) + f(x_i)) \cdot h$, or the trapezoid approximation. This means, the error over this interval - actual minus estimate - satisfies:
We choose $A$ to be $-h/2$, any constant is possible, for then the term $f(t+x_i)(t+A)\Big|_0^h$ becomes $(1/2)(f(x_i+h) + f(x_i)) \cdot h$, or the trapezoid approximation. This means, the error over this interval - actual minus estimate - satisfies:
$$
@@ -392,7 +415,7 @@ Again we added a constant of integration, $B$, to $v$. The error becomes:
$$
\text{error}_i = -(\frac{(t+A)^2}{2} + B)f'(t+x_i)\big|_0^h + \int_0^h (\frac{(t+A)^2}{2} + B) \cdot f''(t+x_i) dt.
\text{error}_i = -\left(\frac{(t+A)^2}{2} + B\right)f'(t+x_i)\Big|_0^h + \int_0^h \left(\frac{(t+A)^2}{2} + B\right) \cdot f''(t+x_i) dt.
$$
With $A=-h/2$, $B$ is chosen so $(t+A)^2/2 + B = 0$ at endpoints, or $B=-h^2/8$. The error becomes
@@ -406,14 +429,14 @@ Now, we assume the $\lvert f''(t)\rvert$ is bounded by $K$ for any $a \leq t \le
$$
\lvert \text{error}_i \rvert \leq K \int_0^h \lvert (\frac{(t-h/2)^2}{2} - \frac{h^2}{8}) \rvert dt.
\lvert \text{error}_i \rvert \leq K \int_0^h \lVert \left(\frac{(t-h/2)^2}{2} - \frac{h^2}{8}\right) \rVert dt.
$$
But what is the function in the integrand? Clearly it is a quadratic in $t$. Expanding gives $1/2 \cdot (t^2 - ht)$. This is negative over $[0,h]$ (and $0$ at these endpoints, so the integral above is just:
$$
\frac{1}{2}\int_0^h (ht - t^2)dt = \frac{1}{2} (\frac{ht^2}{2} - \frac{t^3}{3})\big|_0^h = \frac{h^3}{12}
\frac{1}{2}\int_0^h (ht - t^2)dt = \frac{1}{2} \left(\frac{ht^2}{2} - \frac{t^3}{3}\right)\Big|_0^h = \frac{h^3}{12}
$$
This gives the bound: $\vert \text{error}_i \rvert \leq K h^3/12$. The *total* error may be less, but is not more than the value found by adding up the error over each of the $n$ intervals. As our bound does not depend on the $i$, we have this sum satisfies: