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Update integration_by_parts.qmd

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some typos.
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Fang Liu 2023-05-19 16:49:16 +08:00
parent 55c795bb9b
commit 81f1bdda33
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quarto/integrals/integration_by_parts.qmd
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@ -192,7 +192,7 @@ Combining gives the answer:
$$ $$
\int_a^b x^2 e^x dx \int_a^b x^2 e^x dx
= (x^2 \cdot e^x)\big|_a^b - 2( (xe^x - e^x)\big|_a^b ) = = (x^2 \cdot e^x)\big|_a^b - 2( (xe^x - e^x)\big|_a^b ) =
e^x(x^2 - 2x - 1) \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: 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:
@ -220,7 +220,7 @@ Now we let $u = \cos(x)$ and again $dv=e^x dx$:
$$ $$
\int e^x \sin(x)dx = \sin(x) e^x - \int \cos(x) e^x dx = \sin(x)e^x - \cos(x)e^x - \int (-\sin(x))e^x dx. \int e^x \sin(x)dx = \sin(x) e^x - \int \cos(x) e^x dx = \sin(x)e^x - \cos(x)e^x + \int (-\sin(x))e^x dx.
$$ $$
But simplifying this gives: But simplifying this gives:
@ -245,8 +245,8 @@ Positive integer powers of trigonometric functions can be addressed by this tech
\begin{align*} \begin{align*}
\int \cos(x)^n dx &= \cos(x)^{n-1} \cdot (\sin(x)) - \int (\sin(x)) ((n-1)\sin(x) \cos(x)^{n-2}) dx \\ \int \cos(x)^n dx &= \cos(x)^{n-1} \cdot (\sin(x)) + \int (\sin(x)) ((n-1)\sin(x) \cos(x)^{n-2}) dx \\
&= \sin(x) \cos(x)^{n-1} + (n-1)\int \sin^2(x) \cos(x)^{n-1} dx\\ &= \sin(x) \cos(x)^{n-1} + (n-1)\int \sin^2(x) \cos(x)^{n-2} dx\\
&= \sin(x) \cos(x)^{n-1} + (n-1)\int (1 - \cos(x)^2) \cos(x)^{n-2} dx\\ &= \sin(x) \cos(x)^{n-1} + (n-1)\int (1 - \cos(x)^2) \cos(x)^{n-2} dx\\
&= \sin(x) \cos(x)^{n-1} + (n-1)\int \cos(x)^{n-2}dx - (n-1)\int \cos(x)^n dx. &= \sin(x) \cos(x)^{n-1} + (n-1)\int \cos(x)^{n-2}dx - (n-1)\int \cos(x)^n dx.
\end{align*} \end{align*}
@ -259,7 +259,7 @@ $$
\int \cos(x)^n dx = \frac{1}{n}\sin(x) \cos(x)^{n-1} + \frac{n-1}{n}\int \cos(x)^{n-2}dx. \int \cos(x)^n dx = \frac{1}{n}\sin(x) \cos(x)^{n-1} + \frac{n-1}{n}\int \cos(x)^{n-2}dx.
$$ $$
This is called a reduction formula as it reduces the problem from an integral with a power of $n$ to one with a power of $n - 2$, so could be repeated until the remaining indefinite integral required knowing either $\int \cos(x) dx$ (which is $-\sin(x)$) or $\int \cos(x)^2 dx$, which by a double angle formula application, is $x/2 - \sin(2x)/4$. This is called a reduction formula as it reduces the problem from an integral with a power of $n$ to one with a power of $n - 2$, so could be repeated until the remaining indefinite integral required knowing either $\int \cos(x) dx$ (which is $-\sin(x)$) or $\int \cos(x)^2 dx$, which by a double angle formula application, is $x/2 + \sin(2x)/4$.
`SymPy` is quite able to do this repeated bookkeeping. For example with $n=10$: `SymPy` is quite able to do this repeated bookkeeping. For example with $n=10$:
@ -275,7 +275,7 @@ integrate(cos(𝒙)^10, 𝒙)
The visual interpretation of integration by parts breaks area into two pieces, the one labeled "B" looks like it would be labeled "A" for an inverse function for $f$. Indeed, integration by parts gives a means to possibly find antiderivatives for inverse functions. The visual interpretation of integration by parts breaks area into two pieces, the one labeled "B" looks like it would be labeled "A" for an inverse function for $f$. Indeed, integration by parts gives a means to possibly find antiderivatives for inverse functions.
Let $uv = x f^{-1}(x)$. Then we have $[uv]' = u'v + uv' = f^{-1}(x) + x [f^{-1}(x)]'$. So, up to a constant $uv = \int [uv]'dx = \int f^{-1}(x) + \int x [f^{-1}(x)]'$. Re-expressing gives: Let $uv = x f^{-1}(x)$. Then we have $[uv]' = u'v + uv' = f^{-1}(x) + x [f^{-1}(x)]'$. So, up to a constant $uv = \int [uv]'dx = \int f^{-1}(x)dx + \int x [f^{-1}(x)]'dx$. Re-expressing gives:
@ -352,7 +352,7 @@ $$
\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 = -(\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.
$$ $$
With $A=-h/2$, $B$ is chosen so $(t+A)^2/2 + B = 0$, or $B=-h^2/8$. The error becomes With $A=-h/2$, $B$ is chosen so $(t+A)^2/2 + B = 0$ at endpoints, or $B=-h^2/8$. The error becomes
$$ $$
@ -411,7 +411,7 @@ let
end end
``` ```
We added a rectangle for a Riemann sum for $t_i = \pi/3$ and $t_{i+1} = \pi/3 + \pi/8$. The height of this rectangle if $y(t_i)$, the base is of length $x(t_i) - x(t_{i+1})$ *given* the orientation of how the circular curve is parameterized (counter clockwise here). We added a rectangle for a Riemann sum for $t_i = \pi/3$ and $t_{i+1} = \pi/3 + \pi/8$. The height of this rectangle is $y(t_i)$, the base is of length $x(t_i) - x(t_{i+1})$ *given* the orientation of how the circular curve is parameterized (counter clockwise here).
Taking this Riemann sum approach, we can approximate the area under the curve parameterized by $(u(t), v(t))$ over the time range $[t_i, t_{i+1}]$ as a rectangle with height $y(t_i)$ and base $x(t_{i}) - x(t_{i+1})$. Then we get, as expected: Taking this Riemann sum approach, we can approximate the area under the curve parameterized by $(u(t), v(t))$ over the time range $[t_i, t_{i+1}]$ as a rectangle with height $y(t_i)$ and base $x(t_{i}) - x(t_{i+1})$. Then we get, as expected: