Calculus II Reference

Unit 1 41 topics

Integration Techniques

Integration Using Substitution, Special Techniques for Integration, Integration Using Trigonometric Identities, and more.

1.1 Integration Using Substitution

1.1.1
Integrating Algebraic Functions Using Substitution
Let $u$ be the inner expression whose derivative also appears as a factor. Rewrite everything in $u$ , integrate, then substitute back.
e.g. $\int 2x(x^2+1)^3\,dx$ : $u = x^2+1$ gives $\tfrac14(x^2+1)^4 + C$
$\displaystyle \int f(g(x))\,g'(x)\,dx$
$\displaystyle = \int f(u)\,du, \quad u = g(x)$
1.1.2
Integrating Linear Rational Functions Using Substitution
For a fraction with a linear denominator, let $u = ax + b$ , so $dx = \tfrac1a\,du$ . If the numerator is also linear, write it in terms of $u$ and split the fraction.
e.g. $\int \frac{3}{2x+1}\,dx = \tfrac32 \ln\lvert 2x+1\rvert + C$
$\displaystyle \int \frac{1}{ax+b}\,dx$
$\displaystyle = \frac1a \ln\lvert ax+b\rvert + C$
1.1.3
Integration Using Substitution
Choose $u$ , find $du$ , and rewrite the whole integrand in $u$ with no $x$ left. A missing constant factor can be fixed by multiplying and dividing. Always substitute back at the end.
e.g. $\int x\sqrt{x^2+4}\,dx$ : $u = x^2+4$ , $du = 2x\,dx$ , so $\tfrac12\int u^{1/2}\,du = \tfrac13(x^2+4)^{3/2} + C$
$\displaystyle u = g(x)$
$\displaystyle du = g'(x)\,dx$
1.1.4
Calculating Definite Integrals Using Substitution
Change the limits to $u$ -values when you substitute. Then evaluate in $u$ with no need to substitute back.
e.g. $\int_0^1 2x(x^2+1)\,dx$ : $u$ runs from $1$ to $2$ , so $\int_1^2 u\,du = 2 - \tfrac12 = \tfrac32$
$\displaystyle \int_a^b f(g(x))\,g'(x)\,dx$
$\displaystyle = \int_{g(a)}^{g(b)} f(u)\,du$
1.1.5
Further Integration of Algebraic Functions Using Substitution
If an extra $x$ is left over after substituting, solve $u = g(x)$ for $x$ and replace it too. Then expand and integrate term by term.
e.g. $\int x\sqrt{x+1}\,dx$ : $u = x+1$ , $x = u - 1$ , so $\int (u^{3/2} - u^{1/2})\,du = \tfrac25 (x+1)^{5/2} - \tfrac23 (x+1)^{3/2} + C$
1.1.6
Integrating Exponential Functions Using Linear Substitution
With a linear exponent, integrate as usual and divide by the coefficient of $x$ .
e.g. $\int e^{3x-1}\,dx = \tfrac13 e^{3x-1} + C$
$\displaystyle \int e^{ax+b}\,dx = \frac1a e^{ax+b} + C$
1.1.7
Integrating Exponential Functions Using Substitution
Let $u$ be the exponent. The derivative of the exponent must appear as a factor, up to a constant.
e.g. $\int x e^{x^2}\,dx$ : $u = x^2$ , $du = 2x\,dx$ , so the answer is $\tfrac12 e^{x^2} + C$
$\displaystyle \int e^{g(x)}\,g'(x)\,dx = e^{g(x)} + C$
1.1.8
Integrating Trigonometric Functions Using Substitution
Let $u$ be the angle, or let $u$ be $\sin x$ or $\cos x$ when its derivative is a factor. Watch the sign: $\int \sin u\,du = -\cos u + C$ .
e.g. $\int \sin^2 x\cos x\,dx$ : $u = \sin x$ gives $\tfrac13\sin^3 x + C$
$\displaystyle \int \cos(ax+b)\,dx$
$\displaystyle = \frac1a \sin(ax+b) + C$
$\displaystyle \int \sin(ax+b)\,dx$
$\displaystyle = -\frac1a \cos(ax+b) + C$
1.1.9
Integrating Logarithmic Functions Using Substitution
When $\ln x$ appears with a factor of $\tfrac1x$ , let $u = \ln x$ . A fraction whose numerator is the derivative of its denominator integrates to a natural log.
e.g. $\int \frac{\ln x}{x}\,dx = \tfrac12(\ln x)^2 + C$
$\displaystyle u = \ln x \Rightarrow du = \frac1x\,dx$
$\displaystyle \int \frac{g'(x)}{g(x)}\,dx$
$\displaystyle = \ln\lvert g(x)\rvert + C$
1.1.10
Integration by Substitution With Inverse Trigonometric Functions
Match the integrand to an inverse trig form, using substitution to make the variable part a perfect square. Here $a > 0$ .
e.g. $\int \frac{dx}{1 + 4x^2}$ : $u = 2x$ gives $\tfrac12\arctan(2x) + C$
$\displaystyle \int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C$
$\displaystyle \int \frac{dx}{a^2 + x^2}$
$\displaystyle = \frac1a \arctan\frac{x}{a} + C$
1.1.11
Integrating Hyperbolic Functions Using Substitution
Let $u$ be the inner function, replace $dx$ using $du$ , integrate the basic hyperbolic form, then substitute back.
e.g. $\int x\cosh(x^2)\,dx$ : $u = x^2$ , $du = 2x\,dx$ , so $\tfrac12 \sinh(x^2) + C$
$\displaystyle \int f(g(x))\,g'(x)\,dx = \int f(u)\,du$
$\displaystyle \int \sinh(ax)\,dx = \tfrac1a\cosh(ax) + C$
$\displaystyle \int \tanh x\,dx = \ln(\cosh x) + C$
1.1.12
Integration by Substitution With Inverse Hyperbolic Functions
Choose $u$ so the integrand matches an inverse hyperbolic form in $u$ , with $du$ already present. Here $a > 0$ ; the $\cosh^{-1}$ form needs $u > a$ and the $\tanh^{-1}$ form needs $\lvert u \rvert < a$ .
e.g. $\int \frac{e^x}{\sqrt{e^{2x} + 1}}\,dx$ : $u = e^x$ , $du = e^x\,dx$ , so $\sinh^{-1}(e^x) + C$
$\displaystyle \int \frac{du}{\sqrt{u^2 + a^2}} = \sinh^{-1}\frac{u}{a} + C$
$\displaystyle \int \frac{du}{\sqrt{u^2 - a^2}} = \cosh^{-1}\frac{u}{a} + C$
$\displaystyle \int \frac{du}{a^2 - u^2} = \frac1a \tanh^{-1}\frac{u}{a} + C$
1.1.13
Integration by Substitution With Inverse Reciprocal Hyperbolic Functions
Choose $u$ so the integrand becomes $\frac{du}{u\sqrt{a^2 \mp u^2}}$ or $\frac{du}{a^2 - u^2}$ , then use the matching inverse reciprocal form (stated for $u > 0$ ).
e.g. $\int \frac{dx}{x\sqrt{1 + 9x^2}}$ : $u = 3x$ turns it into $\int \frac{du}{u\sqrt{1 + u^2}}$ , so $-\operatorname{csch}^{-1}(3x) + C$ for $x > 0$
$\displaystyle \int \frac{du}{u\sqrt{a^2 - u^2}}$
$\displaystyle = -\frac1a \operatorname{sech}^{-1}\frac{u}{a} + C$
$\displaystyle \int \frac{du}{u\sqrt{a^2 + u^2}}$
$\displaystyle = -\frac1a \operatorname{csch}^{-1}\frac{u}{a} + C$

1.2 Special Techniques for Integration

1.2.1
Integrating Functions Using Polynomial Division
If the numerator's degree is at least the denominator's, divide first. Integrate the quotient and the remainder fraction separately.
e.g. $\int \frac{x^2+1}{x+1}\,dx = \int \left(x - 1 + \tfrac{2}{x+1}\right)dx = \tfrac{x^2}{2} - x + 2\ln\lvert x+1\rvert + C$
$\displaystyle \frac{N(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}$
1.2.2
Integrating Functions by Completing the Square
For a quadratic denominator that does not factor, complete the square, then use the arctan form with $u = x + \tfrac b2$ .
e.g. $\int \frac{dx}{x^2 + 2x + 5} = \int \frac{dx}{(x+1)^2 + 4} = \tfrac12\arctan\tfrac{x+1}{2} + C$
$\displaystyle x^2 + bx + c$
$\displaystyle = \left(x + \tfrac b2\right)^2 + c - \tfrac{b^2}{4}$
$\displaystyle \int \frac{du}{u^2 + a^2}$
$\displaystyle = \frac1a \arctan\frac{u}{a} + C$
1.2.3
Integration Using Hyperbolic Functions and Completing the Square
Complete the square in the quadratic, let $u = x + h$ , and use the matching inverse hyperbolic form.
e.g. $\int \frac{dx}{\sqrt{x^2 + 2x + 5}} = \int \frac{dx}{\sqrt{(x+1)^2 + 4}} = \sinh^{-1}\tfrac{x+1}{2} + C$
$\displaystyle x^2 + bx + c$
$\displaystyle = \left(x + \tfrac{b}{2}\right)^2 + c - \tfrac{b^2}{4}$
1.2.4
Integration Using Inverse Reciprocal Hyperbolic Functions and Completing the Square
Complete the square so the integrand reads $\frac{du}{u\sqrt{a^2 \mp u^2}}$ or $\frac{du}{a^2 - u^2}$ with $u = x + h$ . The factor outside the root must be the same $u$ .
e.g. $-x^2 - 2x + 3 = 4 - (x+1)^2$ , so $\int \frac{dx}{(x+1)\sqrt{-x^2 - 2x + 3}} = -\tfrac12 \operatorname{sech}^{-1}\tfrac{x+1}{2} + C$ for $-1 < x < 1$
$\displaystyle \int \frac{du}{u\sqrt{a^2 - u^2}}$
$\displaystyle = -\frac1a \operatorname{sech}^{-1}\frac{u}{a} + C$

1.3 Integration Using Trigonometric Identities

1.3.1
Integration Using Basic Trigonometric Identities
Rewrite the integrand with quotient and reciprocal identities until it matches a known antiderivative.
e.g. $\int \frac{\sin x}{\cos^2 x}\,dx = \int \sec x\tan x\,dx = \sec x + C$
$\displaystyle \tan x = \frac{\sin x}{\cos x}$
$\displaystyle \sec x = \frac{1}{\cos x},\quad \csc x = \frac{1}{\sin x}$
1.3.2
Integration Using the Pythagorean Identities
Replace a square you cannot integrate with one you can, or split off one factor so substitution works.
e.g. $\int \tan^2 x\,dx = \int (\sec^2 x - 1)\,dx = \tan x - x + C$
$\displaystyle \sin^2 x + \cos^2 x = 1$
$\displaystyle 1 + \tan^2 x = \sec^2 x$
1.3.3
Integration Using the Double-Angle Formulas
To integrate $\sin^2 x$ or $\cos^2 x$ , lower the power with these identities first.
e.g. $\int \sin^2 x\,dx = \tfrac{x}{2} - \tfrac{\sin 2x}{4} + C$
$\displaystyle \sin^2 x = \frac{1 - \cos 2x}{2}$
$\displaystyle \cos^2 x = \frac{1 + \cos 2x}{2}$
$\displaystyle \sin x\cos x = \frac12 \sin 2x$
1.3.4
Integrating Products of Trigonometric Functions
For $\sin mx\cos nx$ , $\sin mx\sin nx$ , or $\cos mx\cos nx$ with $m \ne n$ , rewrite the product as a sum, then integrate each term.
e.g. $\int \sin 3x\cos x\,dx = \tfrac12\int(\sin 2x + \sin 4x)\,dx = -\tfrac14\cos 2x - \tfrac18\cos 4x + C$
$\displaystyle \sin A\cos B$
$\displaystyle = \tfrac12[\sin(A-B) + \sin(A+B)]$
$\displaystyle \sin A\sin B$
$\displaystyle = \tfrac12[\cos(A-B) - \cos(A+B)]$
$\displaystyle \cos A\cos B$
$\displaystyle = \tfrac12[\cos(A-B) + \cos(A+B)]$

1.4 Integration Using Hyperbolic Identities

1.4.1
Integration Using Basic Hyperbolic Identities
Rewrite the integrand with the exponential definitions or quotient identities, simplify, then integrate.
e.g. $\int e^x\sinh x\,dx = \int \tfrac{e^{2x} - 1}{2}\,dx = \tfrac14 e^{2x} - \tfrac12 x + C$
$\displaystyle \sinh x = \frac{e^x - e^{-x}}{2}$
$\displaystyle \cosh x = \frac{e^x + e^{-x}}{2}$
$\displaystyle \cosh x + \sinh x = e^x$
$\displaystyle \tanh x = \frac{\sinh x}{\cosh x}$
1.4.2
Integration Using the Hyperbolic Pythagorean Identities
Swap a squared hyperbolic function for an equivalent form you can integrate. Note the signs differ from the trig versions.
e.g. $\int \tanh^2 x\,dx = \int (1 - \operatorname{sech}^2 x)\,dx = x - \tanh x + C$
$\displaystyle \cosh^2 x - \sinh^2 x = 1$
$\displaystyle 1 - \tanh^2 x = \operatorname{sech}^2 x$
$\displaystyle \coth^2 x - 1 = \operatorname{csch}^2 x$
1.4.3
Integration Using the Hyperbolic Double-Angle Formulas
Lower the power of $\sinh^2 x$ or $\cosh^2 x$ with the double-angle formulas, then integrate.
e.g. $\int \sinh^2 x\,dx = \int \tfrac{\cosh 2x - 1}{2}\,dx = \tfrac14\sinh 2x - \tfrac12 x + C$
$\displaystyle \sinh 2x = 2\sinh x\cosh x$
$\displaystyle \cosh 2x = \cosh^2 x + \sinh^2 x$
$\displaystyle \sinh^2 x = \frac{\cosh 2x - 1}{2}$
$\displaystyle \cosh^2 x = \frac{\cosh 2x + 1}{2}$

1.5 Integration by Parts

1.5.1
Introduction to Integration by Parts
Use for a product of two different kinds of functions. Pick $u$ to get simpler when differentiated, using LIATE order: logs, inverse trig, algebraic, trig, exponential. The rest is $dv$ .
e.g. $\int xe^x\,dx$ : $u = x$ , $dv = e^x\,dx$ , so $= xe^x - \int e^x\,dx = xe^x - e^x + C$
$\displaystyle \int u\,dv = uv - \int v\,du$
1.5.2
Using Integration by Parts to Calculate Integrals With Logarithms
Let $u = \ln x$ , since it simplifies to $\tfrac1x$ when differentiated. Let $dv$ be the rest, even if that is just $dx$ .
e.g. $\int x\ln x\,dx$ : $u = \ln x$ , $v = \tfrac{x^2}{2}$ , so $= \tfrac{x^2}{2}\ln x - \int \tfrac{x}{2}\,dx = \tfrac{x^2}{2}\ln x - \tfrac{x^2}{4} + C$
$\displaystyle \int \ln x\,dx = x\ln x - x + C$
1.5.3
Applying the Integration By Parts Twice
If the new integral still needs parts, apply it again. Choose the same kind of function as $u$ the second time, or you will undo the first step.
e.g. $x^2e^x - 2(xe^x - e^x) + C = e^x(x^2 - 2x + 2) + C$
$\displaystyle \int x^2e^x\,dx = x^2e^x - 2\int xe^x\,dx$
1.5.4
The Tabular Method of Integration by Parts
Use when $u$ is a polynomial. Column D: differentiate $u$ until you reach 0. Column I: integrate $dv$ the same number of times. Multiply each D entry by the next row's I entry, with signs $+, -, +, \ldots$ , and add.
e.g. $\int x^2\sin x\,dx$ : D: $x^2, 2x, 2, 0$ ; I: $\sin x, -\cos x, -\sin x, \cos x$ ; result $-x^2\cos x + 2x\sin x + 2\cos x + C$
$\displaystyle \int f g\,dx = fG_1 - f'G_2 + f''G_3 - \cdots$
1.5.5
Integration by Parts in Cyclic Cases
For products like $e^x\sin x$ , neither factor ever simplifies. Integrate by parts twice, keeping the same type of $u$ . The original integral $I$ reappears, so solve for $I$ algebraically.
e.g. $\int e^x\cos x\,dx = \tfrac12 e^x(\sin x + \cos x) + C$
$\displaystyle I = \int e^x\sin x\,dx$
$\displaystyle I = e^x\sin x - e^x\cos x - I$
$\displaystyle \Rightarrow\; I = \tfrac12 e^x(\sin x - \cos x) + C$

1.6 Integration Using Partial Fractions

1.6.1
Expressing Rational Functions as Sums of Partial Fractions
The numerator's degree must be less than the denominator's (long divide first if not). Factor the denominator. Each distinct linear factor gets a constant over it. Clear denominators and plug in each root to find the constants.
e.g. $\tfrac{5x - 1}{(x - 1)(x + 1)}$ : $5x - 1 = A(x + 1) + B(x - 1)$ ; $x = 1$ gives $A = 2$ , $x = -1$ gives $B = 3$
$\displaystyle \frac{px + q}{(x - a)(x - b)} = \frac{A}{x - a} + \frac{B}{x - b}$
1.6.2
Expressing Rational Functions with Repeated Factors as Sums of Partial Fractions
A factor $(x - a)^k$ gets $k$ terms, one for each power from 1 to $k$ . Plug in roots for some constants, then match coefficients for the rest.
e.g. $\tfrac{1}{x^2(x - 1)} = -\tfrac1x - \tfrac{1}{x^2} + \tfrac{1}{x - 1}$
$\displaystyle \frac{A_1}{x - a} + \frac{A_2}{(x - a)^2} + \cdots + \frac{A_k}{(x - a)^k}$
1.6.3
Expressing Rational Functions with Irreducible Quadratic Factors as Sums of Partial
A quadratic factor that cannot be factored over the reals ( $b^2 - 4ac < 0$ ) gets a linear numerator $Bx + C$ . Match coefficients to find the constants.
e.g. $\tfrac{2x^2 + x + 1}{x(x^2 + 1)} = \tfrac1x + \tfrac{x + 1}{x^2 + 1}$
$\displaystyle \frac{A}{x - a} + \frac{Bx + C}{x^2 + px + q}$
1.6.4
Integrating Rational Functions Using Partial Fractions
Decompose into partial fractions, then integrate each term. Each linear term gives a natural log.
e.g. $\int \tfrac{5x - 1}{x^2 - 1}\,dx = \int\left(\tfrac{2}{x - 1} + \tfrac{3}{x + 1}\right)dx = 2\ln\lvert x - 1\rvert + 3\ln\lvert x + 1\rvert + C$
$\displaystyle \int \frac{A}{x - a}\,dx = A\ln\lvert x - a\rvert + C$
1.6.5
Integrating Rational Functions with Repeated Factors
Terms over $(x - a)^n$ with $n \ge 2$ integrate by the power rule, not to a log. Only the first-power term gives a log.
e.g. $\int \tfrac{dx}{x^2(x - 1)} = \int\left(-\tfrac1x - \tfrac{1}{x^2} + \tfrac{1}{x - 1}\right)dx = -\ln\lvert x\rvert + \tfrac1x + \ln\lvert x - 1\rvert + C$
$\displaystyle \int \frac{dx}{(x - a)^n} = \frac{(x - a)^{1 - n}}{1 - n} + C$
$\displaystyle \int \frac{dx}{(x - a)^2} = -\frac{1}{x - a} + C$
1.6.6
Integrating Rational Functions with Irreducible Quadratic Factors
Split $\tfrac{Bx + C}{x^2 + k^2}$ into an $x$ part, which gives a log, and a constant part, which gives arctangent. Complete the square first if the quadratic has an $x$ term.
e.g. $\int\left(\tfrac1x + \tfrac{x + 1}{x^2 + 1}\right)dx = \ln\lvert x\rvert + \tfrac12\ln(x^2 + 1) + \arctan x + C$
$\displaystyle \int \frac{x\,dx}{x^2 + k^2} = \tfrac12\ln(x^2 + k^2) + C$
$\displaystyle \int \frac{dx}{x^2 + k^2} = \tfrac1k\arctan\tfrac{x}{k} + C$

1.7 Improper Integrals

The region under y = 1/x² goes on forever, yet its area adds up to exactly 1. An improper integral is a limit: integrate to b, then let b run to infinity.

1.7.1
Improper Integrals
An integral with an infinite limit is defined as a limit. It converges if the limit is a finite number and diverges otherwise. $\int_1^\infty \tfrac{1}{x^p}\,dx$ converges only when $p > 1$ .
e.g. $\int_1^\infty \tfrac{1}{x^2}\,dx = \lim_{b \to \infty}\left(1 - \tfrac1b\right) = 1$
$\displaystyle \int_a^\infty f(x)\,dx$
$\displaystyle = \lim_{b \to \infty}\int_a^b f(x)\,dx$
1.7.2
Improper Integrals Involving Exponential Functions
Integrate up to $b$ , then let $b \to \infty$ . Decaying exponentials vanish: $e^{-kb} \to 0$ for $k > 0$ .
e.g. $\int_0^\infty e^{-2x}\,dx = \lim_{b \to \infty}\left(\tfrac12 - \tfrac12 e^{-2b}\right) = \tfrac12$
$\displaystyle \int_0^\infty e^{-kx}\,dx = \frac1k \quad (k > 0)$
1.7.3
Improper Integrals Involving Arctangent
Integrands like $\tfrac{1}{x^2 + k^2}$ integrate to arctangent, which levels off at $\pm\tfrac{\pi}{2}$ , so these integrals converge.
e.g. $\int_0^\infty \tfrac{dx}{x^2 + 4} = \lim_{b \to \infty}\tfrac12\arctan\tfrac{b}{2} = \tfrac{\pi}{4}$
$\displaystyle \int \frac{dx}{1 + x^2} = \arctan x + C$
$\displaystyle \lim_{b \to \infty}\arctan b = \frac{\pi}{2}$
1.7.4
Improper Integrals Over the Real Line
Split at any convenient point $c$ , often 0. The integral converges only if both halves converge. Do not take one symmetric limit $\int_{-b}^{b}$ , since that can hide divergence.
e.g. $\int_{-\infty}^{\infty}\tfrac{dx}{1 + x^2} = \tfrac{\pi}{2} + \tfrac{\pi}{2} = \pi$
$\displaystyle \int_{-\infty}^{\infty} f\,dx = \int_{-\infty}^{c} f\,dx + \int_{c}^{\infty} f\,dx$
1.7.5
Improper Integrals of the Second Kind
The integrand blows up at an endpoint. Replace that endpoint with $t$ and take a one-sided limit. $\int_0^1 \tfrac{1}{x^p}\,dx$ converges only when $p < 1$ .
e.g. $\int_0^1 \tfrac{dx}{\sqrt{x}} = \lim_{t \to 0^+}\left(2 - 2\sqrt{t}\right) = 2$
$\displaystyle \int_a^b f(x)\,dx$
$\displaystyle = \lim_{t \to a^+}\int_t^b f(x)\,dx$
1.7.6
Improper Integrals of the Second Kind: Discontinuities at Interior Points
If $f$ blows up at a point $c$ inside $[a, b]$ , split there and take one-sided limits on each piece. If either piece diverges, the whole integral diverges. Always check for a vertical asymptote before using the Fundamental Theorem.
e.g. $\int_{-1}^{1}\tfrac{dx}{x^2}$ diverges because $\int_0^1 \tfrac{dx}{x^2}$ diverges. The shortcut answer $-2$ is wrong.
$\displaystyle \int_a^b f\,dx = \int_a^c f\,dx + \int_c^b f\,dx$

Unit 2 27 topics

Applications of Integration

Applications of Integration, Volumes of Solids With Known Cross Sections, Volumes of Revolution, and more.

2.8 Applications of Integration

2.8.1
Integrating Rates of Change
The integral of a rate of change gives the net change in the amount. Add the starting amount to get the final amount. For velocity, $\int v\,dt$ is displacement and $\int \lvert v \rvert\,dt$ is total distance.
e.g. Tank has 10 gal, water flows in at $r(t) = 4t$ gal/min: after 3 min, $10 + \int_0^3 4t\,dt = 10 + 18 = 28$ gal
$\displaystyle Q(b) = Q(a) + \int_a^b Q'(t)\,dt$
2.8.2
Integrating Density Functions
Integrating a density (amount per unit length) over an interval gives the total amount. For a density per unit area that depends on the distance $r$ from a center, multiply by the ring circumference $2\pi r$ .
e.g. A 2 m rod with density $\rho(x) = 3x^2$ kg/m has mass $\int_0^2 3x^2\,dx = 8$ kg
$\displaystyle \text{total} = \int_a^b \rho(x)\,dx$
$\displaystyle \text{radial: } \int_0^R 2\pi r\,\rho(r)\,dr$
2.8.3
The Average Value of a Function
The average value of $f$ on $[a, b]$ is its integral divided by the length of the interval.
e.g. $f(x) = x^2$ on $[0, 3]$ : $\tfrac13\int_0^3 x^2\,dx = \tfrac13 \cdot 9 = 3$
$\displaystyle f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx$
2.8.4
The Area Between Curves Expressed as Functions of X
Integrate top minus bottom with respect to $x$ . The limits are usually the $x$ -values where the curves intersect.
e.g. $y = x$ and $y = x^2$ on $[0, 1]$ : $\int_0^1 (x - x^2)\,dx = \tfrac12 - \tfrac13 = \tfrac16$
$\displaystyle A = \int_a^b \left[f(x) - g(x)\right]dx$
$\displaystyle f(x) \ge g(x) \text{ on } [a, b]$
2.8.5
The Area Between Curves Expressed as Functions of Y
When the curves are easier to write as $x$ in terms of $y$ , integrate right minus left with respect to $y$ . The limits are $y$ -values.
e.g. $x = y^2$ and $x = y + 2$ meet at $y = -1, 2$ : $\int_{-1}^{2} (y + 2 - y^2)\,dy = \tfrac92$
$\displaystyle A = \int_c^d \left[x_R(y) - x_L(y)\right]dy$
2.8.6
Finding Areas Between Curves that Intersect at More Than Two Points
Find every intersection. Between each pair, check which curve is on top, then add the separate areas so that no part cancels.
e.g. $y = x^3$ and $y = x$ meet at $x = -1, 0, 1$ : $\int_{-1}^0 (x^3 - x)\,dx + \int_0^1 (x - x^3)\,dx = \tfrac14 + \tfrac14 = \tfrac12$
$\displaystyle A = \int_a^b \lvert f(x) - g(x)\rvert\,dx$
2.8.7
The Arc Length of a Planar Curve
The length of a smooth curve $y = f(x)$ from $x = a$ to $x = b$ . Square the derivative, add 1, take the root, and integrate. If $x = g(y)$ , use the $y$ version.
e.g. $y = \tfrac23 x^{3/2}$ on $[0, 3]$ : $y' = \sqrt x$ , $L = \int_0^3 \sqrt{1 + x}\,dx = \tfrac23(8 - 1) = \tfrac{14}3$
$\displaystyle L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx$
$\displaystyle L = \int_c^d \sqrt{1 + [g'(y)]^2}\,dy$

2.9 Volumes of Solids With Known Cross Sections

2.9.1
Volumes of Solids with Square Cross Sections
Slice perpendicular to the axis. Each slice is a square whose side $s$ is the length across the base region at that point. Integrate the area of the square.
e.g. Base under $y = \sqrt{x}$ on $[0, 4]$ , squares perpendicular to the $x$ -axis: $s = \sqrt{x}$ , $V = \int_0^4 x\,dx = 8$
$\displaystyle A(x) = [s(x)]^2$
$\displaystyle V = \int_a^b [s(x)]^2\,dx$
2.9.2
Volumes of Solids with Rectangular Cross Sections
One side $s$ of each rectangle lies in the base region. The other side $h$ is given, often a constant or a multiple of $s$ . Integrate base times height.
e.g. Base under $y = x$ on $[0, 2]$ , height 3 times the base: $A = 3x^2$ , $V = \int_0^2 3x^2\,dx = 8$
$\displaystyle A(x) = s(x)\,h(x)$
$\displaystyle V = \int_a^b s(x)\,h(x)\,dx$
2.9.3
Volumes of Solids with Triangular Cross Sections
Find the side $s$ that lies in the base, then use the area formula for the stated triangle. Read carefully whether $s$ is a leg or the hypotenuse.
e.g. Base under $y = 2x$ on $[0, 3]$ , equilateral triangles: $A = \tfrac{\sqrt3}{4}(2x)^2 = \sqrt3\,x^2$ , $V = \sqrt3\int_0^3 x^2\,dx = 9\sqrt3$
$\displaystyle A = \tfrac{\sqrt3}{4}s^2 \;\;\text{equilateral}$
$\displaystyle A = \tfrac12 s^2 \;\;\text{isos. right, leg } s$
$\displaystyle A = \tfrac14 s^2 \;\;\text{isos. right, hyp. } s$
$\displaystyle V = \int_a^b A(x)\,dx$
2.9.4
Volumes of Solids with Circular Cross Sections
The length $s$ across the base is the diameter, so the radius is $s/2$ . For semicircles, take half the circle's area.
e.g. Diameter $s = x$ on $[0, 2]$ : $V = \int_0^2 \tfrac{\pi}{4}x^2\,dx = \tfrac{\pi}{4}\cdot\tfrac83 = \tfrac{2\pi}{3}$
$\displaystyle A = \pi\left(\tfrac{s}{2}\right)^2 = \tfrac{\pi}{4}s^2$
$\displaystyle A_{\text{semicircle}} = \tfrac{\pi}{8}s^2$
$\displaystyle V = \int_a^b A(x)\,dx$

2.10 Volumes of Revolution

Spinning the region under y = √x around the x-axis makes a solid built from thin disks of radius f(x). Adding up the disk volumes πr² dx gives the integral.

2.10.1
Volumes of Revolution Using the Disc Method: Rotation About the Coordinate Axes
Use when the region touches the axis of rotation, so each slice is a solid disc. Rotating about the $x$ -axis, integrate in $x$ . Rotating about the $y$ -axis, write the curve as $x = g(y)$ and integrate in $y$ .
e.g. $y = \sqrt{x}$ , $0 \le x \le 4$ , about the $x$ -axis: $V = \pi\int_0^4 x\,dx = 8\pi$
$\displaystyle V = \pi\int_a^b [f(x)]^2\,dx$
$\displaystyle V = \pi\int_c^d [g(y)]^2\,dy$
2.10.2
Volumes of Revolution Using the Disc Method: Rotation About Other Axes
The radius is the distance from the curve to the axis, not the curve's height. About $y = k$ , $R = \lvert f(x) - k\rvert$ and integrate in $x$ . About $x = h$ , $R = \lvert g(y) - h\rvert$ and integrate in $y$ .
e.g. Region between $y = x^2$ and $y = 1$ , about $y = 1$ : $R = 1 - x^2$ , $V = \pi\int_{-1}^{1}(1 - x^2)^2\,dx = \tfrac{16\pi}{15}$
$\displaystyle V = \pi\int_a^b [R(x)]^2\,dx$
$\displaystyle R(x) = \lvert f(x) - k\rvert$
2.10.3
Volumes of Revolution Using the Washer Method: Rotation About the Coordinate Axes
Use when there is a gap between the region and the axis. Each slice is a washer with outer radius $R$ and inner radius $r$ . Square each radius separately: $R^2 - r^2$ , not $(R - r)^2$ .
e.g. Between $y = x$ and $y = x^2$ on $[0, 1]$ , about the $x$ -axis: $V = \pi\int_0^1 (x^2 - x^4)\,dx = \pi\left(\tfrac13 - \tfrac15\right) = \tfrac{2\pi}{15}$
$\displaystyle V = \pi\int_a^b \left(R^2 - r^2\right)dx$
2.10.4
Volumes of Revolution Using the Washer Method: Rotation About Other Axes
Measure both radii from the axis of rotation. $R$ goes to the curve farther from the axis, $r$ to the nearer one. For a vertical axis $x = h$ , use $x$ in terms of $y$ and integrate in $y$ .
e.g. Between $y = x$ and $y = x^2$ on $[0, 1]$ , about $y = -1$ : $R = x + 1$ , $r = x^2 + 1$ , $V = \pi\int_0^1 (2x - x^2 - x^4)\,dx = \tfrac{7\pi}{15}$
$\displaystyle R = \lvert y_{\text{far}} - k\rvert$
$\displaystyle r = \lvert y_{\text{near}} - k\rvert$
$\displaystyle V = \pi\int_a^b \left(R^2 - r^2\right)dx$
2.10.5
The Shell Method: Rotating a Region About the X-Axis
Shells are horizontal, so integrate in $y$ . The radius is $y$ and the height is the region's width at that $y$ , with $x$ written as a function of $y$ .
e.g. Region under $y = \sqrt x$ , $0 \le x \le 4$ : $h = 4 - y^2$ , $V = 2\pi\int_0^2 y(4 - y^2)\,dy = 2\pi(8 - 4) = 8\pi$
$\displaystyle V = 2\pi\int_c^d y\,h(y)\,dy$
2.10.6
The Shell Method: Rotating a Region Between Two Curves About the X-Axis
Write both curves as $x$ in terms of $y$ . The shell height is right curve minus left curve, and the radius is $y$ (or $\lvert y - k \rvert$ about $y = k$ ).
e.g. Between $x = y$ and $x = y^2$ , $0 \le y \le 1$ : $V = 2\pi\int_0^1 y(y - y^2)\,dy = 2\pi\left(\tfrac13 - \tfrac14\right) = \tfrac{\pi}{6}$
$\displaystyle V = 2\pi\int_c^d y\,[x_R(y) - x_L(y)]\,dy$
2.10.7
The Shell Method: Rotation About the Y-Axis
Shells are vertical, so integrate in $x$ . The radius is $x$ (or $\lvert x - k \rvert$ about $x = k$ ) and the height is top curve minus bottom curve.
e.g. Under $y = x^2$ , $0 \le x \le 2$ : $V = 2\pi\int_0^2 x \cdot x^2\,dx = 2\pi(4) = 8\pi$
$\displaystyle V = 2\pi\int_a^b x\,[f(x) - g(x)]\,dx$

2.11 Surface Areas of Revolution

2.11.1
Surface Areas of Revolution: Rotation About the X-Axis
Each slice sweeps a band of radius $f(x)$ and slant length $ds$ . Requires $f(x) \ge 0$ and $f'$ continuous.
e.g. $f(x) = x$ on $[0, 1]$ (a cone): $S = 2\pi\int_0^1 x\sqrt2\,dx = \pi\sqrt2$
$\displaystyle S = 2\pi\int_a^b f(x)\sqrt{1 + [f'(x)]^2}\,dx$
2.11.2
Surface Areas of Revolution: Rotation About the Y-Axis
The radius of each band is now $x$ , the distance to the $y$ -axis. Requires $x \ge 0$ on $[a, b]$ .
e.g. $y = 2x$ on $[0, 1]$ : $S = 2\pi\int_0^1 x\sqrt5\,dx = \pi\sqrt5$
$\displaystyle S = 2\pi\int_a^b x\sqrt{1 + [f'(x)]^2}\,dx$
2.11.3
Surface Areas of Revolution for Parametric Curves
Use the parametric arc length element. About the $x$ -axis the radius is $y(t) \ge 0$ . About the $y$ -axis use $x(t)$ instead. The curve must be traced once.
e.g. $x = \cos t$ , $y = \sin t$ , $0 \le t \le \pi$ , about the $x$ -axis: $S = 2\pi\int_0^\pi \sin t\,dt = 4\pi$
$\displaystyle S = 2\pi\int_\alpha^\beta y(t)\,ds$
$\displaystyle ds = \sqrt{x'(t)^2 + y'(t)^2}\,dt$

2.12 Connecting Position, Velocity and Acceleration Using Integrals

2.12.1
Calculating Velocity Using Integration
Velocity is an antiderivative of acceleration. Use the initial velocity to find $C$ .
e.g. $a(t) = 6t$ , $v(0) = 2$ : $v(t) = 3t^2 + 2$
$\displaystyle v(t) = \int a(t)\,dt$
$\displaystyle v(b) = v(a) + \int_a^b a(t)\,dt$
2.12.2
Determining Characteristics of Moving Objects Using Integration
Integrate to get velocity or position, then read the motion. At rest when $v = 0$ . Moving right when $v > 0$ and left when $v < 0$ . Speeding up when $v$ and $a$ have the same sign, slowing down when opposite.
e.g. $a(t) = 2$ , $v(0) = -4$ : $v(t) = 2t - 4$ , at rest at $t = 2$ , moving left and slowing down before that
$\displaystyle \text{speeding up: } v, a \text{ same sign}$
$\displaystyle \text{slowing down: opposite signs}$
2.12.3
Calculating the Position Function of a Particle Using Integration
Position is an antiderivative of velocity. Use the initial position to find $C$ .
e.g. $v(t) = 4t - 3$ , $s(0) = 5$ : $s(t) = 2t^2 - 3t + 5$
$\displaystyle s(t) = \int v(t)\,dt$
$\displaystyle s(b) = s(a) + \int_a^b v(t)\,dt$
2.12.4
Calculating the Displacement of a Particle Using Integration
Displacement is the net change in position: the integral of velocity. It can be negative or zero, because motion in opposite directions cancels.
e.g. $v(t) = t - 2$ on $[0, 4]$ : $\int_0^4 (t - 2)\,dt = 8 - 8 = 0$
$\displaystyle \int_a^b v(t)\,dt = s(b) - s(a)$
2.12.5
Calculating the Total Distance Traveled by a Particle
Total distance is the integral of speed, which is never negative. Split the interval where $v$ changes sign and add the absolute values.
e.g. $v(t) = t - 2$ on $[0, 4]$ : $\int_0^2 (2 - t)\,dt + \int_2^4 (t - 2)\,dt = 2 + 2 = 4$
$\displaystyle \text{distance} = \int_a^b \lvert v(t)\rvert\,dt$
2.12.6
Average Position, Velocity, and Acceleration
Each is the average value of that function on $[a, b]$ . Average velocity is also change in position over time, and average acceleration is change in velocity over time.
e.g. $v(t) = 3t^2$ on $[0, 2]$ : average velocity $\tfrac12\int_0^2 3t^2\,dt = \tfrac12 \cdot 8 = 4$
$\displaystyle \bar v = \frac{1}{b-a}\int_a^b v(t)\,dt$
$\displaystyle = \frac{s(b) - s(a)}{b - a}$
$\displaystyle \bar a = \frac{v(b) - v(a)}{b - a}$
$\displaystyle \bar s = \frac{1}{b-a}\int_a^b s(t)\,dt$

Unit 3 23 topics

Parametric & Polar Equations

Parametric Equations, Vector-Valued Functions, The Planar Motion of a Particle, and Polar Coordinates.

3.13 Parametric Equations

A parametric curve gives x and y separately as functions of t. Plotting points in order of t shows the path and its direction, which an ordinary y = f(x) graph cannot show.

3.13.1
Differentiating Parametric Curves
For $x = x(t)$ , $y = y(t)$ , the slope is $dy/dt$ divided by $dx/dt$ . It is defined only where $dx/dt \ne 0$ .
e.g. $x = t^2$ , $y = t^3$ : $\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}$
$\displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
3.13.2
Calculating Tangent and Normal Lines with Parametric Equations
Plug $t_0$ into $x(t)$ and $y(t)$ to get the point, and into $dy/dx$ to get the slope $m$ . The normal line has slope $-1/m$ .
e.g. $x = t^2$ , $y = t^3$ at $t = 1$ : point $(1, 1)$ , $m = \tfrac32$ . Tangent $y - 1 = \tfrac32(x - 1)$ , normal $y - 1 = -\tfrac23(x - 1)$
$\displaystyle y - y_0 = m(x - x_0)$
$\displaystyle m = \frac{dy}{dx}\Big\rvert_{t = t_0}$
3.13.3
Second Derivatives of Parametric Equations
Differentiate $dy/dx$ with respect to $t$ , then divide by $dx/dt$ again. Do not just take $d^2y/dt^2$ over $d^2x/dt^2$ .
e.g. $x = t^2$ , $y = t^3$ : $\frac{dy}{dx} = \tfrac{3t}{2}$ , so $\frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t}$
$\displaystyle \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$
3.13.4
The Arc Length of a Parametric Curve
Integrate the speed $\sqrt{x'(t)^2 + y'(t)^2}$ over the interval. The curve must be traced exactly once on $[a, b]$ .
e.g. $x = 3\cos t$ , $y = 3\sin t$ , $0 \le t \le 2\pi$ : integrand $= 3$ , so $L = 6\pi$
$\displaystyle L = \int_a^b \sqrt{x'(t)^2 + y'(t)^2}\,dt$
3.13.5
Calculating Areas Bounded by Parametric Functions
Substitute $y = y(t)$ and $dx = x'(t)\,dt$ into $\int y\,dx$ . Set the $t$ -limits so $x$ runs left to right, or take the absolute value.
e.g. $x = t^2$ , $y = t^3$ , $0 \le t \le 1$ : $A = \int_0^1 t^3 (2t)\,dt = \tfrac25$
$\displaystyle A = \int_\alpha^\beta y(t)\,x'(t)\,dt$

3.14 Vector-Valued Functions

3.14.1
Defining Vector-Valued Functions
A function that takes a number $t$ and outputs a vector. As $t$ changes, the tip of the vector traces a curve. The domain is where every component is defined.
e.g. $\mathbf r(t) = \langle t, t^2 \rangle$ : $\mathbf r(2) = \langle 2, 4 \rangle$ , and the curve is $y = x^2$
$\displaystyle \mathbf r(t) = \langle x(t), y(t) \rangle$
$\displaystyle = x(t)\,\mathbf i + y(t)\,\mathbf j$
3.14.2
Differentiating Vector-Valued Functions
Differentiate each component separately. $\mathbf r'(t)$ is a tangent vector to the curve.
e.g. $\mathbf r(t) = \langle t^2, \sin t \rangle \Rightarrow \mathbf r'(t) = \langle 2t, \cos t \rangle$
$\displaystyle \mathbf r'(t) = \langle x'(t), y'(t) \rangle$
3.14.3
Integrating Vector-Valued Functions
Integrate each component separately. An indefinite integral gets a constant vector $\mathbf C$ , found from an initial condition.
e.g. $\int_0^1 \langle 2t, 3t^2 \rangle\,dt = \langle 1, 1 \rangle$
$\displaystyle \int \mathbf r(t)\,dt = \left\langle \int x\,dt, \int y\,dt \right\rangle$
$\displaystyle \qquad + \mathbf C$

3.15 The Planar Motion of a Particle

3.15.1
Calculating Velocity for Plane Motion Using Differentiation
Velocity is the derivative of position, component by component. Speed is the length of the velocity vector.
e.g. $\mathbf r(t) = \langle t^2, 3t \rangle$ : $\mathbf v(2) = \langle 4, 3 \rangle$ , speed $= 5$
$\displaystyle \mathbf v(t) = \langle x'(t), y'(t) \rangle$
$\displaystyle \text{speed} = \sqrt{x'(t)^2 + y'(t)^2}$
3.15.2
Calculating Acceleration for Plane Motion Using Differentiation
Acceleration is the derivative of velocity, the second derivative of position, component by component.
e.g. $\mathbf r(t) = \langle t^3, 2t^2 \rangle$ : $\mathbf v = \langle 3t^2, 4t \rangle$ , $\mathbf a = \langle 6t, 4 \rangle$ , $\mathbf a(1) = \langle 6, 4 \rangle$
$\displaystyle \mathbf a(t) = \mathbf v'(t)$
$\displaystyle = \langle x''(t), y''(t) \rangle$
3.15.3
Finding Velocity Vectors in Two Dimensions Using Integration
Integrate acceleration component by component. Use the initial velocity to find the constant in each component.
e.g. $\mathbf a = \langle 2, 6t \rangle$ , $\mathbf v(0) = \langle 1, -1 \rangle$ : $\mathbf v(t) = \langle 2t + 1, 3t^2 - 1 \rangle$
$\displaystyle \mathbf v(t) = \mathbf v(0) + \int_0^t \mathbf a(s)\,ds$
3.15.4
Finding Displacement Vectors in Two Dimensions Using Integration
Displacement is the integral of velocity, component by component. Add it to the starting position to get the ending position. Distance traveled is the integral of speed instead.
e.g. $\mathbf v = \langle 2t, 3t^2 \rangle$ on $[0, 2]$ : $\Delta\mathbf r = \langle 4, 8 \rangle$ . From $\langle 1, 1 \rangle$ it ends at $\langle 5, 9 \rangle$
$\displaystyle \Delta\mathbf r = \int_a^b \mathbf v(t)\,dt$
$\displaystyle \mathbf r(b) = \mathbf r(a) + \Delta\mathbf r$
3.15.5
Applying Newton's Second Law in the Plane
Force and acceleration are vectors: apply $F = ma$ to each component. Differentiate position twice to get force, or divide force by mass and integrate to get velocity and position.
e.g. $m = 3$ , $\mathbf r(t) = \langle t^3, 2t^2 \rangle$ : $\mathbf a = \langle 6t, 4 \rangle$ , so $\mathbf F(1) = \langle 18, 12 \rangle$
$\displaystyle \mathbf{F}(t) = m\,\mathbf{a}(t) = m\,\mathbf{r}''(t)$
$\displaystyle \langle F_x, F_y \rangle = \langle m x'', m y'' \rangle$

3.16 Polar Coordinates

A polar point is a distance r from the pole at an angle θ. The conversion formulas come straight from the right triangle with legs x and y and hypotenuse r.

