Appendix G Some Formulas Worth Knowing
It is worth knowing all the Differentiation Rules from the Reference Pages at the back of the text, and all the Basic Forms for integrals, items 1 to 20 in the Table of Integrals in those Reference Pages. Appendix A is a guide to checking your knowledge of these.
Beyond that, here are some important formulas, mostly integrals.
There are some useful formulas not given here but better reviewed in the notes where their usage is explained. In particular see Section 7.4 on integrating rational functions and 7.8 on approximate (numerical) integration.
\begin{align*}
\cosh x \amp= \frac{e^x + e^{-x}}{2}, \; \sinh x = \frac{e^x - e^{-x}}{2}.\\
\tanh x \amp= \frac{\sinh x}{\cosh x}, \; \text{etc.}\\
\cos^2(ax) \amp= \frac12[1 + \cos(2ax)]\\
\sin^2(ax) \amp= \frac12[1 - \cos(2ax)]\\
\sin(ax) \cos(ax) \amp= \frac12 \sin(2ax)\\
\frac{1}{1-x} \amp= \sum_{n=0}^\infty x^n = 1 + x + x^2 +x^3 \cdots, |x| \lt; 1
\end{align*}
The area between curves \(y=g(x) \leq f(x)\text{,}\) \(a \leq x \leq b\).
This has infinitesimal area differential
\begin{equation*}
dA = (f(x)-g(x)) dx
\end{equation*}
giving area integral
\begin{equation*}
A = \int dA = \int_{x=a}^b (f(x)-g(x)) dx
\end{equation*}
The solid produced by rotating curve \(y=f(x)\) about the horizontal axis.
This solid can be sliced perpendicular to the \(x\)-axis into disks with volume differential
\begin{equation*}
dV = (\text{disk area}) dx = \pi r^2 dx = \pi y^2 dx
\end{equation*}
giving volume integral
\begin{equation*}
V = \int dV = \int_{x=a}^b \pi r^2 dx = \int_{x=a}^b \pi y^2 dx = \int_{x=a}^b \pi [f(x)]^2 dx
\end{equation*}
As always I recommend using the versions in terms of variables like \(x\) and \(y\) or geometric quantities like the radius \(r\text{,}\) without using function names like \(f(x)\text{:}\) you will know the formula for \(y\text{,}\) or be able to work out the relevant formula for the radius \(r\text{.}\)
The solid produced by rotating the region \(g(x) \leq y \leq f(x)\text{,}\) \(a \leq x \leq b\) about the horizontal axis.
This can be sliced perpendicular to the \(x\)-axis into annuli with volume differential
\begin{equation*}
dV = (\text{annulus area}) dx = \pi R^2 - \pi r^2 dx, \; R=f(x) > r=g(x)
\end{equation*}
giving volume integral
\begin{equation*}
V = \int dV = \pi \int_{x=a}^b R^2 - r^2 dx
\end{equation*}
The solid produced by rotating the region \(g(x) \leq y \leq f(x)\text{,}\) \(0 \leq a \leq x \leq b\) about the vertical axis.
This can be sliced into cylindrical shells of volume differential
\begin{equation*}
dV = \text{(shell circumference)(shell height)} dx = 2 \pi r [f(x)-g(x)] dx = 2 \pi x [f(x)-g(x)] dx
\end{equation*}
(since the radius of each shell is \(r=x \geq 0\)) giving volume integral
\begin{equation*}
V = \int dV = \int_{x=a}^b 2 \pi x [f(x)-g(x)] dx
\end{equation*}
The average value of the function \(f(x)\) on interval \(a \leq x \leq b\).
\begin{equation*}
\bar{f} = f_{ave} = \frac{1}{b-a} \int_{x=a}^b f(x) \, dx
\end{equation*}
Substitution in a Definite Integral with \(u=g(x)\).
\begin{equation*}
\int_{x=a}^b f(u(x)) \frac{du}{dx} dx = \int_{u=c}^d f(u) du, \; c = g(a),\, d=g(b)
\end{equation*}
Note that the limits of integration change, from \(x\) values to \(u\) values.
Integration by Parts.
\begin{equation*}
\int u \, dv = uv - \int v \, du \text{ or } \int u(x) \frac{dv}{dx} = u(x)v(x) - \int v(x) \frac{du}{dx} \, dx
\end{equation*}
With definite integrals, this becomes
\begin{equation*}
\int _{x=a}^b u \, dv = [uv]_{x=a}^b- \int_{x=a}^b v \, du
\end{equation*}
Note that the limits of integration are still \(x\) values, as \(du\) and \(dv\) are shorthands \(du=\ds\frac{du}{dx} dx\) and \(dv=\ds\frac{dv}{dx} dx\text{:}\) the dummy variable in the integration is still \(x\text{.}\)
For Integrals Involving the Square Root of a Quadratic.
- With \(\sqrt{a^2-x^2}\text{,}\) try \(x=a\sin\theta\) so that \(\sqrt{a^2-x^2}=a\cos\theta\text{,}\) \(dx=a\cos\theta \, d\theta\text{.}\)
- With \(\sqrt{a^2+x^2}\text{,}\) try \(x=a\tan\theta\) so that \(\sqrt{a^2+x^2}=a\sec\theta\text{,}\) \(dx=a\sec^2\theta \, d\theta\text{.}\)
- With \(\sqrt{x^2-a^2}\text{,}\) try \(x=a\sec\theta\) so that \(\sqrt{x^2-a^2}=a\tan\theta\text{,}\) \(dx=a\tan\theta\sec\theta \, d\theta\text{.}\)
For more general quadratics, first extract the coefficient of \(x^2\) as a common factor and then complete the square
\begin{equation*}
x^2 + bx + c = [x+b/2]^2 + c-(b/2)^2
\end{equation*}
Then the substitution \(u=x+b/2\) is useful, and \(a^2=c-(b/2)^2\) or \((b/2)^2-c\text{,}\) whichever is positive.
