Mathematics

100 Best Calculus Formulas, Theorems, and Examples Every Student Should Know

100 calculus formulas and theorems, from limits and derivatives to series, multivariable and vector calculus and differential equations, each with an example.

Calculus is the mathematics of change and accumulation. Derivatives measure how fast something changes; integrals add up small pieces into a whole; and the fundamental theorem of calculus says these two operations undo each other. Almost everything else in a calculus course, from optimization to series to differential equations, is built from those ideas.

This article collects 100 formulas and theorems that a student meets across single-variable calculus, series, multivariable calculus, vector calculus and a first look at differential equations. For each one you get the statement, the conditions under which it holds (theorems are only as good as their hypotheses), and a short worked example. Every example was recomputed with SymPy before publication.

The basic differentiation and integration rules also appear, more briefly, in 100 mathematics formulas every student should know. This article goes further: it states the theorems behind the rules and covers series, several variables and vector fields. All angles are in radians.

Limits and continuity

1. The epsilon–delta definition of a limit

lim⁡x→af(x)=L  ⟺  ∀ε>0  ∃δ>0:0<∣x−a∣<δ⇒∣f(x)−L∣<ε\lim_{x \to a} f(x) = L \iff \forall \varepsilon > 0\; \exists \delta > 0 : 0 < |x - a| < \delta \Rightarrow |f(x) - L| < \varepsilon

Example: To show lim⁡x→2(3x+1)=7\lim_{x \to 2}(3x + 1) = 7: ∣3x+1−7∣=3∣x−2∣|3x + 1 - 7| = 3|x - 2|, so choosing δ=ε/3\delta = \varepsilon / 3 guarantees ∣f(x)−7∣<ε|f(x) - 7| < \varepsilon.

2. Limit laws

Limits of sums, differences, products and quotients are the sums, differences, products and quotients of the limits, provided each limit exists and no denominator limit is zero.

Example: lim⁡x→2(x2+3x)=4+6=10\lim_{x \to 2}(x^2 + 3x) = 4 + 6 = 10.

3. Squeeze theorem

If g(x)≤f(x)≤h(x)g(x) \le f(x) \le h(x) near aa and lim⁡g=lim⁡h=L\lim g = \lim h = L, then lim⁡f=L\lim f = L.

Example: −x2≤x2sin⁡(1/x)≤x2-x^2 \le x^2\sin(1/x) \le x^2, and both bounds go to 0, so lim⁡x→0x2sin⁡(1/x)=0\lim_{x \to 0} x^2\sin(1/x) = 0 even though sin⁡(1/x)\sin(1/x) oscillates wildly.

4. Limits at infinity of rational functions

Divide numerator and denominator by the highest power of xx in the denominator.

Example: lim⁡x→∞3x2+15x2−x=lim⁡x→∞3+1/x25−1/x=35\displaystyle\lim_{x \to \infty}\frac{3x^2 + 1}{5x^2 - x} = \lim_{x \to \infty}\frac{3 + 1/x^2}{5 - 1/x} = \frac{3}{5}.

5. The limit that defines e

lim⁡n→∞(1+1n)n=e≈2.71828\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e \approx 2.71828

Example: Compounding 100% interest nn times a year turns $1 into (1+1/n)n(1 + 1/n)^n dollars; as n→∞n \to \infty (continuous compounding) the amount approaches $2.72.

6. Exponential and logarithm limits

lim⁡x→0ex−1x=1,lim⁡x→0ln⁡(1+x)x=1\lim_{x \to 0}\frac{e^x - 1}{x} = 1, \qquad \lim_{x \to 0}\frac{\ln(1 + x)}{x} = 1

Example: lim⁡x→0e3x−1x=3⋅lim⁡x→0e3x−13x=3\displaystyle\lim_{x \to 0}\frac{e^{3x} - 1}{x} = 3 \cdot \lim_{x \to 0}\frac{e^{3x} - 1}{3x} = 3.

7. One-sided limits

lim⁡x→af(x)\lim_{x \to a} f(x) exists exactly when both one-sided limits exist and are equal.

Example: For f(x)=∣x∣/xf(x) = |x|/x, the left limit at 0 is −1-1 and the right limit is 11, so the two-sided limit does not exist.

8. Continuity and removable discontinuities

ff is continuous at aa when lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a).