3.16.1
Differentiating Curves Given in Polar Form
Treat $\theta$ as the parameter with $x = r\cos\theta$ and $y = r\sin\theta$ . Use the product rule on each, then divide.
e.g. $r = \theta$ at $\theta = \tfrac\pi2$ : $\frac{dy}{d\theta} = 1$ , $\frac{dx}{d\theta} = -\tfrac\pi2$ , so $\frac{dy}{dx} = -\tfrac2\pi$
$\displaystyle x = r\cos\theta,\; y = r\sin\theta$
$\displaystyle \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$
$\displaystyle = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}$
3.16.2
Further Differentiation of Curves Given in Polar Form
Write $x$ and $y$ as functions of $\theta$ first, simplify with trig identities, then differentiate. Evaluate $dy/dx$ at the given $\theta$ .
e.g. $r = 2\cos\theta$ : $x = 2\cos^2\theta$ , $y = \sin 2\theta$ , so $\frac{dy}{dx} = \frac{2\cos 2\theta}{-2\sin 2\theta} = -\cot 2\theta$ . At $\theta = \tfrac\pi8$ : $-1$
$\displaystyle x = f(\theta)\cos\theta$
$\displaystyle y = f(\theta)\sin\theta$
3.16.3
Horizontal and Vertical Tangents to Polar Curves
Horizontal tangent where $dy/d\theta = 0$ and $dx/d\theta \ne 0$ . Vertical tangent where $dx/d\theta = 0$ and $dy/d\theta \ne 0$ . Convert $\theta$ to the point $(r\cos\theta, r\sin\theta)$ .
e.g. $r = 2\cos\theta$ : horizontal at $\theta = \tfrac\pi4$ , point $(1, 1)$ . Vertical at $\theta = 0$ , point $(2, 0)$
$\displaystyle \text{Horiz: } \tfrac{dy}{d\theta} = 0,\; \tfrac{dx}{d\theta} \ne 0$
$\displaystyle \text{Vert: } \tfrac{dx}{d\theta} = 0,\; \tfrac{dy}{d\theta} \ne 0$
3.16.4
Horizontal and Vertical Tangents to Polar Curves in Non-Differentiable Cases
If $dy/d\theta$ and $dx/d\theta$ are both 0, the test is inconclusive. Find the limit of $dy/dx$ as $\theta \to \theta_0$ (factor or use L'Hôpital). A limit of 0 means horizontal; an infinite limit means vertical.
e.g. $r = 1 + \cos\theta$ at $\theta = \pi$ : both are 0, and $\frac{dy}{dx} = \frac{(2\cos\theta - 1)(1 + \cos\theta)}{-\sin\theta(1 + 2\cos\theta)} \to 0$ , so horizontal at the pole
$\displaystyle \lim_{\theta \to \theta_0} \frac{dy/d\theta}{dx/d\theta}$
3.16.5
Tangent and Normal Lines to Polar Curves
Find the point $(x_0, y_0) = (r\cos\theta_0, r\sin\theta_0)$ and the slope $m = dy/dx$ at $\theta_0$ . Use point-slope form. The normal slope is $-1/m$ .
e.g. $r = 2\sin\theta$ at $\theta = \tfrac\pi6$ : point $(\tfrac{\sqrt3}2, \tfrac12)$ , $m = \tan 2\theta = \sqrt3$ , tangent $y = \sqrt3x - 1$
$\displaystyle y - y_0 = m(x - x_0)$
3.16.6
Finding the Area of a Polar Region
The area swept out by $r = f(\theta)$ from $\theta = \alpha$ to $\theta = \beta$ . The region is a fan of thin sectors, not a strip under a curve.
e.g. $r = 2\cos\theta$ , $-\tfrac\pi2 \le \theta \le \tfrac\pi2$ : $A = \tfrac12\int 4\cos^2\theta\,d\theta = \pi$
$\displaystyle A = \frac12\int_\alpha^\beta r^2\,d\theta$
3.16.7
Finding the Limits of Integration For a Given Polar Region
Find the $\theta$ values where the region starts and ends, usually where $r = 0$ or where two curves meet. The interval should sweep the region exactly once.
e.g. One petal of $r = \cos 2\theta$ : $r = 0$ at $\theta = \pm\tfrac\pi4$ , so $A = \tfrac12\int_{-\pi/4}^{\pi/4} \cos^2 2\theta\,d\theta = \tfrac\pi8$
3.16.8
The Total Area Bounded by a Single Polar Curve
Integrate over one full trace of the curve, or find one piece and multiply by symmetry. Going past one trace counts area twice.
e.g. $r = \sin 3\theta$ traces once on $[0, \pi]$ : $A = \tfrac12\int_0^\pi \sin^2 3\theta\,d\theta = \tfrac\pi4$
$\displaystyle A = \frac12\int_\alpha^\beta r^2\,d\theta$
3.16.9
The Area Bounded by Two Polar Curves
Set the curves equal to find the limits. Subtract the squares of outer $R$ and inner $r$ , not the square of the difference.
e.g. Inside $r = 2\cos\theta$ , outside $r = 1$ : meet at $\theta = \pm\tfrac\pi3$ , $A = \tfrac12\int_{-\pi/3}^{\pi/3}(4\cos^2\theta - 1)\,d\theta = \tfrac\pi3 + \tfrac{\sqrt3}2$
$\displaystyle A = \frac12\int_\alpha^\beta \left(R^2 - r^2\right)d\theta$
3.16.10
The Arc Length of a Polar Curve
Add $r^2$ and $(dr/d\theta)^2$ , take the root, and integrate over one trace of the curve.
e.g. $r = 2\cos\theta$ , $0 \le \theta \le \pi$ : $r^2 + r'^2 = 4$ , so $L = \int_0^\pi 2\,d\theta = 2\pi$
$\displaystyle L = \int_\alpha^\beta \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta$

Unit 4 46 topics

Sequences & Series

Sequences, Monotonic Sequences, Infinite Series, and more.

4.17 Sequences

Arithmetic sequences grow by equal steps, like linear functions. Geometric sequences grow by equal factors, like exponential functions.

4.17.1
Limits of Sequences
A sequence converges to $L$ if its terms get arbitrarily close to $L$ as $n \to \infty$ . If there is no finite limit, it diverges. If $a_n = f(n)$ and $f(x) \to L$ , then $a_n \to L$ .
e.g. $a_n = \frac{3n + 1}{n + 2} \to 3$
$\displaystyle \lim_{x \to \infty} f(x) = L \;\Rightarrow\; \lim_{n \to \infty} a_n = L$
4.17.2
Convergence of Geometric Sequences
$r^n$ goes to 0 when $\lvert r \rvert < 1$ and stays 1 when $r = 1$ . It diverges when $r > 1$ or $r \le -1$ .
e.g. $\left(\tfrac23\right)^n \to 0$ . $(-1)^n$ diverges (it bounces between $-1$ and $1$ )
$\displaystyle \lim_{n\to\infty} r^n = 0 \text{ if } \lvert r \rvert < 1$
$\displaystyle \text{converges} \iff -1 < r \le 1$
4.17.3
Further Convergence of Geometric Sequences
Combine the powers to rewrite the term as $a \cdot r^n$ , then apply the geometric rule to $r$ .
e.g. $\frac{2^{n+1}}{5^n} = 2\left(\tfrac25\right)^n \to 0$ . $\left(\tfrac x2\right)^n$ converges for $-2 < x \le 2$
$\displaystyle \frac{b^{n+k}}{c^n} = b^k\left(\frac{b}{c}\right)^n$
4.17.4
Limits of Sequences With Factorials
$n!$ grows faster than any $b^n$ , so $b^n/n! \to 0$ . For ratios of factorials, cancel first: $(n+1)! = (n+1)\,n!$ .
e.g. $\frac{(n+1)!}{n \cdot n!} = \frac{n+1}{n} \to 1$ , and $\frac{5^n}{n!} \to 0$
$\displaystyle (n+1)! = (n+1) \cdot n!$
$\displaystyle \lim_{n\to\infty} \frac{b^n}{n!} = 0$
4.17.5
Determining Limits of Sequences Using Relative Magnitudes
As $n \to \infty$ , these grow in this order ( $p > 0$ , $b > 1$ ). A slower one over a faster one goes to 0.
e.g. $\frac{n^{10}}{2^n} \to 0$ and $\frac{\ln n}{\sqrt n} \to 0$
$\displaystyle \ln n \ll n^p \ll b^n \ll n! \ll n^n$
4.17.6
Further Determining Limits of Sequences Using Relative Magnitudes
In a sum, the fastest-growing term dominates. Keep the dominant term on top and bottom, then take the limit.
e.g. $\frac{2^n + n^3}{3 \cdot 2^n - n} \approx \frac{2^n}{3 \cdot 2^n} \to \tfrac13$

4.18 Monotonic Sequences

4.18.1
Monotonic Sequences
Increasing if $a_{n+1} \ge a_n$ for all $n$ , decreasing if $a_{n+1} \le a_n$ . Monotonic means one or the other. A monotonic sequence that is bounded converges.
e.g. $a_n = \frac{n}{n+1}$ : $a_{n+1} - a_n = \frac{1}{(n+1)(n+2)} > 0$ , so increasing
$\displaystyle a_{n+1} - a_n > 0 \;\Rightarrow\; \text{increasing}$
$\displaystyle a_{n+1} - a_n < 0 \;\Rightarrow\; \text{decreasing}$
4.18.2
Identifying Monotonic Sequences Using Differentiation
Let $f(n) = a_n$ . If $f'(x) > 0$ for $x \ge k$ , the sequence increases from $n = k$ on. If $f'(x) < 0$ , it decreases.
e.g. $a_n = \frac{\ln n}{n}$ : $f'(x) = \frac{1 - \ln x}{x^2} < 0$ for $x > e$ , so decreasing for $n \ge 3$
$\displaystyle f'(x) > 0 \;\Rightarrow\; \text{increasing}$
$\displaystyle f'(x) < 0 \;\Rightarrow\; \text{decreasing}$
4.18.3
Identifying Monotonic Sequences Using Ratios
For positive terms, compare $a_{n+1}/a_n$ to 1. Best for factorials and powers.
e.g. $a_n = \frac{2^n}{n!}$ : ratio $= \frac{2}{n+1} < 1$ for $n \ge 2$ , so decreasing from $n = 2$ on
$\displaystyle \frac{a_{n+1}}{a_n} > 1 \;\Rightarrow\; \text{increasing}$
$\displaystyle \frac{a_{n+1}}{a_n} < 1 \;\Rightarrow\; \text{decreasing}$

4.19 Infinite Series

The partial sums 1/2, 3/4, 7/8, … climb toward 1 but never pass it, so the infinite series converges to 1. A geometric series converges exactly when |r| < 1.

4.19.1
Infinite Series and Partial Sums
A series adds the terms of a sequence. The $n$ th partial sum $S_n$ adds the first $n$ terms. The sum of the series is the limit of $S_n$ .
e.g. $\sum_{n=1}^\infty \tfrac{1}{2^n}$ : $S_3 = \tfrac78$ , $S_n = 1 - \tfrac{1}{2^n} \to 1$
$\displaystyle S_n = a_1 + a_2 + \cdots + a_n$
$\displaystyle \sum_{n=1}^\infty a_n = \lim_{n\to\infty} S_n$
4.19.2
Convergent and Divergent Infinite Series
A series converges if $S_n$ has a finite limit and diverges otherwise. nth-term test: if $a_n \not\to 0$ , the series diverges. If $a_n \to 0$ , the test says nothing (the harmonic series $\sum \tfrac1n$ diverges).
e.g. $\sum \frac{n}{n+1}$ diverges because the terms go to 1
$\displaystyle \lim_{n\to\infty} a_n \ne 0 \;\Rightarrow\; \text{diverges}$
4.19.3
Properties of Infinite Series
If $\sum a_n = A$ and $\sum b_n = B$ both converge, sums and constant multiples work term by term.
e.g. $\sum_{n=1}^\infty \left(\tfrac{3}{2^n} + \tfrac{1}{3^n}\right) = 3(1) + \tfrac12 = \tfrac72$
$\displaystyle \sum (a_n \pm b_n) = A \pm B$
$\displaystyle \sum c\,a_n = cA$
4.19.4
Further Properties of Infinite Series
Adding or removing finitely many terms does not change whether a series converges, only its sum. Convergent plus divergent diverges. A nonzero multiple of a divergent series diverges.
e.g. $\sum_{n=3}^\infty \tfrac{1}{2^n} = 1 - \tfrac12 - \tfrac14 = \tfrac14$
$\displaystyle \sum_{n=1}^\infty a_n = a_1 + \sum_{n=2}^\infty a_n$
4.19.5
Telescoping Series
Write each term as a difference $b_n - b_{n+1}$ (often with partial fractions). The middle terms cancel, leaving $S_n = b_1 - b_{n+1}$ .
e.g. $\sum_{n=1}^\infty \frac{1}{n(n+1)} = \sum \left(\tfrac1n - \tfrac1{n+1}\right)$ , $S_n = 1 - \tfrac1{n+1} \to 1$
$\displaystyle S_n = b_1 - b_{n+1}$
$\displaystyle \text{sum} = b_1 - \lim_{n\to\infty} b_{n+1}$

4.20 Geometric Series

4.20.1
Finding the Sum of an Infinite Geometric Series
With first term $a$ and ratio $r$ , the series converges only when $\lvert r \rvert < 1$ .
e.g. $6 + 3 + \tfrac32 + \cdots = \frac{6}{1 - 1/2} = 12$
$\displaystyle a + ar + ar^2 + \cdots = \frac{a}{1 - r}$
4.20.2
Writing an Infinite Geometric Series in Sigma Notation
Find the first term $a$ and the ratio $r$ . Starting at $n = 0$ uses $r^n$ ; starting at $n = 1$ uses $r^{n-1}$ .
e.g. $5 - \tfrac53 + \tfrac59 - \cdots = \sum_{n=0}^\infty 5\left(-\tfrac13\right)^n$
$\displaystyle \sum_{n=0}^\infty a r^n = \sum_{n=1}^\infty a r^{n-1}$
4.20.3
Sums of Infinite Geometric Series Given in Sigma Notation
Plug in the starting index to get the first term. The base of the power is $r$ . If $\lvert r \rvert < 1$ , the sum is first term over $1 - r$ ; otherwise the series diverges.
e.g. $\sum_{n=2}^\infty 3\left(\tfrac12\right)^n$ : first term $\tfrac34$ , sum $= \frac{3/4}{1/2} = \tfrac32$
$\displaystyle \sum_{n=k}^\infty a r^n = \frac{a r^k}{1 - r}$
4.20.4
Convergence of Geometric Series
A geometric series with $a \ne 0$ converges when $\lvert r \rvert < 1$ and diverges when $\lvert r \rvert \ge 1$ . If $r$ contains $x$ , solve $\lvert r \rvert < 1$ for $x$ .
e.g. $\sum_{n=0}^\infty \left(\tfrac x3\right)^n$ converges for $-3 < x < 3$ , with sum $\frac{3}{3 - x}$
$\displaystyle \lvert r \rvert < 1 \iff \text{converges}$
4.20.5
Repeating Decimals as Infinite Geometric Series
A block of $k$ repeating digits gives first term $a = \text{block}/10^k$ and ratio $r = 1/10^k$ . Sum it as a geometric series.
e.g. $0.\overline{36} = \frac{0.36}{1 - 0.01} = \frac{36}{99} = \frac{4}{11}$
$\displaystyle a = \frac{\text{block}}{10^k},\quad r = \frac{1}{10^k}$
$\displaystyle \text{value} = \frac{a}{1 - r}$

4.21 Infinite Series Convergence Tests

4.21.1
The Nth Term Test for Divergence
If the terms do not approach 0, the series diverges. If $\lim a_n = 0$ , the test tells you nothing and you need another test.
e.g. $\sum \frac{n}{2n+1}$ : terms $\to \tfrac12 \ne 0$ , so it diverges
$\displaystyle \lim_{n\to\infty} a_n \ne 0$
$\displaystyle \Rightarrow\; \textstyle\sum a_n \text{ diverges}$
4.21.2
The Integral Test
If $f$ is positive, continuous, and decreasing for $x \ge 1$ with $f(n) = a_n$ , the series and the integral both converge or both diverge. The sum does not equal the integral.
e.g. $\sum \frac{1}{n^2}$ : $\int_1^\infty x^{-2}\,dx = 1$ , so the series converges
$\displaystyle \textstyle\sum_{n=1}^{\infty} a_n \;\text{and}\; \int_1^{\infty} f(x)\,dx$
$\displaystyle \text{converge or diverge together}$
4.21.3
The Remainder Estimate for the Integral Test
If $a_k = f(k)$ with $f$ positive, continuous, and decreasing, the error $R_n = s - s_n$ after $n$ terms is trapped between two improper integrals.
e.g. $\sum \tfrac{1}{k^2}$ , $n = 10$ : $\int_{11}^\infty \tfrac{dx}{x^2} = \tfrac{1}{11}$ and $\int_{10}^\infty \tfrac{dx}{x^2} = \tfrac{1}{10}$ , so $\tfrac{1}{11} \le R_{10} \le \tfrac{1}{10}$
$\displaystyle \int_{n+1}^{\infty} f(x)\,dx \le R_n$
$\displaystyle R_n \le \int_n^{\infty} f(x)\,dx$
4.21.4
Harmonic Series and p-Series
A $p$ -series converges only when $p > 1$ . The harmonic series is the case $p = 1$ , and it diverges.
e.g. $\sum \frac{1}{\sqrt n}$ : $p = \tfrac12 \le 1$ , so it diverges
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^p}$
$\displaystyle p > 1:\ \text{converges}$
$\displaystyle p \le 1:\ \text{diverges}$
4.21.5
The Comparison Test
For positive terms with $a_n \le b_n$ : a smaller series under a convergent one converges, and a larger series over a divergent one diverges.
e.g. $\frac{1}{n^2+1} < \frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges, so $\sum \frac{1}{n^2+1}$ converges
$\displaystyle 0 \le a_n \le b_n$
$\displaystyle \textstyle\sum b_n \text{ conv.} \Rightarrow \sum a_n \text{ conv.}$
$\displaystyle \textstyle\sum a_n \text{ div.} \Rightarrow \sum b_n \text{ div.}$
4.21.6
The Limit Comparison Test
For positive terms, compare to a known series $b_n$ with the same dominant behavior. If the limit is a positive finite number, both series do the same thing.
e.g. $a_n = \frac{3n+1}{n^3+2}$ , $b_n = \frac{1}{n^2}$ : $L = 3$ , and $\sum \frac{1}{n^2}$ converges, so $\sum a_n$ converges
$\displaystyle \lim_{n\to\infty} \frac{a_n}{b_n} = L$
$\displaystyle 0 < L < \infty \Rightarrow \text{same result}$
4.21.7
The Alternating Series Test
An alternating series $\sum (-1)^n b_n$ with $b_n > 0$ converges if $b_n$ is decreasing and $b_n \to 0$ . The test can only show convergence, never divergence.
e.g. $\sum \frac{(-1)^{n+1}}{n}$ : $\frac1n$ decreases to 0, so it converges
$\displaystyle b_{n+1} \le b_n \;\text{and}\; \lim_{n\to\infty} b_n = 0$
4.21.8
The Ratio Test
Take the limit of the absolute ratio of consecutive terms. Best for factorials and powers like $2^n$ . If $L = 1$ the test is inconclusive.
e.g. $\sum \frac{2^n}{n!}$ : ratio $= \frac{2}{n+1} \to 0 < 1$ , so it converges
$\displaystyle L = \lim_{n\to\infty} \left\lvert \frac{a_{n+1}}{a_n} \right\rvert$
$\displaystyle L < 1:\ \text{converges absolutely}$
$\displaystyle L > 1:\ \text{diverges}$
$\displaystyle L = 1:\ \text{inconclusive}$
4.21.9
The Root Test
Find $L$ . If $L < 1$ the series converges absolutely. If $L > 1$ or $L = \infty$ it diverges. If $L = 1$ the test says nothing. Best when $a_n$ has an $n$ th power.
e.g. $\sum \left(\tfrac{n}{2n+1}\right)^n$ : $L = \lim \tfrac{n}{2n+1} = \tfrac12 < 1$ , so it converges
$\displaystyle L = \lim_{n \to \infty} \sqrt[n]{\lvert a_n \rvert}$
4.21.10
Absolute and Conditional Convergence
If $\sum \lvert a_n \rvert$ converges, the series converges absolutely, and so it converges. If $\sum a_n$ converges but $\sum \lvert a_n \rvert$ diverges, it converges conditionally.
e.g. $\sum \frac{(-1)^n}{n}$ converges conditionally; $\sum \frac{(-1)^n}{n^2}$ converges absolutely
$\displaystyle \textstyle\sum \lvert a_n \rvert \text{ conv.} \Rightarrow \sum a_n \text{ conv.}$
4.21.11
The Alternating Series Error Bound
For a series that passes the alternating series test, stopping after $n$ terms gives an error no bigger than the first term left out.
e.g. $\sum \frac{(-1)^{n+1}}{n^2}$ : $S_3 = 1 - \tfrac14 + \tfrac19 = \tfrac{31}{36}$ , error $\le \tfrac{1}{16}$
$\displaystyle \lvert S - S_n \rvert \le b_{n+1}$
4.21.12
Determining Convergence Parameters for Infinite Series
Apply a test with the parameter left in, then solve the inequality the test requires. Check boundary values separately.
e.g. $\sum \frac{1}{n^{3k-2}}$ converges when $3k - 2 > 1$ , so $k > 1$
$\displaystyle \textstyle\sum ar^n \text{ converges} \iff \lvert r \rvert < 1$
$\displaystyle \textstyle\sum \frac{1}{n^p} \text{ converges} \iff p > 1$
4.21.13
Selecting Procedures for Analyzing Infinite Series
Try the $n$ th term test first. Spot geometric and $p$ -series on sight, use the ratio test for factorials or powers like $3^n$ , and the alternating series test for alternating signs. Use (limit) comparison with a $p$ -series for rational or root expressions, and the integral test when $f(x)$ is easy to integrate.
e.g. $\sum \frac{n^2}{3^n}$ : ratio test gives $L = \tfrac13 < 1$ , so it converges

4.22 Taylor Polynomials

Each Taylor polynomial matches sin x at 0 and then drifts away. Adding higher-degree terms keeps it on the curve over a wider stretch around 0.

4.22.1
Second-Degree Taylor Polynomials
Approximates $f$ near $x = a$ by matching its value, slope, and concavity at $a$ .
e.g. $e^x$ at $a = 0$ : $P_2(x) = 1 + x + \tfrac{x^2}{2}$
$\displaystyle P_2(x) = f(a) + f'(a)(x-a)$
$\displaystyle + \frac{f''(a)}{2}(x-a)^2$
4.22.2
Analyzing Second-Degree Taylor Polynomials
$P_2$ and $f$ share the same value, first derivative, and second derivative at $a$ . The coefficient of $(x-a)^2$ is half of $f''(a)$ , so double it to recover $f''(a)$ .
e.g. $P_2 = 3 - 2(x-1) + 5(x-1)^2$ : $f(1) = 3$ , $f'(1) = -2$ , $f''(1) = 10$ , concave up at $x = 1$
$\displaystyle P_2^{(k)}(a) = f^{(k)}(a),\ k = 0, 1, 2$
$\displaystyle c_2 = \frac{f''(a)}{2}$
4.22.3
Third-Degree Taylor Polynomials
Add the cubic term to $P_2$ . Its coefficient is the third derivative divided by $3! = 6$ .
e.g. $\sin x$ at $a = 0$ : $P_3(x) = x - \tfrac{x^3}{6}$
$\displaystyle P_3(x) = P_2(x)$
$\displaystyle + \frac{f'''(a)}{6}(x-a)^3$
4.22.4
Higher-Degree Taylor Polynomials
Each term uses the $k$ th derivative at $a$ divided by $k!$ . Higher degree usually means a better approximation near $a$ .
e.g. $e^x$ at $0$ : $P_4 = 1 + x + \tfrac{x^2}{2} + \tfrac{x^3}{6} + \tfrac{x^4}{24}$
$\displaystyle P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$
4.22.5
The Lagrange Error Bound
Bounds the error of $P_n$ at $x$ . $M$ is the largest value of $\lvert f^{(n+1)} \rvert$ between $a$ and $x$ .
e.g. $\cos 0.2 \approx 1 - \tfrac{0.04}{2} = 0.98$ ; $M = 1$ , so error $\le \tfrac{0.2^3}{3!} \approx 0.0013$
$\displaystyle \lvert f(x) - P_n(x) \rvert$
$\displaystyle \le \frac{M\,\lvert x-a \rvert^{n+1}}{(n+1)!}$

4.23 Taylor Series

4.23.1
Radius of Convergence of Power Series Centered at the Origin
$\sum c_n x^n$ converges for $\lvert x \rvert < R$ and diverges for $\lvert x \rvert > R$ . Find $R$ with the ratio test, and check the endpoints $x = \pm R$ separately.
e.g. $\sum \frac{x^n}{3^n}$ : ratio $\to \frac{\lvert x \rvert}{3} < 1$ , so $R = 3$
$\displaystyle \lim_{n\to\infty} \left\lvert \frac{c_{n+1}x^{n+1}}{c_n x^n} \right\rvert < 1$
4.23.2
Radius of Convergence of Power Series
A series centered at $a$ converges for $\lvert x - a \rvert < R$ . The interval is $(a - R, a + R)$ plus any endpoints that pass a separate check. $R = 0$ means only $x = a$ ; $R = \infty$ means all $x$ .
e.g. $\sum \frac{(x-2)^n}{n}$ : $R = 1$ ; at $x = 3$ it diverges, at $x = 1$ it converges, so the interval is $[1, 3)$
$\displaystyle \textstyle\sum c_n (x-a)^n$
$\displaystyle \text{converges for } \lvert x - a \rvert < R$
4.23.3
Maclaurin Series
A Taylor series centered at $0$ .
e.g. $e^x = 1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots$ for all $x$
$\displaystyle f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$
4.23.4
Taylor Series
The infinite version of the Taylor polynomial, centered at $x = a$ . It equals $f(x)$ only inside its interval of convergence.
e.g. $\frac1x$ at $a = 1$ : $f^{(n)}(1) = (-1)^n n!$ , so $\frac1x = \sum (-1)^n (x-1)^n$ for $0 < x < 2$
$\displaystyle f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$
4.23.5
Representing Functions as Power Series
Start from the geometric series. Substitute for $x$ , multiply by powers of $x$ , or differentiate or integrate term by term, and track the new interval.
e.g. $\frac{1}{1+x^2} = \sum (-1)^n x^{2n}$ for $\lvert x \rvert < 1$
$\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$
$\displaystyle \lvert x \rvert < 1$
4.23.6
Recognizing Standard Maclaurin Series
Match a given series to a known one, then evaluate that function. The $\ln(1+x)$ series holds for $-1 < x \le 1$ .
e.g. $\sum_{n=0}^{\infty} \frac{2^n}{n!} = e^2$
$\displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$
$\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n,\ \lvert x \rvert < 1$
$\displaystyle \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n}$
4.23.7
Recognizing Standard Maclaurin Series for Trigonometric Functions
Sine uses odd powers and cosine uses even powers, with alternating signs. Both hold for all $x$ .
e.g. $\sum_{n=0}^{\infty} \frac{(-1)^n \pi^{2n}}{(2n)!} = \cos \pi = -1$
$\displaystyle \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$
$\displaystyle \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$
4.23.8
Differentiating Taylor Series
Differentiate term by term. The radius of convergence stays the same, but recheck the endpoints.
e.g. $\frac{d}{dx}\left(x - \tfrac{x^3}{3!} + \tfrac{x^5}{5!} - \cdots\right) = 1 - \tfrac{x^2}{2!} + \tfrac{x^4}{4!} - \cdots = \cos x$
$\displaystyle \frac{d}{dx}\sum c_n x^n = \sum n c_n x^{n-1}$
4.23.9
Approximating Integrals Using Taylor Series
Replace the integrand with its series, integrate term by term, and add the first few terms. If the result alternates, the first omitted term bounds the error.
e.g. $\int_0^1 e^{-x^2}dx = \int_0^1 \left(1 - x^2 + \tfrac{x^4}{2} - \cdots\right)dx \approx 1 - \tfrac13 + \tfrac{1}{10} = \tfrac{23}{30}$
$\displaystyle \int_0^b \sum c_n x^n\,dx$
$\displaystyle = \sum \frac{c_n b^{n+1}}{n+1}$

Unit 5 23 topics

Differential Equations

Introduction to Differential Equations, Qualitative Techniques for Differential Equations, Modeling Exponential Growth and Decay With Differential Equations, and more.

5.24 Introduction to Differential Equations

5.24.1
Introduction to Differential Equations
A differential equation relates a function to its derivatives. Its order is the highest derivative that appears. A solution is a function, not a number.
e.g. $\frac{dy}{dx} = 2x$ is first order; $y'' + y = 0$ is second order
5.24.2
Verifying Solutions of Differential Equations
Compute the needed derivatives of the proposed function and substitute them into the equation. It is a solution only if both sides are equal for every $x$ .
e.g. $y = e^{3x}$ solves $y' = 3y$ , since $y' = 3e^{3x} = 3y$
5.24.3
Solving First-Order ODEs Using Direct Integration
If the derivative depends only on $x$ , integrate both sides. Keep the $+ C$ : it gives a whole family of solutions.
e.g. $y' = 6x^2 \Rightarrow y = 2x^3 + C$
$\displaystyle \frac{dy}{dx} = f(x)$
$\displaystyle \Rightarrow\; y = \int f(x)\,dx + C$
5.24.4
Solving First-Order ODEs Using Separation of Variables
Move all $y$ terms (with $dy$ ) to one side and all $x$ terms (with $dx$ ) to the other, then integrate both sides. One $+ C$ on the $x$ side is enough. Solve for $y$ if you can.
e.g. $\frac{dy}{dx} = xy$ : $\ln\lvert y\rvert = \tfrac{x^2}{2} + C$ , so $y = Ae^{x^2/2}$
$\displaystyle \frac{dy}{dx} = g(x)\,h(y)$
$\displaystyle \Rightarrow\; \int \frac{dy}{h(y)} = \int g(x)\,dx$
5.24.5
Solving First-Order IVPs Using Separation of Variables
Separate and integrate, then plug in the initial condition to find $C$ . Use it also to pick the right sign when you take a square root.
e.g. $\frac{dy}{dx} = \frac{x}{y}$ , $y(0) = 2$ : $\tfrac{y^2}{2} = \tfrac{x^2}{2} + 2$ , so $y = \sqrt{x^2 + 4}$
5.24.6
Modeling With First-Order ODEs
Translate the words into a rate equation. "Proportional to" means multiply by a constant $k$ . Use a negative sign for a quantity that is decreasing.
e.g. Water drains at a rate proportional to the square root of the depth $h$ : $\frac{dh}{dt} = -k\sqrt{h}$
$\displaystyle \frac{dy}{dt} = ky \quad \text{(proportional to } y\text{)}$
5.24.7
Further Modeling With First-Order ODEs
In Newton's law of cooling, the rate of temperature change is proportional to the gap between the object and its surroundings. For mixing or flow problems, the net rate is rate in minus rate out.
e.g. Coffee at 180°F in a 70°F room: $\frac{dT}{dt} = -k(T - 70)$ with $k > 0$
$\displaystyle \frac{dT}{dt} = -k(T - T_s)$
$\displaystyle \frac{dA}{dt} = \text{rate in} - \text{rate out}$

5.25 Qualitative Techniques for Differential Equations

5.25.1
Qualitative Analysis of First-Order ODEs
Read behavior from $\frac{dy}{dx}$ without solving. Where it is positive, solutions increase; where negative, they decrease. Differentiate $\frac{dy}{dx}$ implicitly to get $\frac{d^2y}{dx^2}$ for concavity.
e.g. $\frac{dy}{dx} = y(3 - y)$ : solutions increase for $0 < y < 3$ and decrease for $y > 3$ or $y < 0$
$\displaystyle \frac{dy}{dx} > 0 \Rightarrow y \text{ increasing}$
$\displaystyle \frac{d^2y}{dx^2} > 0 \Rightarrow \text{concave up}$
5.25.2
Equilibrium Solutions of First-Order ODEs
An equilibrium is a constant solution $y = c$ , where $\frac{dy}{dx} = 0$ for every $x$ . It is stable if nearby solutions move toward it and unstable if they move away.
e.g. $\frac{dy}{dx} = y(3 - y)$ : $y = 0$ is unstable and $y = 3$ is stable
$\displaystyle \frac{dy}{dx} = f(y),\; f(c) = 0$
$\displaystyle \Rightarrow\; y = c \text{ is an equilibrium}$

5.26 Modeling Exponential Growth and Decay With Differential Equations

Equal differences mean linear; equal ratios mean exponential. Any exponential growth eventually passes any linear growth.