For a Separable Differential Equation \(\ds\frac{dy}{dx} = f(x)g(y)\).
Write in differential form
\begin{equation*}
\frac{1}{g(y)} dy = f(x) dx
\end{equation*}
and integrate, but beware of division by zero: check the case \(g(y)=0\text{.}\)
The Arc Length of the Curve \(y=f(x)\text{,}\) \(a \leq x \leq b\).
The arc length differential is
\begin{equation*}
ds = \sqrt{dx^2+dy^2} = \sqrt{1+\left(\frac{dy}{dx}\right)^2} \; dx
\end{equation*}
so the total arc length is
\begin{equation*}
L = \int ds = \int_{x=a}^b \sqrt{1+\left(\frac{dy}{dx}\right)^2} \; dx
\end{equation*}
The parametric curve \(x=f(t)\text{,}\) \(y=f(t)\).
This has slope at point \(P(x(t),y(t))\) given by
\begin{equation*}
m = \frac{dy}{dx} = \frac{dy/dt}{dx/dt}
\end{equation*}
but beware of division by zero, whihc happens at points where \(\ds\frac{dx}{dt}=0\text{.}\) There, the above formula needs to be simplified before using it.
The arc length of the above parametric curve..
For \(\alpha \leq t \leq \beta\) the arc length differential is
\begin{equation*}
ds = \sqrt{dx^2+dy^2} = \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} \; dt
\end{equation*}
so the total arc length is
\begin{equation*}
L = \int ds = \int_{t=\alpha}^\beta \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} \; dt
\end{equation*}
Polar coordinates.
To convert between polar coordinates \(r,\theta\) and cartesian (rectangular) coordinates \(x,y\text{,}\) use
\begin{equation*}
x = r \cos \theta, \; y = r \sin \theta,
\end{equation*}
\begin{equation*}
r^2 = x^2 + y^2, \; \tan \theta = \frac{y}{x}
\end{equation*}
with special handling of points on the \(y\)-axis: the ones with \(x=0\text{.}\)
The polar curve \(r=f(\theta)\).
This is a case of parametric curve, with angle \(\theta\) as the parameter and
\begin{equation*}
x = r \cos \theta = f(\theta) \cos \theta, \; y = r \sin \theta = g(\theta) \sin \theta
\end{equation*}
The slope of polar curve \(r=f(\theta)\).
This is given by combining formulas above, but simplifies to
\begin{equation*}
m= \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\ds\frac{dr}{d\theta}\sin \theta+r\cos\theta}{\ds\frac{dr}{d\theta}\cos\theta-r\sin\theta}
\end{equation*}
The area of the region inside polar curve \(r=f(\theta)\text{,}\) \(a \leq \theta \leq b\).
This has area differential
\begin{equation*}
dA = \frac{1}{2}r^2 d\theta = \frac{1}{2} (f(\theta))^2 d\theta
\end{equation*}
so the area is
\begin{equation*}
A = \int dA = \int_{\theta=a}^b \frac{1}{2}r^2 d\theta = \frac{1}{2} \int_{\theta=a}^b \left( f(\theta) \right)^2 d\theta.
\end{equation*}
The arc length of the above polar curve.
This can be derived using formulas above for parametric curves but simplifies nicely: the arc length differential is
\begin{equation*}
ds = \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta
\end{equation*}
so the total arc length is
\begin{equation*}
L = \int ds = \int_{\theta=a}^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta.
\end{equation*}
Integral error bound for the series \(S=\sum a_n\) with \(a_n = f(n)\text{,}\) \(f\) positive and decreasing.
The error \(R_N\) for the partial sum \(S_N = \sum_{n=1}^N a_n\) as an approximation of \(S=\sum a_n\) has a size limit
\begin{equation*}
0 \leq R_N = S-S_N \leq \int_N^\infty f(n) \, dn
\end{equation*}
The Taylor Series.
\begin{equation*}
\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n = f(a) + f'(a) (x-a) + \frac{f''(a)}{2} (x-a)^2+ \frac{f^{(3)}(a)}{3!} (x-a)^3 \cdots
\end{equation*}
and the \(N\)-the degree Taylor polynomial
\begin{equation*}
T_N(x) = \sum_{n=0}^N \frac{f^{(n)}(a)}{n!} (x-a)^n = f(a) + f'(a) (x-a) + \frac{f''(a)}{2} (x-a)^2 \cdots \frac{f^{(N)}(a)}{N!} (x-a)^N,
\end{equation*}
the remainder
\begin{equation*}
R_N(x) = f(x) - T_N(x)
\end{equation*}
has an upper size limit
\begin{equation*}
|R_N(x)| \leq \frac{M}{(N+1)!} |x-a|^{N+1}
\end{equation*}
if you have a number \(M\) such that \(|f^{(n)}(x)| \leq M\text{.}\) If the later is only true for some \(x\text{,}\) \(|x-a| \leq d\text{,}\) then the limit on \(R_N(x)\) is also only for these \(x\) values.
