## Partial fractions
-On CAS: Action $\rightarrow$ Transformation $\rightarrow$ `expand/combine`
+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
$${dy \over dx} = {{dy \over dt} \div {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$
$$\int^b_a f(x) \> dx = F(b) - F(a)$$
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 $A$.