5.26.1
Exponential Growth and Decay Models With First-Order ODEs
If the rate is proportional to the amount, the solution is exponential. $k > 0$ means growth and $k < 0$ means decay. $y_0$ is the amount at $t = 0$ .
e.g. $\frac{dy}{dt} = 0.05y$ , $y(0) = 200$ : $y = 200e^{0.05t}$
$\displaystyle \frac{dy}{dt} = ky \;\Rightarrow\; y = y_0 e^{kt}$
5.26.2
Exponential Growth and Decay Models With First-Order ODEs: Calculating Unknown Times
Set the model equal to the target amount, divide by $y_0$ , and take ln of both sides.
e.g. $200e^{0.05t} = 400 \Rightarrow t = \tfrac{\ln 2}{0.05} \approx 13.9$
$\displaystyle y_0 e^{kt} = y$
$\displaystyle \Rightarrow\; t = \frac1k \ln\frac{y}{y_0}$
5.26.3
Exponential Growth and Decay Models With First-Order ODEs: Half-Life Problems
The half-life $T$ is the time for the amount to halve. It fixes the decay constant $k$ .
e.g. 64 g with a 5-day half-life: after 15 days, $64\left(\tfrac12\right)^3 = 8$ g
$\displaystyle k = -\frac{\ln 2}{T}$
$\displaystyle y = y_0 e^{kt} = y_0\left(\tfrac12\right)^{t/T}$

5.27 Modeling Logistic Growth With Differential Equations

Logistic growth starts out nearly exponential, is fastest at half the carrying capacity, and levels off as P approaches L.

5.27.1
Logistic Growth Models With First-Order ODEs
Growth is nearly exponential when $P$ is small and slows as $P$ nears the carrying capacity $L$ . $k > 0$ is the growth constant. In the form $kP(L - P)$ , $L$ is still the carrying capacity.
e.g. $\tfrac{dP}{dt} = 0.2P\left(1 - \tfrac{P}{500}\right)$ : $L = 500$ ; at $P = 100$ , $\tfrac{dP}{dt} = 0.2(100)(0.8) = 16$
$\displaystyle \frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right)$
5.27.2
Qualitative Analysis of the Logistic Growth Equation
Equilibria are $P = 0$ and $P = L$ . For $0 < P < L$ , $P$ increases; for $P > L$ , it decreases. Either way $P \to L$ . Growth is fastest at $P = \tfrac{L}{2}$ , where the graph changes from concave up to concave down.
e.g. $\tfrac{dP}{dt} = 0.3P\left(1 - \tfrac{P}{80}\right)$ , $P(0) = 10$ : increases toward 80, growing fastest at $P = 40$
$\displaystyle \lim_{t \to \infty}P(t) = L \quad (P_0 > 0)$
$\displaystyle \text{fastest growth at } P = \tfrac{L}{2}$
5.27.3
Solving the Logistic Growth Equation
Separate variables and use partial fractions, or use the known solution. Find $A$ from the initial value $P_0$ .
e.g. $L = 500$ , $P_0 = 100$ , $k = 0.2$ : $A = 4$ , $P(t) = \tfrac{500}{1 + 4e^{-0.2t}}$ ; $P = 250$ when $t = 5\ln 4 \approx 6.93$
$\displaystyle P(t) = \frac{L}{1 + Ae^{-kt}}$
$\displaystyle A = \frac{L - P_0}{P_0}$

5.28 Slope Fields

Each short dash shows the slope the differential equation demands at that point. A solution curve is one that runs along the dashes everywhere it goes.

5.28.1
Slope Fields for Directly Integrable Differential Equations
For $\frac{dy}{dx} = f(x)$ , the slope depends only on $x$ . All segments in the same vertical column are parallel, and solution curves are vertical shifts of each other.
e.g. $\frac{dy}{dx} = 2x$ : slope $0$ along $x = 0$ , slope $2$ along $x = 1$ ; solutions are $y = x^2 + C$
5.28.2
Slope Fields for Autonomous Differential Equations
For $\frac{dy}{dx} = f(y)$ , the slope depends only on $y$ . All segments in the same horizontal row are parallel, and solution curves are horizontal shifts of each other. Rows where $f(y) = 0$ are flat.
e.g. $\frac{dy}{dx} = y - 1$ : slope $0$ along $y = 1$ , slope $1$ along $y = 2$ , slope $-1$ along $y = 0$
5.28.3
Slope Fields for Nonautonomous Differential Equations
For $\frac{dy}{dx} = f(x, y)$ , plug each grid point into $f$ to get its slope. The curve where $f(x, y) = 0$ has horizontal segments.
e.g. $\frac{dy}{dx} = x - y$ : slope $0$ at $(1, 1)$ , $2$ at $(2, 0)$ , $-1$ at $(0, 1)$ ; flat along $y = x$
5.28.4
Analyzing Slope Fields for Directly Integrable Differential Equations
If segments in each vertical column are parallel, the equation has the form $\frac{dy}{dx} = f(x)$ . Note where slopes are zero, positive, and negative to identify $f$ .
e.g. Slope $0$ on the $y$ -axis, positive for $x > 0$ , negative for $x < 0$ : fits $\frac{dy}{dx} = x$
5.28.5
Analyzing Slope Fields for Autonomous Differential Equations
If segments in each horizontal row are parallel, the equation has the form $\frac{dy}{dx} = f(y)$ . Flat rows are equilibria. Arrows pointing toward a flat row mark it stable.
e.g. Flat along $y = 2$ , rising below it, falling above it: fits $\frac{dy}{dx} = 2 - y$ , and solutions approach $y = 2$
5.28.6
Analyzing Slope Fields for Nonautonomous Differential Equations
When slopes change in both directions, test a few points in each candidate equation. Look first for the line or curve of zero slopes and the sign of the slope on each side of it.
e.g. Flat along $y = -x$ and rising above it: fits $\frac{dy}{dx} = x + y$

5.29 Numerical Solutions of Differential Equations

Euler's method walks along the tangent line for one step, recomputes the slope, and repeats. On a curve that bends upward like eˣ, each step falls a little further below the true solution.

5.29.1
Euler's Method: Calculating One Step
Approximate the solution of $\tfrac{dy}{dx} = f(x, y)$ by following the tangent line for one step of size $h$ .
e.g. $\tfrac{dy}{dx} = x + y$ , $y(0) = 1$ , $h = 0.1$ : $y(0.1) \approx 1 + 0.1(0 + 1) = 1.1$
$\displaystyle x_1 = x_0 + h$
$\displaystyle y_1 = y_0 + h\,f(x_0, y_0)$
5.29.2
Euler's Method: Calculating Multiple Steps
Repeat the one-step rule, using each new point to compute the next slope. Smaller $h$ gives better accuracy. If the solution is concave up, Euler underestimates; if concave down, it overestimates.
e.g. $\tfrac{dy}{dx} = y$ , $y(0) = 1$ , $h = 0.5$ : $y(0.5) \approx 1.5$ , $y(1) \approx 1.5 + 0.5(1.5) = 2.25$ (true value $e \approx 2.718$ )
$\displaystyle y_{n+1} = y_n + h\,f(x_n, y_n)$
$\displaystyle x_{n+1} = x_n + h$

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Topic numbers match the course outline, so a topic here lines up with the same number in the lesson list.

1.1 Integration Using Substitution

Integrating Algebraic Functions Using Substitution

Integrating Linear Rational Functions Using Substitution

Integration Using Substitution

Calculating Definite Integrals Using Substitution

Further Integration of Algebraic Functions Using Substitution

Integrating Exponential Functions Using Linear Substitution

Integrating Exponential Functions Using Substitution

Integrating Trigonometric Functions Using Substitution

Integrating Logarithmic Functions Using Substitution

Integration by Substitution With Inverse Trigonometric Functions

Integrating Hyperbolic Functions Using Substitution

Integration by Substitution With Inverse Hyperbolic Functions

Integration by Substitution With Inverse Reciprocal Hyperbolic Functions

1.2 Special Techniques for Integration

Integrating Functions Using Polynomial Division

Integrating Functions by Completing the Square

Integration Using Hyperbolic Functions and Completing the Square

Integration Using Inverse Reciprocal Hyperbolic Functions and Completing the Square

1.3 Integration Using Trigonometric Identities

Integration Using Basic Trigonometric Identities

Integration Using the Pythagorean Identities

Integration Using the Double-Angle Formulas

Integrating Products of Trigonometric Functions

1.4 Integration Using Hyperbolic Identities

Integration Using Basic Hyperbolic Identities

Integration Using the Hyperbolic Pythagorean Identities

Integration Using the Hyperbolic Double-Angle Formulas

1.5 Integration by Parts

Introduction to Integration by Parts

Using Integration by Parts to Calculate Integrals With Logarithms

Applying the Integration By Parts Twice

The Tabular Method of Integration by Parts

Integration by Parts in Cyclic Cases

1.6 Integration Using Partial Fractions

Expressing Rational Functions as Sums of Partial Fractions

Expressing Rational Functions with Repeated Factors as Sums of Partial Fractions

Expressing Rational Functions with Irreducible Quadratic Factors as Sums of Partial

Integrating Rational Functions Using Partial Fractions

Integrating Rational Functions with Repeated Factors

Integrating Rational Functions with Irreducible Quadratic Factors

1.7 Improper Integrals

Improper Integrals

Improper Integrals Involving Exponential Functions

Improper Integrals Involving Arctangent

Improper Integrals Over the Real Line

Improper Integrals of the Second Kind

Improper Integrals of the Second Kind: Discontinuities at Interior Points

2.8 Applications of Integration

Integrating Rates of Change

Integrating Density Functions

The Average Value of a Function

The Area Between Curves Expressed as Functions of X

The Area Between Curves Expressed as Functions of Y

Finding Areas Between Curves that Intersect at More Than Two Points

The Arc Length of a Planar Curve

2.9 Volumes of Solids With Known Cross Sections

Volumes of Solids with Square Cross Sections

Volumes of Solids with Rectangular Cross Sections

Volumes of Solids with Triangular Cross Sections

Volumes of Solids with Circular Cross Sections

2.10 Volumes of Revolution

Volumes of Revolution Using the Disc Method: Rotation About the Coordinate Axes

Volumes of Revolution Using the Disc Method: Rotation About Other Axes

Volumes of Revolution Using the Washer Method: Rotation About the Coordinate Axes

Volumes of Revolution Using the Washer Method: Rotation About Other Axes

The Shell Method: Rotating a Region About the X-Axis

The Shell Method: Rotating a Region Between Two Curves About the X-Axis

The Shell Method: Rotation About the Y-Axis

2.11 Surface Areas of Revolution

Surface Areas of Revolution: Rotation About the X-Axis

Surface Areas of Revolution: Rotation About the Y-Axis

Surface Areas of Revolution for Parametric Curves

2.12 Connecting Position, Velocity and Acceleration Using Integrals

Calculating Velocity Using Integration

Determining Characteristics of Moving Objects Using Integration

Calculating the Position Function of a Particle Using Integration

Calculating the Displacement of a Particle Using Integration

Calculating the Total Distance Traveled by a Particle

Average Position, Velocity, and Acceleration