add notes on integration

uni/mmme/1026_maths_for_engineering/calculus.md

@@ -173,7 +173,7 @@ There may be more than one $y$ value for each $x$ value.
 
 </details>
 
-# Differentation
+# Differentiation
 
 The derivative of the function $f(x)$ is denoted by:
 
@@ -333,7 +333,7 @@ $$f'(a) = 0 \text{ and } f''(a) = 0 \text { and } f'''(a) \ne 0$$
 
 ![](./images/vimscrot-2021-12-27T15:38:29,395666506+00:00.png)
 
-# Taylor series
+# Approximating with the Taylor series
 
 The expansion
 
@@ -445,3 +445,331 @@ $$f(x) = f(a) + \frac 1 2 f''(a)(x-a)^2 + \cdots$$
   have a maximum
 - If $f''(a) = 0$ then we must include a higer order terms to determine what happens
   have a minimum
+
+# Integration
+
+Integration is the reverse of [differentiation](#differentiation).
+
+Take velocity and displacement as an example:
+
+$$\int\! v \mathrm dt = s + c$$
+
+where $c$ is the constant of integration, which is required for
+[indefinite integrals](#indefinite-integrals).A
+
+## Definite Integrals
+
+The definite integral of a function $f(x)$ in the range $a \le x \le b$ is denoted be:
+
+$$\int^b_a \! f(x) \,\mathrm dx$$
+
+If $f(x) = F'(x)$ ($f(x)$ is the derivative of $F(x)$) then
+
+$$\int^b_a \! f(x) \,\mathrm dx = \left[F(x)\right]^b_a = F(b) - F(a)$$
+
+## Area and Integration
+
+Approximate the area under a smooth curve using a large number of narrow rectangles.
+
+![](./images/vimscrot-2021-12-28T15:18:59,911868873+00:00.png)
+
+Area under curve $\approx \sum_{n} f(x_n)\Delta x_n$.
+
+As the rectangles get more numerous and narrow, the approximation approaches the real area.
+
+The limiting value is denoted
+
+$$\approx \sum_{n} f(x_n)\Delta x_n \rightarrow \int^b_a\! f(x) \mathrm dx$$
+
+This explains the notation used for integrals.
+
+<details>
+<summary>
+
+#### Example 1
+
+Calculate the area between these two curves:
+
+\begin{align*}
+y &= f_1(x) = 2 - x^2 \\
+y &= f_2(x) = x
+\end{align*}
+
+</summary>
+
+![](./images/vimscrot-2021-12-28T15:25:12,556743251+00:00.png)
+
+1. Find the crossing points $P$ and $Q$
+
+   \begin{align*}
+   f_1(x) &= f_2(x) \\
+   x &= 2-x^2 \\
+   x &= 1 \\
+   x &= -2
+   \end{align*}
+
+2. Since $f_1(x) \ge f_2(x)$ between $P$ and $Q$
+
+  \begin{align*}
+  A &= \int^1_{-2}\! (f_1(x) - f_2(x)) \mathrm dx \\
+    &= \int^1_{-2}\! (2 - x^2  - x) \mathrm dx \\
+    &= \left[ 2x - \frac 13 x^3 - \frac 12 x^2 \right]^1_{-2} \\
+    &= \left(2 - \frac 13 - \frac 12 \right) - \left( -4 + \frac 83 - \frac 42 \right) \\
+    &= \frac 92
+  \end{align*}
+
+</details>
+
+## Techniques for Integration
+
+Integration requires multiple techniques and methods to do correctly because it is a PITA.
+
+These are best explained by examples so try to follow those rather than expect and explanation.
+
+### Integration by Substitution
+
+Integration but substitution lets us integrate functions of functions.
+
+<details>
+<summary>
+
+#### Example 1
+
+Find
+
+$$I = \int\!(5x - 1)^3 \mathrm dx$$
+
+</summary>
+
+1. Let $w(x) = 5x - 1$
+2. 
+   \begin{align*}
+   \frac{\mathrm d}{\mathrm dx} w &= 5 \\
+   \frac 15 \mathrm dw &= \mathrm dx
+   \end{align*}
+
+3. The integral is then
+
+   \begin{align*}
+   I &= \int\! w^3 \frac 15 \mathrm dw \\
+     &= \frac 15 \cdot \frac 14 \cdot w^4 + c \\
+     &= \frac{1}{20}w^4 + c
+   \end{align*}
+
+4. Finally substitute $w$ out
+
+  $$I = \frac{(5x-1)^4}{20} + c$$
+
+</details>
+
+<details>
+<summary>
+
+#### Example 2
+
+Find
+
+$$I = \int\! \cos x \sqrt{\sin x + 1} \mathrm dx$$
+
+</summary>
+
+1. Let
+
+   $$w(x) = \sin x + 1$$
+
+2. Then
+
+   \begin{align*}
+   \frac{\mathrm d}{\mathrm dx} w = \cos x \\
+   \mathrm dw = \cos x \mathrm dx \\
+   \end{align*}
+
+3. The integral is now
+
+   \begin{align*}
