math: add integration

docs/mcv4u7.md

@@ -2,6 +2,76 @@
 
 The course code for this page is **MCV4U7**.
 
+## Integration
+
+Integration is an operation that finds the **net** area under a curve, and is the opposite operation of differentiation. As such, it is also known as **anti-differentiation**.
+
+The area under a curve between the interval of x-values $[a,b]$ is:
+$$A=\lim_{x\to\infty}\sum^n_{i=1}f(x_i)\Delta x$$
+
+which can be simplified to, where $dx$ indicates that integration should be performed with respect to $x$:
+$$A=\int^b_a f(x)dx$$
+
+While $\Sigma$ refers to a finite sum, $\int$ refers to the sum of a limit.
+
+As integration is the opposite operation of differentiation, they can cancel each other out.
+$$\frac{d}{dx}\int f(x)dx=f(x)$$
+
+The **integral** or **anti-derivative** of a function is capitalised by convention. Where $C$ is an unknown constant:
+$$\int f(x)dx=F(x)+C$$
+
+When integrating, there is always an unknown constant $C$ as there are infinitely many possible functions that have the same rate of change but have different vertical translations.
+
+!!! definition
+    - $C$ is known as the **constant of integration**.
+    - $f(x)$ is the **integrand**.
+
+### Integration rules
+
+$$
+\begin{align*}
+&\int 1dx &= &&x+C \\
+&\int (ax^n)dx, n≠-1 &=&&\frac{a}{n+1}x^{n+1} + C \\
+&\int (x^{-1})dx&=&&\ln|x|+C \\
+&\int (ax+b)^{-1}dx&=&&\frac{\ln|ax+b|}{a}+C \\
+&\int (ae^{kx})dx &= &&\frac{a}{k}e^{kx} + C \\
+&\int (\sin kx)dx &= &&\frac{-\cos kx}{k}+C \\
+&\int (\cos kx)dx &= &&\frac{\sin kx}{k}+C \\
+\end{align*}
+$$
+
+Similar to differentiation, integration allows for constant multiples to be brought out and terms to be considered individually.
+
+$$
+\begin{align*}
+&\int k\cdot f(x)dx&=&&k\int f(x)dx \\
+&\int[f(x)\pm g(x)]dx&=&&\int f(x)dx \pm \int g(x)dx
+\end{align*}
+$$
+
+### Indefinite integration
+
+The indefinite integral of a function contains every possible anti-derivative — that is, it contains the constant of integration $C$.
+$$\int f(x)dx=F(x)+C$$
+
+### Substitution rule
+
+Similar to limit evaluation, the substitution of complex expressions involving $x$ and $dx$ with $u$ and $du$ can be done.
+
+??? example
+    To solve $\int (x\sqrt{x-1})dx$:
+    $$
+    let\ u=x-1 \\
+    ∴ \frac{du}{dx}=1 \\
+    ∴ du=dx \\
+    \begin{align*}
+    \int (x\sqrt{x-1})dx &\to \int(u+1)(u^\frac{1}{2})du \\
+    &= \int(u^\frac{3}{2}+u^\frac{1}{2})du \\
+    &= \frac{2}{5}u^\frac{5}{2}+\frac{2}{3}u^\frac{3}{2}+C \\
+    &= \frac{2}{5}(x-1)^\frac{5}{2} + \frac{2}{3}(x-1)^\frac{3}{2} + C
+    \end{align*}
+    $$
+
 ## Resources
 
  - [IB Math Analysis and Approaches Syllabus](/resources/g11/ib-math-syllabus.pdf)