Reference sheet

Calculus reference sheet

The main definitions, formulas and conditions of the calculus course on one page, unit by unit, with links to the lessons that explain each topic in full.

For Calculus

This sheet follows the ten units of the course in order. Each unit starts with the big idea in a few sentences, then gives the results you will look up most often, with the conditions they need. When you want the full explanation, a proof or more examples, follow the lesson links under each topic.

Angles are in radians throughout. A formula holds wherever both sides are defined unless a condition says otherwise.

Limits

A limit describes the value f(x)f(x) approaches as xx gets close to a point, whatever happens at the point itself. Derivatives, integrals and series are all limits, so the rules here come back in every later unit.

What a limit means

We write lim⁡x→af(x)=L\lim_{x\to a} f(x) = L when f(x)f(x) can be made as close to LL as we like by taking x≠ax \ne a close enough to aa. The value f(a)f(a) plays no part. The precise version says that for every ε>0\varepsilon > 0 there is a δ>0\delta > 0 such that for every xx in the domain

0<∣x−a∣<δ  ⟹  ∣f(x)−L∣<ε.0 < \lvert x - a \rvert < \delta \implies \lvert f(x) - L \rvert < \varepsilon.

The left-hand limit uses only x<ax < a and the right-hand limit only x>ax > a. The two-sided limit exists exactly when both one-sided limits exist and are equal. For example ∣x∣x\frac{\lvert x \rvert}{x} tends to −1-1 from the left of 00 and to 11 from the right, so it has no limit at 00.

Lessons Intuitive definition, One-sided limits, Two-sided limits, The formal definition

Limit laws

If lim⁡x→af(x)=L\lim_{x\to a} f(x) = L and lim⁡x→ag(x)=M\lim_{x\to a} g(x) = M with LL and MM finite, limits can be taken piece by piece. All limits below are as x→ax \to a, and the same laws hold for one-sided limits and for limits at ±∞\pm\infty.

LawFormula
Sum and differencelim⁡ [f(x)±g(x)]=L±M\lim\,[f(x) \pm g(x)] = L \pm M
Constant multiplelim⁡ c f(x)=cL\lim\, c\,f(x) = cL
Productlim⁡ f(x) g(x)=LM\lim\, f(x)\,g(x) = LM
Quotientlim⁡f(x)g(x)=LM\lim \dfrac{f(x)}{g(x)} = \dfrac{L}{M} when M≠0M \ne 0
Powerlim⁡ [f(x)]n=Ln\lim\, [f(x)]^n = L^n for a positive integer nn
Rootlim⁡f(x)n=Ln\lim \sqrt[n]{f(x)} = \sqrt[n]{L}, with L≥0L \ge 0 when nn is even
Compositionlim⁡h(f(x))=h(L)\lim h(f(x)) = h(L) when hh is continuous at LL

Lessons Limit laws

How to evaluate a limit

Try these in order.

  1. Substitute. If ff is continuous at aa, for example a polynomial or a rational function defined at aa, the limit is f(a)f(a).
  2. Simplify. If substitution gives 00\tfrac{0}{0}, factor and cancel, or multiply by a conjugate. This is allowed because a limit only looks at x≠ax \ne a.
  3. Squeeze. If g(x)≤f(x)≤h(x)g(x) \le f(x) \le h(x) near aa and gg and hh have the same limit LL, then ff has limit LL too.
  4. Use a known limit from the table below, or l’Hôpital’s rule from the Applications of Differentiation unit.

For example, for x≠0x \ne 0 near 00, multiplying by the conjugate gives

x+4−2x=1x+4+2⟶14.\frac{\sqrt{x + 4} - 2}{x} = \frac{1}{\sqrt{x + 4} + 2} \longrightarrow \frac{1}{4}.

Lessons Limit theorems

Special limits

LimitValue
lim⁡x→0sin⁡xx\displaystyle\lim_{x\to 0} \frac{\sin x}{x}11
lim⁡x→01−cos⁡xx\displaystyle\lim_{x\to 0} \frac{1 - \cos x}{x}00
lim⁡x→0ex−1x\displaystyle\lim_{x\to 0} \frac{e^x - 1}{x}11
lim⁡x→0ln⁡(1+x)x\displaystyle\lim_{x\to 0} \frac{\ln(1 + x)}{x}11
lim⁡x→±∞(1+ax)x\displaystyle\lim_{x\to \pm\infty} \left(1 + \frac{a}{x}\right)^xeae^a
lim⁡x→∞ln⁡xxp\displaystyle\lim_{x\to\infty} \frac{\ln x}{x^p} and lim⁡x→∞xpex\displaystyle\lim_{x\to\infty} \frac{x^p}{e^x} for p>0p > 000

The first limit needs radians. The last row says logarithms grow slower than any positive power of xx, and powers grow slower than exe^x.

Lessons Special limits

Infinity and asymptotes

For a rational function, only the leading terms matter as x→±∞x \to \pm\infty. With leading terms anxna_n x^n on top and bmxmb_m x^m below,

DegreesLimit as x→±∞x \to \pm\infty
n<mn < m00
n=mn = manbm\dfrac{a_n}{b_m}
n>mn > m∞\infty or −∞-\infty
  • The line x=ax = a is a vertical asymptote when f(x)→±∞f(x) \to \pm\infty as x→ax \to a from at least one side.
  • The line y=Ly = L is a horizontal asymptote when f(x)→Lf(x) \to L as x→∞x \to \infty or as x→−∞x \to -\infty.
  • The line y=mx+by = mx + b is an oblique asymptote when f(x)−(mx+b)→0f(x) - (mx + b) \to 0 as x→∞x \to \infty or as x→−∞x \to -\infty. For a rational function whose top degree is one more than its bottom degree, polynomial division finds it.

Lessons Limits at infinity, Infinite limits, Asymptotes

Continuity

A continuous function has no breaks, so its value at a point is exactly what nearby values lead you to expect. That simple property is enough to prove that equations have solutions and that maximum and minimum values exist.

Continuity at a point

A function ff is continuous at aa when

lim⁡x→af(x)=f(a).\lim_{x\to a} f(x) = f(a).

That asks for three things. The value f(a)f(a) is defined, the limit exists, and the two are equal. On a closed interval [a,b][a, b] only one-sided limits are used at the endpoints. Polynomials, rational functions, roots, trigonometric, exponential and logarithmic functions are continuous on their domains, and so are sums, products, quotients and compositions of continuous functions.

DiscontinuityWhat happens at aaExample at 00
Removablethe limit exists but differs from f(a)f(a), or f(a)f(a) is undefinedsin⁡xx\dfrac{\sin x}{x}
Jumpthe one-sided limits are finite and different⌊x⌋\lfloor x \rfloor
Infinitea one-sided limit is ±∞\pm\infty1x\dfrac{1}{x}
Oscillatinga one-sided limit fails to exist in another waysin⁡1x\sin \dfrac{1}{x}

A removable discontinuity can be filled in. Giving sin⁡xx\frac{\sin x}{x} the value 11 at 00 makes it continuous everywhere.

Lessons Continuous functions, Discontinuities, Graphs of continuous functions

The intermediate value and extreme value theorems

Both theorems need ff continuous on a closed bounded interval [a,b][a, b].

  • Intermediate value theorem. Every number strictly between f(a)f(a) and f(b)f(b) equals f(c)f(c) for some cc in (a,b)(a, b). In particular, if f(a)f(a) and f(b)f(b) have opposite signs, ff has a root in (a,b)(a, b). The theorem promises a root, not a unique one.
  • Extreme value theorem. The function ff takes an absolute maximum value and an absolute minimum value on [a,b][a, b]. On an interval that is not closed this can fail, as f(x)=xf(x) = x on [0,1)[0, 1) has no maximum.

Lessons Continuity theorems, The intermediate value theorem

Uniform continuity

A function is uniformly continuous on a set when one δ\delta works at every point at once. For every ε>0\varepsilon > 0 there is δ>0\delta > 0 such that ∣f(x)−f(y)∣<ε\lvert f(x) - f(y) \rvert < \varepsilon for all xx and yy in the set with ∣x−y∣<δ\lvert x - y \rvert < \delta. Every continuous function on a closed bounded interval is uniformly continuous there, while 1x\frac{1}{x} on (0,1)(0, 1) is continuous but not uniformly continuous.

Lessons Uniform continuity

Derivatives

The derivative measures how fast a function changes at a single point. Geometrically it is the slope of the tangent line, and physically it is an instantaneous rate such as velocity.

Definition

The derivative of ff at aa is

f′(a)=lim⁡h→0f(a+h)−f(a)h,f'(a) = \lim_{h\to 0} \frac{f(a + h) - f(a)}{h},

when this limit exists and is finite. The tangent line at (a,f(a))(a, f(a)) is y=f(a)+f′(a)(x−a)y = f(a) + f'(a)(x - a). Other notations are dydx\frac{dy}{dx} and ddxf(x)\frac{d}{dx} f(x). A differentiable function is continuous, but a continuous function need not be differentiable. The function ∣x∣\lvert x \rvert has a corner at 00 and no derivative there.

Lessons Velocity and tangent, Derivatives and rates of change, The derivative as a function

Rules

Here ff and gg are differentiable and cc is a constant.

RuleFormula
Constant(c)′=0(c)' = 0
Power(xr)′=rxr−1(x^r)' = r x^{r-1} for any real rr when x>0x > 0, and for every xx when rr is a positive integer
Sum and constant multiple(f+g)′=f′+g′(f + g)' = f' + g' and (cf)′=cf′(c f)' = c f'
Product(fg)′=f′g+fg′(fg)' = f'g + fg'
Quotient(fg)′=f′g−fg′g2\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2} where g≠0g \ne 0
Chain(f(g(x)))′=f′(g(x)) g′(x)\big(f(g(x))\big)' = f'(g(x))\, g'(x)

In Leibniz notation the chain rule reads dydx=dydu dudx\frac{dy}{dx} = \frac{dy}{du}\,\frac{du}{dx}, which is how you handle a function inside a function.

Lessons Elementary rules, Product rule, Quotient rule, Chain rule

Derivatives to know

With the chain rule each row also works for an inner function u(x)u(x), as in (sin⁡u)′=u′cos⁡u(\sin u)' = u' \cos u.

FunctionDerivativeHolds for
exe^xexe^xall xx
axa^xaxln⁡aa^x \ln aa>0a > 0
ln⁡∣x∣\ln \lvert x \rvert1x\dfrac{1}{x}x≠0x \ne 0
sin⁡x\sin xcos⁡x\cos xall xx
cos⁡x\cos x−sin⁡x-\sin xall xx
tan⁡x\tan xsec⁡2x\sec^2 xcos⁡x≠0\cos x \ne 0
sec⁡x\sec xsec⁡xtan⁡x\sec x \tan xcos⁡x≠0\cos x \ne 0
arcsin⁡x\arcsin x11−x2\dfrac{1}{\sqrt{1 - x^2}}∣x∣<1\lvert x \rvert < 1
arctan⁡x\arctan x11+x2\dfrac{1}{1 + x^2}all xx
sinh⁡x\sinh xcosh⁡x\cosh xall xx
cosh⁡x\cosh xsinh⁡x\sinh xall xx

The hyperbolic functions are sinh⁡x=ex−e−x2\sinh x = \frac{e^x - e^{-x}}{2} and cosh⁡x=ex+e−x2\cosh x = \frac{e^x + e^{-x}}{2}, and they satisfy cosh⁡2x−sinh⁡2x=1\cosh^2 x - \sinh^2 x = 1.

Lessons Trigonometric, Exponential and logarithmic, Hyperbolic

Implicit, inverse and logarithmic differentiation

  • Implicit differentiation. Differentiate both sides of an equation in xx and yy, remembering that each yy term picks up a factor y′y', then solve for y′y'. From x2+y2=25x^2 + y^2 = 25 you get 2x+2yy′=02x + 2yy' = 0, so y′=−xyy' = -\frac{x}{y} where y≠0y \ne 0.
  • Inverse functions. If ff is one-to-one and differentiable on an interval and f′(f−1(x))≠0f'(f^{-1}(x)) \ne 0, then
(f−1)′(x)=1f′(f−1(x)).(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}.
  • Logarithmic differentiation. For products, quotients and variable powers, take ln⁡\ln first. For y=xxy = x^x with x>0x > 0 this gives ln⁡y=xln⁡x\ln y = x \ln x, so y′y=ln⁡x+1\frac{y'}{y} = \ln x + 1 and y′=xx(ln⁡x+1)y' = x^x(\ln x + 1).
  • Higher derivatives. The second derivative f′′f'' is the derivative of f′f'. For a position s(t)s(t) the velocity is s′s' and the acceleration is s′′s''.

Lessons Higher derivatives, Implicit differentiation, Inverse functions, Logarithmic differentiation

Applications of Differentiation

Once you can differentiate, the sign and size of f′f' and f′′f'' tell you where a function rises, falls and bends, where it is largest, and how well a straight line can stand in for it.

Finding maximum and minimum values

A critical point is a number cc in the domain where f′(c)=0f'(c) = 0 or f′(c)f'(c) does not exist. By Fermat’s theorem, a local maximum or minimum at an interior point can only happen at a critical point.

Closed interval method. For ff continuous on [a,b][a, b], compute ff at the critical points inside (a,b)(a, b) and at aa and bb. The largest value is the absolute maximum and the smallest is the absolute minimum.

To classify a critical point cc where ff is continuous,

TestLocal maximumLocal minimumNo conclusion
First derivativef′f' changes from ++ to −- at ccf′f' changes from −- to ++ at ccif f′f' keeps its sign there is no extremum
Second derivative, with f′(c)=0f'(c) = 0f′′(c)<0f''(c) < 0f′′(c)>0f''(c) > 0f′′(c)=0f''(c) = 0

Lessons Extreme values, The closed interval method

The mean value theorem

If ff is continuous on [a,b][a, b] and differentiable on (a,b)(a, b), then for some cc in (a,b)(a, b)

f′(c)=f(b)−f(a)b−a.f'(c) = \frac{f(b) - f(a)}{b - a}.

At some moment the instantaneous rate equals the average rate. Rolle’s theorem is the case f(a)=f(b)f(a) = f(b). The consequences are used everywhere. On an interval, f′>0f' > 0 makes ff increasing, f′<0f' < 0 makes it decreasing, and f′=0f' = 0 makes it constant.

Lessons The mean value theorem, Increasing and decreasing functions

The shape of a graph

Where f′′>0f'' > 0 the graph is concave up and bends upward, and where f′′<0f'' < 0 it is concave down. An inflection point is a point where ff is continuous and the concavity changes. A zero of f′′f'' is only a candidate, since x4x^4 has f′′(0)=0f''(0) = 0 and no inflection point.

To sketch a graph, work through the domain, intercepts and symmetry, then the asymptotes, then the sign of f′f' for rise, fall and extrema, then the sign of f′′f'' for concavity, and finally join the points in a way that agrees with all of it.

Lessons Concavity and inflection points, Sketching graphs

Rates, optimization and growth

  • Related rates. Write one equation linking the changing quantities, differentiate it with respect to time, and only then put in the values at the moment you care about.
  • Optimization. Write the quantity as a function of one variable using the constraint, find its domain, and use the closed interval method or the derivative tests. Check that the answer makes sense in the original problem.
  • Growth and decay. The equation y′=kyy' = ky with y(0)=y0y(0) = y_0 has the solution y=y0ekty = y_0 e^{kt}. The doubling time for k>0k > 0 is ln⁡2k\frac{\ln 2}{k}, and the half-life for k<0k < 0 is ln⁡2∣k∣\frac{\ln 2}{\lvert k \rvert}.
  • Rates in the sciences. Velocity, reaction rate, population growth rate and marginal cost are all derivatives. The marginal cost C′(x)C'(x) approximates the cost of one more unit.

Lessons Related rates, Optimization, Rates in the sciences, Growth and decay

L’Hôpital’s rule

If f(x)/g(x)f(x)/g(x) has the form 00\tfrac{0}{0} or ∞∞\tfrac{\infty}{\infty} as x→ax \to a, where ff and gg are differentiable near aa with g′(x)≠0g'(x) \ne 0 there, then

lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x),\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)},

provided the limit on the right exists or is ±∞\pm\infty. It also works for one-sided limits and for a=±∞a = \pm\infty. Check the form before every use. Rewrite 0⋅∞0 \cdot \infty as a quotient, ∞−∞\infty - \infty as one fraction, and the powers 1∞1^\infty, 000^0 and ∞0\infty^0 by taking logarithms first.

Lessons Indeterminate forms and l’Hôpital’s rule

Linear approximation

Near aa a differentiable function is close to its tangent line,

f(x)≈f(a)+f′(a)(x−a).f(x) \approx f(a) + f'(a)(x - a).

For example 4.1≈2+14(0.1)=2.025\sqrt{4.1} \approx 2 + \frac{1}{4}(0.1) = 2.025. If ∣f′′∣≤K\lvert f'' \rvert \le K between aa and xx, the error is at most K2(x−a)2\frac{K}{2}(x - a)^2. In the language of differentials, dy=f′(x) dxdy = f'(x)\,dx approximates the actual change Δy\Delta y.

Lessons Linear approximations and differentials

Integrals

Integration adds up infinitely many small pieces, such as thin rectangles under a curve. The fundamental theorem of calculus connects it to differentiation, so most integrals are computed by finding an antiderivative.

Antiderivatives

A function FF is an antiderivative of ff on an interval when F′=fF' = f there. Any two antiderivatives on the same interval differ by a constant, which is why each entry below gets a + C+\,C. You can always check an answer by differentiating it.

FunctionAntiderivativeHolds for
xrx^rxr+1r+1\dfrac{x^{r+1}}{r+1}r≠−1r \ne -1
1x\dfrac{1}{x}ln⁡∣x∣\ln \lvert x \rvertx≠0x \ne 0
exe^xexe^xall xx
axa^xaxln⁡a\dfrac{a^x}{\ln a}a>0a > 0, a≠1a \ne 1
sin⁡x\sin x−cos⁡x-\cos xall xx
cos⁡x\cos xsin⁡x\sin xall xx
sec⁡2x\sec^2 xtan⁡x\tan xcos⁡x≠0\cos x \ne 0
tan⁡x\tan xln⁡∣sec⁡x∣\ln \lvert \sec x \rvertcos⁡x≠0\cos x \ne 0
sec⁡x\sec xln⁡∣sec⁡x+tan⁡x∣\ln \lvert \sec x + \tan x \rvertcos⁡x≠0\cos x \ne 0
1a2+x2\dfrac{1}{a^2 + x^2}1aarctan⁡xa\dfrac{1}{a} \arctan \dfrac{x}{a}a>0a > 0
1a2−x2\dfrac{1}{\sqrt{a^2 - x^2}}arcsin⁡xa\arcsin \dfrac{x}{a}a>0a > 0, ∣x∣<a\lvert x \rvert < a

Lessons Indefinite integrals, Integration rules

The definite integral

Split [a,b][a, b] into nn pieces of width Δx=b−an\Delta x = \frac{b - a}{n} and pick a point xi∗x_i^* in each. For a continuous ff, or a bounded one with finitely many discontinuities,

∫abf(x) dx=lim⁡n→∞∑i=1nf(xi∗) Δx.\int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x.

When f≥0f \ge 0 this is the area under the graph. In general it is the area above the axis minus the area below. Integrals are linear, they add over adjacent intervals, reversing the limits changes the sign, and for a≤ba \le b a larger function never has a smaller integral.

Lessons Sigma notation, The area problem, Riemann sums, Definite integrals, Properties

The fundamental theorem of calculus

Let ff be continuous on [a,b][a, b].

Part 1. The accumulation function g(x)=∫axf(t) dtg(x) = \int_a^x f(t)\,dt is differentiable on (a,b)(a, b) with g′(x)=f(x)g'(x) = f(x). With a differentiable upper limit u(x)u(x) that stays in [a,b][a, b], the chain rule gives

ddx∫au(x)f(t) dt=f(u(x)) u′(x).\frac{d}{dx} \int_a^{u(x)} f(t)\,dt = f(u(x))\,u'(x).

Part 2. If FF is any antiderivative of ff on [a,b][a, b], then

∫abf(x) dx=F(b)−F(a).\int_a^b f(x)\,dx = F(b) - F(a).

Read the other way, integrating a rate of change gives the net change. For a velocity v(t)v(t) the integral of vv is the displacement and the integral of ∣v∣\lvert v \rvert is the distance travelled.

Lessons Part 1, Part 2 and net change

Choosing a technique

Look at the integrand and pick the first row that fits.

You seeTryKey step
a function times the derivative of its inside, f(g(x)) g′(x)f(g(x))\,g'(x)substitutionu=g(x)u = g(x), du=g′(x) dxdu = g'(x)\,dx
a product like xexx e^x or xsin⁡xx \sin x, or ln⁡x\ln x aloneintegration by parts∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du
a ratio of polynomialspartial fractionsdivide first if the top degree is not smaller
powers of sin⁡\sin and cos⁡\cos, or of tan⁡\tan and sec⁡\sectrigonometric identitiessave one factor, convert the rest
a2−x2\sqrt{a^2 - x^2}, a2+x2\sqrt{a^2 + x^2} or x2−a2\sqrt{x^2 - a^2}trigonometric substitutionx=asin⁡θx = a\sin\theta, atan⁡θa\tan\theta or asec⁡θa\sec\theta
a rational function of sin⁡x\sin x and cos⁡x\cos xhalf angle substitutiont=tan⁡x2t = \tan \frac{x}{2}

In a definite integral, change the limits along with the variable. For example, with u=x2u = x^2,

∫02xex2 dx=12∫04eu du=e4−12.\int_0^2 x e^{x^2}\,dx = \frac{1}{2}\int_0^4 e^u\,du = \frac{e^4 - 1}{2}.

For integration by parts, choose uu to be the factor that gets simpler when differentiated. With u=xu = x and dv=ex dxdv = e^x\,dx,

∫xex dx=xex−ex+C.\int x e^x\,dx = x e^x - e^x + C.

The identities used most often are

sin⁡2x+cos⁡2x=11+tan⁡2x=sec⁡2xsin⁡2x=1−cos⁡2x2cos⁡2x=1+cos⁡2x2\begin{aligned} \sin^2 x + \cos^2 x &= 1 \\[0.4em] 1 + \tan^2 x &= \sec^2 x \\[0.4em] \sin^2 x &= \frac{1 - \cos 2x}{2} \\[0.8em] \cos^2 x &= \frac{1 + \cos 2x}{2} \end{aligned}

Some functions, such as e−x2e^{-x^2} and sin⁡xx\frac{\sin x}{x}, have no elementary antiderivative at all. Their integrals come from series or numerical methods.

Lessons Substitution, By parts, Rational functions, Trigonometric integrals, Trigonometric substitutions, Hyperbolic and half angle, Other techniques

Improper integrals

An integral over an infinite interval, or of a function that blows up, is defined as a limit. For ff continuous on [a,∞)[a, \infty),

∫a∞f(x) dx=lim⁡t→∞∫atf(x) dx,\int_a^\infty f(x)\,dx = \lim_{t\to\infty} \int_a^t f(x)\,dx,

and it converges when this limit is finite. If both ends are a problem, or there is a problem inside the interval, split the integral and require each piece to converge. Two families serve as yardsticks. The integral

∫1∞dxxp\int_1^\infty \frac{dx}{x^p}

converges exactly when p>1p > 1, and the integral of the same function over (0,1](0, 1] converges exactly when p<1p < 1. To decide convergence without computing, compare. If 0≤g≤f0 \le g \le f and the integral of ff converges, so does the integral of gg.

Lessons Unbounded intervals, Singularities and comparison

Applications

Every application comes from the same idea. Cut the quantity into thin slices, approximate each slice, and add them up with an integral. The functions here are continuous.

QuantityFormulaConditions
Area between curves∫ab[f(x)−g(x)] dx\displaystyle\int_a^b \big[f(x) - g(x)\big]\,dxf≥gf \ge g on [a,b][a, b]
Average value1b−a∫abf(x) dx\displaystyle\frac{1}{b - a}\int_a^b f(x)\,dxa<ba < b
Volume by disksπ∫abf(x)2 dx\displaystyle\pi\int_a^b f(x)^2\,dxabout the xx-axis
Volume by washersπ∫ab[R(x)2−r(x)2] dx\displaystyle\pi\int_a^b \big[R(x)^2 - r(x)^2\big]\,dx0≤r≤R0 \le r \le R
Volume by shells2π∫abx f(x) dx\displaystyle 2\pi\int_a^b x\,f(x)\,dxabout the yy-axis, 0≤a0 \le a, f≥0f \ge 0
Arc length∫ab1+f′(x)2 dx\displaystyle\int_a^b \sqrt{1 + f'(x)^2}\,dxf′f' continuous
Surface area2π∫abf(x)1+f′(x)2 dx\displaystyle 2\pi\int_a^b f(x)\sqrt{1 + f'(x)^2}\,dxabout the xx-axis, f≥0f \ge 0, f′f' continuous
Work∫abF(x) dx\displaystyle\int_a^b F(x)\,dxforce F(x)F(x) along the motion
ProbabilityP(a≤X≤b)=∫abf(x) dx\displaystyle P(a \le X \le b) = \int_a^b f(x)\,dxdensity f≥0f \ge 0 with total integral 11

Lessons Area between curves, Average value, Disks and washers, Shells, Arc length, Surface area, Centres of mass, Work and force, Economics and biology, Probability

Parametric Equations and Polar Coordinates

Not every curve is the graph of a function. Conics, curves traced by a moving point and curves described by angle and distance need their own descriptions, and calculus works for each of them.

Conics

Parabolas, ellipses and hyperbolas are the curves where a plane cuts a double cone. In standard position,

ConicEquationFociNotes
Parabolax2=4pyx^2 = 4py with p≠0p \ne 0(0,p)(0, p)directrix y=−py = -p
Ellipsex2a2+y2b2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 with a>b>0a > b > 0(±c,0)(\pm c, 0), c2=a2−b2c^2 = a^2 - b^2distances to the foci add up to 2a2a
Hyperbolax2a2−y2b2=1\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 with a,b>0a, b > 0(±c,0)(\pm c, 0), c2=a2+b2c^2 = a^2 + b^2asymptotes y=±baxy = \pm \dfrac{b}{a}x

The eccentricity ee tells them apart. It is e<1e < 1 for an ellipse, e=1e = 1 for a parabola and e>1e > 1 for a hyperbola, and e=cae = \frac{c}{a} for the ellipse and hyperbola above. A general equation Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 that describes a nondegenerate conic is an ellipse when B2−4AC<0B^2 - 4AC < 0, a parabola when B2−4AC=0B^2 - 4AC = 0 and a hyperbola when B2−4AC>0B^2 - 4AC > 0.

Lessons Definition of a conic, Parabolas, Ellipses, Hyperbolas, General conics

Parametric curves

A parametric curve gives both coordinates in terms of a parameter, x=f(t)x = f(t) and y=g(t)y = g(t), so it also has a direction of travel. For example x=cos⁡tx = \cos t, y=sin⁡ty = \sin t for 0≤t≤2π0 \le t \le 2\pi traces the unit circle once counterclockwise. With f′f' and g′g' continuous on [α,β][\alpha, \beta],

QuantityFormulaConditions
Slopedydx=dy/dtdx/dt\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}dxdt≠0\dfrac{dx}{dt} \ne 0
Enclosed area∫αβx dydt dt\displaystyle\int_\alpha^\beta x\,\frac{dy}{dt}\,dtclosed curve traced once counterclockwise
Arc length∫αβ(dxdt)2+(dydt)2 dt\displaystyle\int_\alpha^\beta \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dtcurve traced once
Surface area2π∫αβy(dxdt)2+(dydt)2 dt\displaystyle 2\pi\int_\alpha^\beta y\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dtabout the xx-axis, y≥0y \ge 0, traced once

The tangent is horizontal where dydt=0\frac{dy}{dt} = 0 and dxdt≠0\frac{dx}{dt} \ne 0, and vertical where dxdt=0\frac{dx}{dt} = 0 and dydt≠0\frac{dy}{dt} \ne 0.

Lessons Parametric curves, Sketching, Slope, Area, Length and surfaces

Polar coordinates

A point with polar coordinates (r,θ)(r, \theta) lies at distance rr from the origin along the ray at angle θ\theta. The two systems are linked by

x=rcos⁡θ,y=rsin⁡θ,r2=x2+y2.x = r\cos\theta, \qquad y = r\sin\theta, \qquad r^2 = x^2 + y^2.

The same point has many polar coordinates, since (r,θ+2π)(r, \theta + 2\pi) and (−r,θ+π)(-r, \theta + \pi) describe it too. Some curves are much simpler in polar form, such as the circle r=ar = a, the cardioid r=a(1+cos⁡θ)r = a(1 + \cos\theta), and the rose r=acos⁡nθr = a\cos n\theta, which has nn petals for odd nn and 2n2n for even nn. A conic with a focus at the origin, eccentricity ee and directrix x=dx = d with d>0d > 0 is

r=ed1+ecos⁡θ.r = \frac{ed}{1 + e\cos\theta}.

Lessons Polar coordinates, Some polar curves, Sketching, Polar conics

Calculus in polar coordinates

For r=f(θ)r = f(\theta) with f′f' continuous, and r′=drdθr' = \frac{dr}{d\theta},

QuantityFormulaConditions
Slopedydx=r′sin⁡θ+rcos⁡θr′cos⁡θ−rsin⁡θ\dfrac{dy}{dx} = \dfrac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}denominator ≠0\ne 0
Area12∫αβr2 dθ\displaystyle\frac{1}{2}\int_\alpha^\beta r^2\,d\thetaregion swept once, β−α≤2π\beta - \alpha \le 2\pi
Arc length∫αβr2+(r′)2 dθ\displaystyle\int_\alpha^\beta \sqrt{r^2 + (r')^2}\,d\thetacurve traced once

For example the cardioid r=1+cos⁡θr = 1 + \cos\theta encloses the area

12∫02π(1+cos⁡θ)2 dθ=3π2.\frac{1}{2}\int_0^{2\pi} (1 + \cos\theta)^2\,d\theta = \frac{3\pi}{2}.

Lessons Slope, Area, Arc length

Infinite Sequences and Series

Adding infinitely many numbers can give a finite total, as 12+14+18+⋯=1\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = 1 does. This unit decides when that happens and then uses it to write functions such as exe^x and sin⁡x\sin x as infinite polynomials.

Sequences

A sequence (an)(a_n) converges to LL when its terms get and stay as close to LL as we like. Precisely, for every ε>0\varepsilon > 0 there is NN with ∣an−L∣<ε\lvert a_n - L \rvert < \varepsilon for all n>Nn > N. The limit laws and the squeeze theorem work for sequences too, and two facts are especially useful.