+   I &= \int\! \sqrt w \,\mathrm dw \\
+     &= \int\! w^{\frac12} \,\mathrm dw \\
+     &= \frac23w^{\frac32} + c
+   \end{align*}
+
+4. Finally substitute $w$ out to get:
+
+   $$I = \frac23 (\sin x + 1)^{\frac32} + c$$
+
+</details>
+
+<details>
+<summary>
+
+#### Example 3
+
+Find
+
+$$I = \int^{\frac\pi2}_0\! \cos x \sqrt{\sin x + 1} \,\mathrm dx$$
+
+</summary>
+
+1. Use the previous example to get to
+
+   $$I = \int^2_1\! \sqrt w \,\mathrm dw = \frac23w^{\frac32} + c$$
+
+2. Since $w(x) = \sin x + 1$ the limits are:
+
+   \begin{align*}
+   x = 0 &\rightarrow w = 1\\
+   x = \frac\pi2 &\rightarrow w = 2
+   \end{align*}
+
+3. This gives us
+
+   $$I = \left[ \frac23w^{\frac32} \right]^2_1 = \frac23 (2^{\frac23} = 1)$$
+
+</details>
+
+<details>
+<summary>
+
+#### Example 4
+
+Find
+
+$$I = \int^1_0\! \sqrt{1 - x^2} \,\mathrm dx$$
+
+</summary>
+
+1. Try a trigonmetrical substitution:
+
+  \begin{align*}
+  x &= \sin w \\
+  \\
+  \frac{\mathrm dx}{\mathrm dw} = \cos w \\
+  \mathrm dx = \cos 2 \,\mathrm dw \\
+  \end{align*}
+
+2.
+
+   \begin{align*}
+   x=0 &\rightarrow w=0 \\
+   x=1 &\rightarrow w=\frac\pi2
+   \end{align*}
+
+3. Therefore
+
+   \begin{align*}
+   I &= \int^{\frac\pi2}_0\! \sqrt{1 - \sin^2 w} \cos w \,\mathrm dw \\
+     &= \int^{\frac\pi2}_0\! \cos^w w \,\mathrm dw
+   \end{align*}
+
+   But $\cos(2w) = 2\cos^2w - 1$ so:
+
+   $$\cos^2w = \frac12 \cos(2w) + \frac12$$
+
+
+   Hence
+
+   \begin{align*}
+   I &= \int^{\frac\pi2}_0\! \frac12 \cos(2w) + \frac12 \,\mathrm dw \\
+     &= \left[ \frac14 \sin(2w) + \frac w2 \right]^{\frac\pi2}_0 \\
+     &= \left( \frac14 \sin\pi + \frac\pi4 \right) - 0 \\
+     &= \frac\pi4
+   \end{align*}
+
+### Integration by Parts
+
+$$uv = \int\! u\frac{\mathrm dv}{\mathrm dx} \,\mathrm dx + \int\! \frac{\mathrm du}{\mathrm dx}v \,\mathrm dx$$
+
+or 
+
+$$\int\! u\frac{\mathrm dv}{\mathrm dx} \,\mathrm dx = uv - \int\! \frac{\mathrm du}{\mathrm dx}v \,\mathrm dx$$
+
+This technique is derived from integrating the product rule.
+
+</details>
+
+<details>
+<summary>
+
+#### Example 1
+
+Find
+
+$$I = \int\! \ln x \,\mathrm dx$$
+
+</summary>
+
+1. Use
+
+   $$\int\! u\frac{\mathrm dv}{\mathrm dx} \,\mathrm dx = uv - \int\! \frac{\mathrm du}{\mathrm dx}v \,\mathrm dx$$
+
+2. Set $u = \ln x$  
+   and $v' = 1$.
+
+3. This means that $u' = \frac1x$ and $v = x$.
+4. 
+
+   \begin{align*}
+   I &= x\ln x - \int\! x\cdot\frac1x \,\mathrm dx + c \\
+     &= x\ln x - \int\! \,\mathrm dx + c \\
+     &= x\ln x - x + c \\
+   \end{align*}
+
+</details>
+
+# Application of Integration
+
+## Differential Equations
+
+Consider the equation
+
+$$\frac{\mathrm dy}{\mathrm dx} = y^2$$
+
+To find $y$, is not a straightforward integration:
+
+$$y = \int\!y^2 \,\mathrm dx$$
+
+The equation above does not solve for $y$ as we can't integrate the right until we know $y$...
+which is what we're trying to find.
+
+This is an example of a first order differential equation.
+The general form is:
+
+$$\frac{\mathrm dy}{\mathrm dx} = F(x, y)$$
+
+### Separable Differential Equations
+
+A first order diferential equation is called *separable* if it is of the form
+
+$$\frac{\mathrm dy}{\mathrm dx} = f(x)g(y)$$
+
+We can solve these by rearranging:
+
+$$\frac1{g(y} \cdot \frac{\mathrm dy}{\mathrm dx} = f(x)$$
+
+$$\int\! \frac1{g(y)} \,\mathrm dy = \int\! f(x) \,\mathrm dx + c$$
+
+
+<details>
+<summary>
+
+#### Example 1
+
+Find $y$ such that
+
+$$\frac{\mathrm dy}{\mathrm dx} = ky$$
+
+where $k$ is a constant.
+
+</summary>
+
+Rearrange to get
+
+\begin{align*}
+\int\! \frac1y \,\mathrm dy &= \int\! k \mathrm dx + c \\
+\ln y &= kx + c
+y &= e^{kx + c} = e^ce^{kx} \\
+  &= Ae^{kx}
+\end{align*}
+
+where $A = e^c$ is an arbitrary constant.
+
+</details>