Example: f(x)=x2−1x−1f(x) = \dfrac{x^2 - 1}{x - 1} is undefined at 1, but equals x+1x + 1 elsewhere, so the limit is 2. Defining f(1)=2f(1) = 2 removes the discontinuity.

9. Intermediate value theorem

If ff is continuous on [a,b][a, b] and NN lies between f(a)f(a) and f(b)f(b), then f(c)=Nf(c) = N for some cc in (a,b)(a, b).

Example: f(x)=x3+x−1f(x) = x^3 + x - 1 has f(0)=−1f(0) = -1 and f(1)=1f(1) = 1, so it has a root between 0 and 1. Bisection, the simplest root-finding algorithm, is this theorem applied repeatedly.

10. Extreme value theorem

A function continuous on a closed, bounded interval [a,b][a, b] attains a maximum and a minimum there.

Example: f(x)=x2f(x) = x^2 on [−1,2][-1, 2] attains its maximum 4 at x=2x = 2 and minimum 0 at x=0x = 0. On the open interval (0,1)(0, 1), f(x)=1/xf(x) = 1/x has no maximum: the hypothesis "closed" matters.

11. L'Hôpital's rule for 0/0

If f(a)=g(a)=0f(a) = g(a) = 0 and f,gf, g are differentiable near aa with g′(x)≠0g'(x) \ne 0, then lim⁡fg=lim⁡f′g′\lim \frac{f}{g} = \lim \frac{f'}{g'} when the right side exists.

Example: lim⁡x→01−cos⁡xx2=lim⁡x→0sin⁡x2x=12\displaystyle\lim_{x \to 0}\frac{1 - \cos x}{x^2} = \lim_{x \to 0}\frac{\sin x}{2x} = \frac{1}{2}.

12. The indeterminate form 1^∞

Take logarithms, find the limit of the logarithm, then exponentiate.

Example: lim⁡x→∞(1+2x)x\displaystyle\lim_{x \to \infty}\left(1 + \frac{2}{x}\right)^x: the log is xln⁡(1+2/x)→2x\ln(1 + 2/x) \to 2, so the limit is e2e^2.

Derivatives

13. Definition of the derivative

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

Example: For f(x)=x2f(x) = x^2 at x=1x = 1, the difference quotient is (1+h)2−1h=2+h→2\frac{(1 + h)^2 - 1}{h} = 2 + h \to 2.

The parabola y = x squared with the point (1, 1) marked. Secant lines through (1, 1) and (2, 4) with slope 3, and through (1, 1) and (1.5, 2.25) with slope 2.5, approach the tangent line with slope 2.
Formula 13: secant slopes 2 + h close in on the tangent slope 2 as h shrinks.

14. Differentiable implies continuous

If f′(a)f'(a) exists, ff is continuous at aa. The converse is false.

Example: ∣x∣|x| is continuous at 0 but not differentiable there: the difference quotient is −1-1 from the left and +1+1 from the right.

15. Power, constant multiple and sum rules

ddx(axn+bxm)=anxn−1+bmxm−1\frac{d}{dx}\left(a x^n + b x^m\right) = a n x^{n-1} + b m x^{m-1}

Example: ddx(4x3−5x+7)=12x2−5\frac{d}{dx}(4x^3 - 5x + 7) = 12x^2 - 5.

16. Product rule

(fg)′=f′g+fg′(fg)' = f'g + fg'

Example: ddx(x2ex)=2xex+x2ex=ex(x2+2x)\frac{d}{dx}(x^2e^x) = 2xe^x + x^2e^x = e^x(x^2 + 2x).

17. Quotient rule

(fg)′=f′g−fg′g2\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}

Example: ddxxx2+1=(x2+1)−x(2x)(x2+1)2=1−x2(x2+1)2\dfrac{d}{dx}\dfrac{x}{x^2 + 1} = \dfrac{(x^2 + 1) - x(2x)}{(x^2 + 1)^2} = \dfrac{1 - x^2}{(x^2 + 1)^2}.

18. Chain rule

ddxf(g(x))=f′(g(x)) g′(x)\frac{d}{dx}f(g(x)) = f'(g(x))\,g'(x)

Example: ddxsin⁡(x2)=cos⁡(x2)⋅2x\frac{d}{dx}\sin(x^2) = \cos(x^2) \cdot 2x.

19. Derivatives of tangent, secant, cotangent and cosecant

(tan⁡x)′=sec⁡2x,(sec⁡x)′=sec⁡xtan⁡x,(cot⁡x)′=−csc⁡2x,(csc⁡x)′=−csc⁡xcot⁡x(\tan x)' = \sec^2 x, \quad (\sec x)' = \sec x\tan x, \quad (\cot x)' = -\csc^2 x, \quad (\csc x)' = -\csc x\cot x

Example: ddxtan⁡(3x)=3sec⁡2(3x)\frac{d}{dx}\tan(3x) = 3\sec^2(3x).

20. Derivatives of inverse trigonometric functions

(arcsin⁡x)′=11−x2,(arctan⁡x)′=11+x2(\arcsin x)' = \frac{1}{\sqrt{1 - x^2}}, \qquad (\arctan x)' = \frac{1}{1 + x^2}

Example: The slope of arctan⁡x\arctan x at x=1x = 1 is 12\tfrac{1}{2}.

21. Exponentials and logarithms in any base

ddxax=axln⁡a,ddxlog⁡ax=1xln⁡a\frac{d}{dx}a^x = a^x\ln a, \qquad \frac{d}{dx}\log_a x = \frac{1}{x\ln a}

Example: ddx2x=2xln⁡2≈0.693⋅2x\frac{d}{dx}2^x = 2^x\ln 2 \approx 0.693 \cdot 2^x. So a quantity that doubles every unit of time is, at each instant, growing at about 69% of its current size per unit of time.

22. Implicit differentiation

Differentiate both sides with respect to xx, treating yy as a function of xx.

Example: For the circle x2+y2=25x^2 + y^2 = 25, 2x+2yy′=02x + 2yy' = 0, so y′=−x/yy' = -x/y. At (3,4)(3, 4) the slope is −34-\tfrac{3}{4}.

23. Logarithmic differentiation

For products, quotients and variable exponents, differentiate ln⁡y\ln y instead of yy.

Example: y=xxy = x^x: 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).

24. Derivative of an inverse function

(f−1)′(b)=1f′(a)where f(a)=b(f^{-1})'(b) = \frac{1}{f'(a)} \quad\text{where } f(a) = b

Example: f(x)=x3+xf(x) = x^3 + x has f(1)=2f(1) = 2 and f′(1)=4f'(1) = 4, so (f−1)′(2)=14(f^{-1})'(2) = \tfrac{1}{4}, without ever solving for f−1f^{-1}.

25. Higher derivatives

f′′f'' is the derivative of f′f', and so on. For position s(t)s(t), s′s' is velocity and s′′s'' is acceleration.

Example: dndxne2x=2ne2x\frac{d^n}{dx^n}e^{2x} = 2^ne^{2x}.

26. Hyperbolic functions

sinh⁡x=ex−e−x2,cosh⁡x=ex+e−x2,(sinh⁡x)′=cosh⁡x,(cosh⁡x)′=sinh⁡x\sinh x = \frac{e^x - e^{-x}}{2}, \quad \cosh x = \frac{e^x + e^{-x}}{2}, \quad (\sinh x)' = \cosh x, \quad (\cosh x)' = \sinh x

Example: A hanging chain or cable under its own weight takes the shape y=acosh⁡(x/a)y = a\cosh(x/a), the catenary.

27. Linear approximation

f(x)≈f(a)+f′(a)(x−a)for x near af(x) \approx f(a) + f'(a)(x - a) \quad\text{for } x \text{ near } a

Example: With f(x)=xf(x) = \sqrt{x} and a=4a = 4: 4.1≈2+0.14=2.025\sqrt{4.1} \approx 2 + \frac{0.1}{4} = 2.025. The true value is 2.024852.02485.

28. Differentials and error propagation

dy=f′(x) dxdy = f'(x)\,dx

Example: A sphere's radius is measured as 10±0.110 \pm 0.1 cm. Since V=43πr3V = \frac{4}{3}\pi r^3, dV=4πr2 dr=4π(100)(0.1)≈125.7 cm3dV = 4\pi r^2\,dr = 4\pi(100)(0.1) \approx 125.7\ \text{cm}^3, about 3% of the volume, three times the 1% error in rr.

Differentiate an equation linking the quantities with respect to time.

Example: A 10 ft ladder slides down a wall. When the base is 6 ft out and moving away at 2 ft/s, x2+y2=100x^2 + y^2 = 100 gives xx˙+yy˙=0x\dot{x} + y\dot{y} = 0. With y=8y = 8: y˙=−6⋅28=−1.5\dot{y} = -\frac{6 \cdot 2}{8} = -1.5 ft/s. The top is falling at 1.5 ft/s.

30. Newton's method

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

Example: For 2\sqrt{2}, use f(x)=x2−2f(x) = x^2 - 2 and x0=1x_0 = 1: x1=1.5x_1 = 1.5, x2=1.41667x_2 = 1.41667, x3=1.414216x_3 = 1.414216. The number of correct digits roughly doubles each step.

31. Mean value theorem

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

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

Example: f(x)=x3−3xf(x) = x^3 - 3x on [0,2][0, 2] has average slope 2−02=1\frac{2 - 0}{2} = 1. Solving 3c2−3=13c^2 - 3 = 1 gives c=2/3≈1.155c = 2/\sqrt{3} \approx 1.155. If a car's average speed over a trip is 60 km/h, the speedometer read exactly 60 at some instant.

The curve y = x cubed minus 3x from x = 0 to 2. A dashed secant joins (0, 0) and (2, 2) with slope 1, and a parallel tangent line touches the curve at x = 2 over root 3, about 1.155.
Theorem 31: somewhere between the endpoints, the tangent is parallel to the secant.

32. Rolle's theorem

The special case of the mean value theorem with f(a)=f(b)f(a) = f(b): then f′(c)=0f'(c) = 0 for some cc in (a,b)(a, b).

Example: f(x)=x2−4xf(x) = x^2 - 4x has f(0)=f(4)=0f(0) = f(4) = 0, and f′(c)=2c−4=0f'(c) = 2c - 4 = 0 at c=2c = 2.

Applications of derivatives

33. Critical points and the first derivative test

Critical points are where f′=0f' = 0 or f′f' is undefined. If f′f' changes from ++ to −-, there is a local maximum; from −- to ++, a local minimum.

Example: f(x)=x3−3xf(x) = x^3 - 3x has f′(x)=3(x−1)(x+1)f'(x) = 3(x - 1)(x + 1). It has a local maximum f(−1)=2f(-1) = 2 and a local minimum f(1)=−2f(1) = -2.

34. Concavity and the second derivative test

f′′>0f'' > 0 means concave up; f′′<0f'' < 0 means concave down. At a critical point, f′′>0f'' > 0 gives a minimum and f′′<0f'' < 0 a maximum. An inflection point is where concavity changes.

Example: For f(x)=x3−3xf(x) = x^3 - 3x, f′′(x)=6xf''(x) = 6x: f′′(1)=6>0f''(1) = 6 > 0 confirms the minimum, and x=0x = 0 is an inflection point.

35. Optimization: the open-top box

Write the quantity as a function of one variable, find critical points, and check the endpoints.

Example: Cut squares of side xx from the corners of a 12×1212 \times 12 sheet and fold up the sides. V(x)=x(12−2x)2V(x) = x(12 - 2x)^2 on [0,6][0, 6]. V′(x)=(12−2x)(12−6x)=0V'(x) = (12 - 2x)(12 - 6x) = 0 gives x=2x = 2, so the largest volume is V(2)=2⋅64=128V(2) = 2 \cdot 64 = 128.

36. Optimization: the cheapest can

Example: A cylinder must hold 1000 cm³ with the least surface area S=2πr2+2πrhS = 2\pi r^2 + 2\pi rh. Using h=1000/(πr2)h = 1000/(\pi r^2), S(r)=2πr2+2000/rS(r) = 2\pi r^2 + 2000/r, and S′(r)=4πr−2000/r2=0S'(r) = 4\pi r - 2000/r^2 = 0 gives r=(500/π)1/3≈5.42r = (500/\pi)^{1/3} \approx 5.42 cm and h≈10.84h \approx 10.84 cm. The optimal can's height equals its diameter.

37. Comparing growth rates

As x→∞x \to \infty, for any n>0n > 0 and p>0p > 0:

ln⁡xxp→0,xnex→0\frac{\ln x}{x^p} \to 0, \qquad \frac{x^n}{e^x} \to 0

Example: Applying L'Hôpital's rule three times, x3ex→6ex→0\frac{x^3}{e^x} \to \frac{6}{e^x} \to 0. Exponentials beat polynomials, which beat logarithms. This ordering is the basis of big-O notation in algorithm analysis.

