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quarto/integrals/integration_by_parts.qmd

@@ -94,7 +94,7 @@ An illustration can clarify.
 Consider the integral $\int_0^\pi x\sin(x) dx$. If we let $u=x$ and $dv=\sin(x) dx$, then $du = 1dx$ and $v=-\cos(x)$. The above then says:
 
 
-$$
+
 \begin{align*}
 \int_0^\pi x\sin(x) dx &= \int_0^\pi u dv\\
 &= uv\big|_0^\pi - \int_0^\pi v du\\
@@ -103,7 +103,7 @@ $$
 &= \pi + \sin(x)\big|_0^\pi\\
 &= \pi.
 \end{align*}
-$$
+
 
 The technique means one part is differentiated and one part integrated. The art is to break the integrand up into a piece that gets easier through differentiation and a piece that doesn't get much harder through integration.
 
@@ -128,7 +128,7 @@ $$
 Putting together gives:
 
 
-$$
+
 \begin{align*}
 \int_1^2 x \log(x) dx
 &= (\log(x) \cdot \frac{x^2}{2}) \big|_1^2 - \int_1^2 \frac{x^2}{2} \frac{1}{x} dx\\
@@ -136,7 +136,7 @@ $$
 &= 2\log(2) - (1 - \frac{1}{4}) \\
 &= 2\log(2) - \frac{3}{4}.
 \end{align*}
-$$
+
 
 ##### Example
 
@@ -144,14 +144,14 @@ $$
 This related problem, $\int \log(x) dx$, uses the same idea, though perhaps harder to see at first glance, as setting `dv=dx` is almost too simple to try:
 
 
-$$
+
 \begin{align*}
 u &= \log(x) & dv &= dx\\
 du &= \frac{1}{x}dx & v &= x
 \end{align*}
-$$
 
-$$
+
+
 \begin{align*}
 \int \log(x) dx
 &= \int u dv\\
@@ -160,7 +160,7 @@ $$
 &= x \log(x) - \int dx\\
 &= x \log(x) - x
 \end{align*}
-$$
+
 
 Were this a definite integral problem, we would have written:
 
@@ -243,14 +243,14 @@ $$
 Positive integer powers of trigonometric functions can be addressed by this technique. Consider $\int \cos(x)^n dx$. We let $u=\cos(x)^{n-1}$ and $dv=\cos(x) dx$. Then $du = (n-1)\cos(x)^{n-2}(-\sin(x))dx$ and $v=\sin(x)$. So,
 
 
-$$
+
 \begin{align*}
 \int \cos(x)^n dx &= \cos(x)^{n-1} \cdot (\sin(x)) - \int (\sin(x)) ((n-1)\sin(x) \cos(x)^{n-2}) dx \\
 &= \sin(x) \cos(x)^{n-1} + (n-1)\int \sin^2(x) \cos(x)^{n-1} dx\\
 &= \sin(x) \cos(x)^{n-1} + (n-1)\int (1 - \cos(x)^2) \cos(x)^{n-2} dx\\
 &= \sin(x) \cos(x)^{n-1} + (n-1)\int \cos(x)^{n-2}dx - (n-1)\int \cos(x)^n dx.
 \end{align*}
-$$
+
 
 We can then solve for the unknown ($\int \cos(x)^{n}dx$) to get this *reduction formula*:
 
@@ -278,13 +278,13 @@ The visual interpretation of integration by parts breaks area into two pieces, t
 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:
 
 
-$$
+
 \begin{align*}
 \int f^{-1}(x) dx
 &= xf^{-1}(x) - \int x [f^{-1}(x)]' dx\\
 &= xf^{-1}(x) - \int f(u) du.\\
 \end{align*}
-$$
+
 
 The last line follows from the $u$-substitution: $u=f^{-1}(x)$ for then $du = [f^{-1}(x)]' dx$ and $x=f(u)$.
 
@@ -292,13 +292,13 @@ The last line follows from the $u$-substitution: $u=f^{-1}(x)$ for then $du = [f
 We use this to find an antiderivative for $\sin^{-1}(x)$:
 
 
-$$
+
 \begin{align*}
 \int \sin^{-1}(x) dx &= x \sin^{-1}(x) - \int \sin(u) du \\
 &=  x \sin^{-1}(x) + \cos(u) \\
 &= x \sin^{-1}(x) + \cos(\sin^{-1}(x)).
 \end{align*}
-$$
+
 
 Using right triangles to simplify, the last value $\cos(\sin^{-1}(x))$ can otherwise be written as $\sqrt{1 - x^2}$.
 
@@ -321,12 +321,12 @@ This [proof](http://www.math.ucsd.edu/~ebender/20B/77_Trap.pdf) for the error es
 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 $v=t+A$:
 
 
-$$
+
 \begin{align*}
 \int_0^h f(t + x_i) dt &= uv \big|_0^h - \int_0^h v du\\
 &= f(t+x_i)(t+A)\big|_0^h - \int_0^h (t + A) f'(t + x_i) dt.
 \end{align*}
-$$
+
 
 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:
 
@@ -338,12 +338,12 @@ $$
 For this, we *again* integrate by parts with
 
 
-$$
+
 \begin{align*}
 u  &= f'(t + x_i)  & dv &= (t + A)dt\\
 du &= f''(t + x_i) & v  &= \frac{(t + A)^2}{2} + B
 \end{align*}
-$$
+
 
 Again we added a constant of integration, $B$, to  $v$. The  error becomes:
 
@@ -417,14 +417,14 @@ We added a rectangle for a Riemann sum for $t_i = \pi/3$ and $t_{i+1} = \pi/3 +
 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:
 
 
-$$
+
 \begin{align*}
 A &\approx \sum_i y(t_i) \cdot (x(t_{i}) - x(t_{i+1}))\\
   &= - \sum_i y(t_i) \cdot (x(t_{i+1}) - x(t_{i}))\\
   &= - \sum_i y(t_i) \cdot \frac{x(t_{i+1}) - x(t_i)}{t_{i+1}-t_i} \cdot (t_{i+1}-t_i)\\
   &\approx -\int_a^b y(t) x'(t) dt.
 \end{align*}
-$$
+
 
 So with a counterclockwise rotation, the actual answer for the area includes a minus sign. If the area is traced out in a *clockwise* manner, there is no minus sign.