  • If an=f(n)a_n = f(n) and f(x)→Lf(x) \to L as x→∞x \to \infty, then an→La_n \to L, so l’Hôpital’s rule applies to sequences.
  • A sequence that is bounded and monotonic converges.
SequenceLimit
rnr^n00 for ∣r∣<1\lvert r \rvert < 1, 11 for r=1r = 1, and divergent otherwise
nn\sqrt[n]{n}11
xnn!\dfrac{x^n}{n!}00 for every real xx
(1+xn)n\left(1 + \dfrac{x}{n}\right)^nexe^x for every real xx

Lessons Definition of a sequence, Convergence, Sequence theorems

Series to know

A series ∑an\sum a_n converges to SS when its partial sums sN=a1+⋯+aNs_N = a_1 + \cdots + a_N approach SS. If it converges, then an→0a_n \to 0. The converse is false, as the harmonic series shows.

SeriesBehaviour
Geometric ∑n=0∞arn\displaystyle\sum_{n=0}^{\infty} a r^n with a≠0a \ne 0converges to a1−r\dfrac{a}{1 - r} when ∣r∣<1\lvert r \rvert < 1, diverges when ∣r∣≥1\lvert r \rvert \ge 1
p-series ∑n=1∞1np\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p}converges when p>1p > 1, diverges when p≤1p \le 1
Harmonic ∑n=1∞1n\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}diverges, though its terms go to 00
Alternating harmonic ∑n=1∞(−1)n−1n\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}converges to ln⁡2\ln 2, but not absolutely
Telescoping ∑n=1∞(1n−1n+1)\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n + 1}\right)converges to 11, since sN=1−1N+1s_N = 1 - \dfrac{1}{N + 1}

Lessons Partial sums, Geometric series, Telescoping and harmonic, Series theorems

Choosing a convergence test

Work down the table and use the first test that fits.

If the seriesUseConclusion
has terms that do not tend to 00divergence testit diverges
is geometric or a p-seriesthe known results aboveread it off
has positive terms that behave like a geometric or p-series for large nncomparison or limit comparisonsame behaviour as the series you compare with
has factorials or powers like 3n3^nratio test, L=lim⁡∣an+1an∣L = \lim \left\lvert \dfrac{a_{n+1}}{a_n} \right\rvertconverges absolutely if L<1L < 1, diverges if L>1L > 1
has a whole term raised to the power nnroot test, L=lim⁡∣an∣nL = \lim \sqrt[n]{\lvert a_n \rvert}converges absolutely if L<1L < 1, diverges if L>1L > 1
alternates in sign, ∑(−1)n−1bn\sum (-1)^{n-1} b_n with bn>0b_n > 0alternating series testconverges if bnb_n decreases to 00
has an=f(n)a_n = f(n) with ff positive, continuous and decreasingintegral testsame behaviour as ∫1∞f(x) dx\int_1^\infty f(x)\,dx

The ratio and root tests say nothing when L=1L = 1. A series converges absolutely when ∑∣an∣\sum \lvert a_n \rvert converges, and absolute convergence implies convergence. A series that converges but not absolutely converges conditionally, and its terms can be rearranged to change the sum.

Lessons Integral test, Comparison tests, Alternating series, Absolute and conditional convergence, Ratio and root tests

How close is a partial sum

To approximate a sum by a partial sum sNs_N, you need a bound on the remainder RN=S−sNR_N = S - s_N.

  • Alternating series. If bnb_n decreases to 00, then ∣RN∣≤bN+1\lvert R_N \rvert \le b_{N+1}, the first term left out.
  • Integral test. Under its hypotheses, ∫N+1∞f(x) dx≤RN≤∫N∞f(x) dx\int_{N+1}^\infty f(x)\,dx \le R_N \le \int_N^\infty f(x)\,dx.
  • Geometric tail. If ∣an+1∣≤r∣an∣\lvert a_{n+1} \rvert \le r\lvert a_n \rvert for all n>Nn > N with r<1r < 1, then ∣RN∣≤∣aN+1∣1−r\lvert R_N \rvert \le \frac{\lvert a_{N+1} \rvert}{1 - r}.

Lessons Estimates of sums

Power series

A power series ∑n=0∞cn(x−a)n\sum_{n=0}^{\infty} c_n (x - a)^n is an infinite polynomial centred at aa. It has a radius of convergence RR, with 0≤R≤∞0 \le R \le \infty, such that it converges absolutely for ∣x−a∣<R\lvert x - a \rvert < R and diverges for ∣x−a∣>R\lvert x - a \rvert > R. The ratio test usually finds RR, and the two endpoints need their own check. Inside the radius you can differentiate and integrate term by term, and the radius stays the same.

Lessons Definition of power series, Pointwise and uniform convergence, Power series theorems

Taylor series

The Taylor series of ff at aa is the power series whose coefficients match all the derivatives of ff at aa,

∑n=0∞f(n)(a)n!(x−a)n=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2+⋯\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 + \cdots

A Maclaurin series is the case a=0a = 0. Stopping after the term of degree NN gives the Taylor polynomial TNT_N. The series equals f(x)f(x) exactly where the remainder RN(x)=f(x)−TN(x)R_N(x) = f(x) - T_N(x) tends to 00, and Taylor’s inequality bounds it. If ∣f(N+1)∣≤M\lvert f^{(N+1)} \rvert \le M between aa and xx, then

∣RN(x)∣≤M(N+1)! ∣x−a∣N+1.\lvert R_N(x) \rvert \le \frac{M}{(N+1)!}\,\lvert x - a \rvert^{N+1}.
Maclaurin seriesValid for
ex=1+x+x22!+x33!+⋯e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdotsall xx
sin⁡x=x−x33!+x55!−⋯\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdotsall xx
cos⁡x=1−x22!+x44!−⋯\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdotsall xx
11−x=1+x+x2+x3+⋯\dfrac{1}{1 - x} = 1 + x + x^2 + x^3 + \cdots−1<x<1-1 < x < 1
ln⁡(1+x)=x−x22+x33−⋯\ln(1 + x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots−1<x≤1-1 < x \le 1
arctan⁡x=x−x33+x55−⋯\arctan x = x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots−1≤x≤1-1 \le x \le 1
(1+x)k=1+kx+k(k−1)2!x2+⋯(1 + x)^k = 1 + kx + \dfrac{k(k - 1)}{2!}x^2 + \cdots−1<x<1-1 < x < 1, any real kk

New series come from these by substitution, differentiation and integration. Replacing xx with −x2-x^2 in the series for 11−x\frac{1}{1 - x} gives the series for 11+x2\frac{1}{1 + x^2}, and integrating that gives the series for arctan⁡x\arctan x. Series also turn hard limits into easy ones, as in

lim⁡x→0sin⁡x−xx3=lim⁡x→0(−13!+x25!−⋯ )=−16.\lim_{x\to 0} \frac{\sin x - x}{x^3} = \lim_{x\to 0} \left(-\frac{1}{3!} + \frac{x^2}{5!} - \cdots\right) = -\frac{1}{6}.

