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quarto/integrals/integration_by_parts.qmd
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@@ -27,56 +27,90 @@ So far we have seen that the *derivative* rules lead to *integration rules*. In
* The sum rule $[au(x) + bv(x)]' = au'(x) + bv'(x)$ gives rise to an integration rule: $\int (au(x) + bv(x))dx = a\int u(x)dx + b\int v(x))dx$. (That is, the linearity of the derivative means the integral has linearity.)
* The chain rule $[f(g(x))]' = f'(g(x)) g'(x)$ gives $\int_a^b f(g(x))g'(x)dx=\int_{g(a)}^{g(b)}f(x)dx$. That is, substitution reverses the chain rule.
Now we turn our attention to the implications of the *product rule*: $[uv]' = u'v + uv'$. The resulting technique is called integration by parts.
::: {.callout-note}
## Integration by parts
The following illustrates integration by parts of the integral $(uv)'$ over $[a,b]$ [original](http://en.wikipedia.org/wiki/Integration_by_parts#Visualization).
By the fundamental theorem of calculus:
$$
[u(x)\cdot v(x)]\big|_a^b = \int_a^b [u(x) v(x)]' dx = \int_a^b u'(x) \cdot v(x) dx + \int_a^b u(x) \cdot v'(x) dx.
$$
Or,
$$
\int_a^b u(x) v'(x) dx = [u(x)v(x)]\big|_a^b - \int_a^b v(x) u'(x)dx.
$$
:::
The following visually illustrates integration by parts:
```{julia}
#| echo: false
#| label: fig-integration-by-parts
#| fig-cap: "Integration by parts figure ([original](http://en.wikipedia.org/wiki/Integration_by_parts#Visualization))"
let
## parts picture
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)
plot!(p, zero, 0, 1)
scatter!(p, [u(a), u(b)], [v(a), v(b)], color=:orange, markersize=5)
## parts picture
u(x) = sin(x*pi/2)
v(x) = x
xs = range(0, stop=1, length=50)
a,b = 1/4, 3/4
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, fillcolor=:orange, alpha=0.25)
annotate!(p, [(0.65, .25, "A"), (0.4, .55, "B")])
annotate!(p, [(u(a),v(a) + .08, "(u(a),v(a))"), (u(b),v(b)+.08, "(u(b),v(b))")])
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)
xs′ = range(a, b, length=50)
plot!(Shape(vcat(u.(xs′), reverse(u.(xs′))),
vcat(zero.(xs′), v.(reverse(xs′)))),
fill=(:red, 0.15),
xlims=(-0.07, 1)
)
plot!(Shape([0,u(a),u(a),0],[0,0,v(a),v(a)]), fill=(:royalblue, 0.5))
scatter!(p, [u(a), u(b)], [v(a), v(b)], color=:mediumorchid3, markersize=5)
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)
])
end
```
The figure is a parametric plot of $(u,v)$ with the points $(u(a), v(a))$ and $(u(b), v(b))$ marked. The difference $u(b)v(b) - u(a)v(a) = u(x)v(x) \mid_a^b$ is shaded. This area breaks into two pieces, $A$ and $B$, partitioned by the curve. If $u$ is increasing and the curve is parameterized by $t \rightarrow u^{-1}(t)$, then $A=\int_{u^{-1}(a)}^{u^{-1}(b)} v(u^{-1}(t))dt$. A $u$-substitution with $t = u(x)$ changes this into the integral $\int_a^b v(x) u'(x) dx$. Similarly, for increasing $v$, it can be seen that $B=\int_a^b u(x) v'(x) dx$. This suggests a relationship between the integral of $u v'$, the integral of $u' v$ and the value $u(b)v(b) - u(a)v(a)$.
@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 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
In terms of formulas, by the fundamental theorem of calculus:
We can compute $A$ by a change of variables with $x=u^{-1}(t)$ (so $u'(x)dx = dt$):
$$
u(x)\cdot v(x)\big|_a^b = \int_a^b [u(x) v(x)]' dx = \int_a^b u'(x) \cdot v(x) dx + \int_a^b u(x) \cdot v'(x) dx.
\begin{align*}
A &= \int_{u(a)}^{u(b)} v(u^{-1}(t)) dt & \text{let } x = u^{-1}(t) \text{ or }u(x) = t \\
&= \int_{u^{-1}(u(a))}^{u^{-1}(u(b))} v(x) u'(x) dx \\
&= \int_a^b v(x) u'(x) dx.
\end{align*}
$$
This is re-expressed as
$B$ is similar with the roles of $u$ and $v$ reversed.
$$
\int_a^b u(x) \cdot v'(x) dx = u(x) \cdot v(x)\big|_a^b - \int_a^b v(x) \cdot u'(x) dx,
$$
Or, more informally, as $\int udv = uv - \int v du$.
This can sometimes be confusingly written as:
Informally, the integration by parts formula is sometimes seen as $\int udv = uv - \int v du$, as well can be somewhat confusingly written as:
$$
@@ -86,8 +120,7 @@ $$
(The confusion coming from the fact that the indefinite integrals are only defined up to a constant.)
How does this help? It allows us to differentiate parts of an integral in hopes it makes the result easier to integrate.
How does this formula help? It allows us to differentiate parts of an integral in hopes it makes the result easier to integrate.
An illustration can clarify.