- *Integrand* is $f$.
- $F(x)$ may be any integral, i.e. $c$ is inconsequential
+#### Properties
+
+$$\int^b_a f(x) \> dx = \int^c_a f(x) \> dx + \int^b_c f(x) \> dx$$
+
+$$\int^a_a f(x) \> dx = 0$$
+
+$$\int^b_a k \cdot f(x) \> dx = k \int^b_a f(x) \> dx$$
+
+$$\int^b_a f(x) \pm g(x) \> dx = \int^b_a f(x) \> dx \pm \int^b_a g(x) \> dx$$
+
+$$\int^b_a f(x) \> dx = - \int^a_b f(x) \> dx$$
+
### Integration by substitution
$$\int f(u) {du \over dx} \cdot dx = \int f(u) \cdot du$$
- $\sin^2x={1 \over 2}(1-\cos 2x)$
- $\cos^2x={1 \over 2}(1+\cos 2x)$
-- $\sin 2x = 2 \sin x \cos x
+- $\sin 2x = 2 \sin x \cos x$
+## Partial fractions
+
+On CAS: Action $\rightarrow$ Transformation $\rightarrow$ `expand/combine`
+or Interactive $\rightarrow$ Transformation $\rightarrow$ `expand` $\rightarrow$ Partial
+
+## Graphing integrals on CAS
+
+In main: Interactive $\rightarrow$ Calculation $\rightarrow$ $\int$ ($\rightarrow$ Definite)
+Restrictions: `Define f(x)=...` $\rightarrow$ `f(x)|x>1` (e.g.)
## Applications of antidifferentiation
To find stationary points of a function, substitute $x$ value of given point into derivative. Solve for ${dy \over dx}=0$. Integrate to find original function.
+## Solids of revolution
+
+Approximate as sum of infinitesimally-thick cylinders
+
+### Rotation about $x$-axis
+
+\begin{align*}
+ V &= \int^{x=b}_{x-a} \pi y^2 \> dx \\
+ &= \pi \int^b_a (f(x))^2 \> dx
+\end{align*}
+
+### Rotation about $y$-axis
+
+\begin{align*}
+ V &= \int^{y=b}_{y=a} \pi x^2 \> dy \\
+ &= \pi \int^b_a (f(y))^2 \> dy
+\end{align*}
+
+### Regions not bound by $y=0$
+
+$$V = \pi \int^b_a f(x)^2 - g(x)^2 \> dx$$
+where $f(x) > g(x)$
+
+## Length of a curve
+
+$$L = \int^b_a \sqrt{1 + ({dy \over dx})^2} \> dx \quad \text{(Cartesian)}$$
+
+$$L = \int^b_a \sqrt{{dx \over dt} + ({dy \over dt})^2} \> dt \quad \text{(parametric)}$$
+
+Evaluate on CAS. Or use Interactive $\rightarrow$ Calculation $\rightarrow$ Line $\rightarrow$ `arcLen`.
+
## Rates
### Related rates
$${da \over db} \quad \text{(change in } a \text{ with respect to } b)$$
-#### Gradient at a point on parametric curve
+### Gradient at a point on parametric curve
$${dy \over dx} = {{dy \over dt} \div {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$
- when two graphs have the same ordinate, $y$-coordinate is double the ordinate
- when two graphs have opposite ordinates, $y$-coordinate is 0 i.e. ($x$-intercept)
- when one of the ordinates is 0, the resulting ordinate is equal to the other ordinate
+
+## Fundamental theorem of calculus
+
+If $f$ is continuous on $[a, b]$, then
+
+$$\int^b_a f(x) \> dx = F(b) - F(a)$$
+
+where $F$ is any antiderivative of $f$
+
+## Differential equations
+
+One or more derivatives
+
+**Order** - highest power inside derivative
+**Degree** - highest power of highest derivative
+e.g. ${\left(dy^2 \over d^2 x\right)}^3$: order 2, degree 3
+
+### Verifying solutions
+
+Start with $y=\dots$, and differentiate. Substitute into original equation.
+
+### Function of the dependent variable
+
+If ${dy \over dx}=g(y)$, then ${dx \over dy} = 1 \div {dy \over dx} = {1 \over g(y)}$. Integrate both sides to solve equation. Only add $c$ on one side. Express $e^c$ as $A$.
+
+### Mixing problems
+
+$$\left({dm \over dt}\right)_\Sigma = \left({dm \over dt}\right)_{\text{in}} - \left({dm \over dt}\)_{\text{out}}$$
+
+### Separation of variables
+
+If ${dy \over dx}=f(x)g(y)$, then:
+
+$$\int f(x) \> dx = \int {1 \over g(y)} \> dy$$
+
+### Using definite integrals to solve DEs
+
+Used for situations where solutions to ${dy \over dx} = f(x)$ is not required.
+
+In some cases, it may not be possible to obtain an exact solution.
+
+Approximate solutions can be found by numerically evaluating a definite integral.
+
+### Using Euler's method to solve a differential equation
+
+$${{f(x+h) - f(x)} \over h } \approx f^\prime (x) \quad \text{for small } h$$
+
+$$\implies f(x+h) \approx f(x) + hf^\prime(x)$$
+