Lessons Taylor’s theorem, Taylor and Maclaurin series, Series theorems, Applications, Binomial series

Fourier Series

A Taylor series builds a function from powers of xx. A Fourier series builds a periodic function from sines and cosines, which is how signals, sound and heat flow are usually described.

Fourier coefficients

For a function ff with period 2L2L that is integrable on [−L,L][-L, L], the Fourier series is

a02+∑n=1∞(ancos⁡nπxL+bnsin⁡nπxL),\frac{a_0}{2} + \sum_{n=1}^{\infty}\left(a_n\cos\frac{n\pi x}{L} + b_n\sin\frac{n\pi x}{L}\right),

with coefficients

an=1L∫−LLf(x)cos⁡nπxL dx(n≥0),bn=1L∫−LLf(x)sin⁡nπxL dx(n≥1).\begin{aligned} a_n &= \frac{1}{L}\int_{-L}^{L} f(x)\cos\frac{n\pi x}{L}\,dx \qquad (n \ge 0), \\[1em] b_n &= \frac{1}{L}\int_{-L}^{L} f(x)\sin\frac{n\pi x}{L}\,dx \qquad (n \ge 1). \end{aligned}

These formulas work because the sines and cosines are orthogonal. Over a period, the integral of the product of two different ones is 00. Symmetry saves work, because an even ff has every bn=0b_n = 0 and an odd ff has every an=0a_n = 0.

For example, the function with period 2π2\pi equal to xx on (−π,π)(-\pi, \pi) is odd, and its series is

2(sin⁡x−sin⁡2x2+sin⁡3x3−⋯ ).2\left(\sin x - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \cdots\right).

Lessons Periodic functions and orthogonality, Fourier coefficients

Convergence

If ff is periodic and piecewise smooth, meaning that in each period ff and f′f' are continuous except at finitely many points where they have finite one-sided limits, then the Fourier series converges at every xx to

f(x+)+f(x−)2,\frac{f(x^+) + f(x^-)}{2},

the midpoint of the jump. Where ff is continuous this is just f(x)f(x). Near a jump the partial sums overshoot by about 9 percent of the jump however many terms you take, which is the Gibbs phenomenon.

Lessons Convergence of Fourier series

Sine and cosine series

A function given only on (0,L)(0, L) can be extended as an odd function, which gives a series of sines only, or as an even function, which gives cosines only. The coefficients are then

bn=2L∫0Lf(x)sin⁡nπxL dxfor the sine series,an=2L∫0Lf(x)cos⁡nπxL dxfor the cosine series.\begin{aligned} b_n &= \frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\,dx \qquad \text{for the sine series,} \\[1em] a_n &= \frac{2}{L}\int_0^L f(x)\cos\frac{n\pi x}{L}\,dx \qquad \text{for the cosine series.} \end{aligned}

Lessons Fourier sine and cosine series

First-Order Differential Equations

A differential equation states how a quantity changes, and solving it recovers the quantity itself. First-order equations already describe population growth, cooling, mixing and simple circuits.

Solutions and initial value problems

An initial value problem asks for a function y(x)y(x) with

y′=f(x,y),y(x0)=y0.y' = f(x, y), \qquad y(x_0) = y_0.

If ff and ∂f∂y\frac{\partial f}{\partial y} are continuous on an open rectangle containing (x0,y0)(x_0, y_0), there is exactly one solution on some interval around x0x_0. Without that condition uniqueness can fail, since y′=3y2/3y' = 3y^{2/3} with y(0)=0y(0) = 0 has both y=0y = 0 and y=x3y = x^3 as solutions. You can always check a proposed solution by substituting it into the equation.

For an autonomous equation y′=f(y)y' = f(y), the constant solutions are the roots of ff. Such an equilibrium y∗y^* is stable when f′(y∗)<0f'(y^*) < 0 and unstable when f′(y∗)>0f'(y^*) > 0.

Lessons Initial value problems, Direction fields and equilibria

Solving first-order equations

Recognize the type, then follow its method.

TypeFormMethod
Separabley′=g(x) h(y)y' = g(x)\,h(y)integrate ∫dyh(y)=∫g(x) dx\displaystyle\int \frac{dy}{h(y)} = \int g(x)\,dx, and keep the constant solutions where h(y)=0h(y) = 0
Lineary′+P(x) y=Q(x)y' + P(x)\,y = Q(x)multiply by μ=e∫P dx\mu = e^{\int P\,dx}, so (μy)′=μQ(\mu y)' = \mu Q
Homogeneousy′=F ⁣(yx)y' = F\!\left(\dfrac{y}{x}\right)put y=vxy = vx, which makes it separable
Bernoulliy′+P(x) y=Q(x) yny' + P(x)\,y = Q(x)\,y^n with n≠0,1n \ne 0, 1put u=y1−nu = y^{1-n}, which makes it linear

For the linear equation with PP and QQ continuous on an interval, the general solution is

y=1μ(x)(∫μ(x) Q(x) dx+C).y = \frac{1}{\mu(x)}\left(\int \mu(x)\,Q(x)\,dx + C\right).

Lessons Separable, Linear, Substitutions

Models

ModelEquationSolution
Logistic growthP′=kP(1−PM)P' = kP\left(1 - \dfrac{P}{M}\right)P=M1+Ae−ktP = \dfrac{M}{1 + Ae^{-kt}} with A=M−P0P0A = \dfrac{M - P_0}{P_0}, for P0>0P_0 > 0
Newton coolingT′=−k(T−Ts)T' = -k(T - T_s)T=Ts+(T0−Ts)e−ktT = T_s + (T_0 - T_s)e^{-kt}
Mixing tankA′=(salt in per unit time)−AV (outflow rate)A' = (\text{salt in per unit time}) - \dfrac{A}{V}\,(\text{outflow rate})linear, solve with an integrating factor
RL circuitL I′+R I=EL\,I' + R\,I = E with constant EEI=ER+(I0−ER)e−Rt/LI = \dfrac{E}{R} + \left(I_0 - \dfrac{E}{R}\right)e^{-Rt/L}

In the logistic model with k,M>0k, M > 0, the carrying capacity MM is a stable equilibrium and the population grows fastest at P=M2P = \frac{M}{2}.

Lessons Logistic growth, Mixing, cooling and circuits

Numerical Methods and Mathematical Computing

Many problems have no exact answer in closed form. Numerical methods give approximate answers, and the point is to know how accurate they are.

Errors and reliability

Two kinds of error come up. Truncation error comes from stopping an infinite process, such as a series or a limit, after finitely many steps. Roundoff error comes from storing numbers with about 16 significant digits. Subtracting nearly equal numbers loses digits, so for small xx compute 1−cos⁡x1 - \cos x as 2sin⁡2x22\sin^2\frac{x}{2} instead. A graph on a screen is a finite sample of points, so it can suggest a root or a maximum, but only a theorem proves it.

Lessons Numerical reliability

Solving equations

Bisection. If ff is continuous on [a,b][a, b] and f(a)f(a) and f(b)f(b) have opposite signs, halve the interval and keep the half where the sign changes. After nn steps the midpoint is within b−a2n+1\frac{b - a}{2^{n+1}} of a root. It always works, but slowly.

Newton’s method. Follow the tangent line to the axis,

xn+1=xn−f(xn)f′(xn).x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.

Near a root rr with f′(r)≠0f'(r) \ne 0 and f′′f'' continuous, the number of correct digits roughly doubles each step. For f(x)=x2−2f(x) = x^2 - 2 starting at 11, the iterates are 1.51.5, 1.416671.41667 and 1.4142161.414216. A poor starting point can make it diverge or cycle.

Lessons Bisection and Newton’s method

Approximating derivatives and integrals

ApproximationFormulaError
Forward difference for f′(x)f'(x)f(x+h)−f(x)h\dfrac{f(x + h) - f(x)}{h}at most h2max⁡∣f′′∣\dfrac{h}{2}\max\lvert f'' \rvert
Central difference for f′(x)f'(x)f(x+h)−f(x−h)2h\dfrac{f(x + h) - f(x - h)}{2h}at most h26max⁡∣f′′′∣\dfrac{h^2}{6}\max\lvert f''' \rvert
Trapezoidal rule TnT_nΔx2[f(x0)+2f(x1)+⋯+2f(xn−1)+f(xn)]\dfrac{\Delta x}{2}\big[f(x_0) + 2f(x_1) + \cdots + 2f(x_{n-1}) + f(x_n)\big]at most K(b−a)312n2\dfrac{K(b - a)^3}{12n^2} with ∣f′′∣≤K\lvert f'' \rvert \le K
Midpoint rule MnM_nΔx[f(xˉ1)+⋯+f(xˉn)]\Delta x \big[f(\bar{x}_1) + \cdots + f(\bar{x}_n)\big]at most K(b−a)324n2\dfrac{K(b - a)^3}{24n^2} with ∣f′′∣≤K\lvert f'' \rvert \le K
Simpson’s rule SnS_n, nn evenΔx3[f(x0)+4f(x1)+2f(x2)+⋯+4f(xn−1)+f(xn)]\dfrac{\Delta x}{3}\big[f(x_0) + 4f(x_1) + 2f(x_2) + \cdots + 4f(x_{n-1}) + f(x_n)\big]at most K(b−a)5180n4\dfrac{K(b - a)^5}{180n^4} with ∣f(4)∣≤K\lvert f^{(4)} \rvert \le K

Here Δx=b−an\Delta x = \frac{b - a}{n}, xi=a+i Δxx_i = a + i\,\Delta x and xˉi\bar{x}_i is the midpoint of [xi−1,xi][x_{i-1}, x_i]. The maxima for the differences are over [x−h,x+h][x - h, x + h]. A smaller hh lowers the truncation error of a difference quotient but raises the roundoff error, so in double precision hh near 10−810^{-8} suits the forward difference and hh near 10−510^{-5} suits the central one. For ∫01e−x2 dx=0.746824…\int_0^1 e^{-x^2}\,dx = 0.746824\ldots with n=4n = 4, the rules give T4=0.742984T_4 = 0.742984, M4=0.748747M_4 = 0.748747 and S4=0.746855S_4 = 0.746855.

For an improper integral, cut it off at a finite point and bound the tail separately. For example, since x2≥4xx^2 \ge 4x for x≥4x \ge 4,

∫4∞e−x2 dx≤∫4∞e−4x dx=e−164≈2.8×10−8.\int_4^\infty e^{-x^2}\,dx \le \int_4^\infty e^{-4x}\,dx = \frac{e^{-16}}{4} \approx 2.8 \times 10^{-8}.

A Taylor series also works. Integrating the series of e−x2e^{-x^2} term by term, the first five terms give 1−13+110−142+1216=0.7474871 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \frac{1}{216} = 0.747487, with an error below the next term 11320\frac{1}{1320}.

Lessons Finite differences, Integral approximation rules, Improper integrals, Taylor’s formula

Solving differential equations numerically

To approximate the solution of y′=f(x,y)y' = f(x, y), y(x0)=y0y(x_0) = y_0, take steps of size hh along the direction field.

MethodStepError after a fixed distance
Euleryn+1=yn+h f(xn,yn)y_{n+1} = y_n + h\,f(x_n, y_n)proportional to hh
Improved Euleraverage the slopes at the start and at the Euler predictionproportional to h2h^2
Runge-Kutta of order 4a weighted average of four slopes across the stepproportional to h4h^4

Halving hh roughly halves the error of Euler’s method and divides the error of Runge-Kutta by 1616. For y′=yy' = y, y(0)=1y(0) = 1 and a single step of h=0.1h = 0.1, the three methods give 1.11.1, 1.1051.105 and 1.10517081.1051708, against the exact value e0.1=1.1051709e^{0.1} = 1.1051709.

Lessons Euler and Runge-Kutta methods