38. Asymptotes

Vertical asymptotes occur where a function grows without bound; horizontal asymptotes are limits at ±∞\pm\infty.

Example: f(x)=2x2x2−1f(x) = \dfrac{2x^2}{x^2 - 1} has vertical asymptotes x=±1x = \pm 1 and horizontal asymptote y=2y = 2.

39. Marginal cost

In economics, the derivative of cost is the marginal cost: the approximate cost of one more unit.

Example: For C(x)=0.01x2+5x+100C(x) = 0.01x^2 + 5x + 100 dollars, C′(x)=0.02x+5C'(x) = 0.02x + 5, so C′(100)=7C'(100) = 7: the 101st unit costs about $7.

40. Velocity and acceleration

v(t)=s′(t)v(t) = s'(t) and a(t)=v′(t)=s′′(t)a(t) = v'(t) = s''(t). The object changes direction where vv changes sign.

Example: s(t)=t3−6t2+9ts(t) = t^3 - 6t^2 + 9t gives v(t)=3(t−1)(t−3)v(t) = 3(t - 1)(t - 3). The object moves forward until t=1t = 1, backward until t=3t = 3, then forward again.

Integrals

41. Riemann sums and the definite integral

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

Example: For ∫02x2 dx\int_0^2 x^2\,dx with 4 midpoint rectangles of width 0.5: 0.5(0.0625+0.5625+1.5625+3.0625)=2.6250.5(0.0625 + 0.5625 + 1.5625 + 3.0625) = 2.625. The exact value is 83≈2.667\tfrac{8}{3} \approx 2.667.

The curve y = x squared from 0 to 2 with four rectangles of width 0.5 whose heights are the function values at the midpoints 0.25, 0.75, 1.25 and 1.75. Text notes the midpoint sum 2.625 and the exact area 8/3, about 2.667.
Formula 41: four midpoint rectangles already come within 0.042 of the exact area.

42. Properties of the definite integral

∫ab(cf+g)=c∫abf+∫abg,∫abf=∫acf+∫cbf,∫baf=−∫abf\int_a^b (cf + g) = c\int_a^b f + \int_a^b g, \qquad \int_a^b f = \int_a^c f + \int_c^b f, \qquad \int_b^a f = -\int_a^b f

Example: Since sin⁡x\sin x is odd, ∫−ππsin⁡x dx=0\int_{-\pi}^{\pi}\sin x\,dx = 0: the areas above and below the axis cancel.

43. Fundamental theorem of calculus, part 1

If ff is continuous, ddx∫axf(t) dt=f(x)\displaystyle\frac{d}{dx}\int_a^x f(t)\,dt = f(x). With a variable upper limit g(x)g(x), multiply by g′(x)g'(x).

Example: ddx∫0x2cos⁡(t2) dt=cos⁡(x4)⋅2x\displaystyle\frac{d}{dx}\int_0^{x^2}\cos(t^2)\,dt = \cos(x^4) \cdot 2x, even though cos⁡(t2)\cos(t^2) has no elementary antiderivative.

44. Fundamental theorem of calculus, part 2

∫abf(x) dx=F(b)−F(a)where F′=f\int_a^b f(x)\,dx = F(b) - F(a) \quad\text{where } F' = f

Example: ∫13x2 dx=273−13=263\int_1^3 x^2\,dx = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}.

45. Antiderivatives to know by heart

∫sec⁡2x dx=tan⁡x+C,∫dx1+x2=arctan⁡x+C,∫dx1−x2=arcsin⁡x+C\int\sec^2 x\,dx = \tan x + C, \quad \int\frac{dx}{1 + x^2} = \arctan x + C, \quad \int\frac{dx}{\sqrt{1 - x^2}} = \arcsin x + C

Example: ∫01dx1+x2=arctan⁡1=π4\int_0^1\frac{dx}{1 + x^2} = \arctan 1 = \frac{\pi}{4}.

46. Substitution

∫f(g(x)) g′(x) dx=∫f(u) du,u=g(x)\int f(g(x))\,g'(x)\,dx = \int f(u)\,du, \quad u = g(x)

Example: With u=x2u = x^2, ∫01xex2 dx=12∫01eu du=e−12\int_0^1 xe^{x^2}\,dx = \frac{1}{2}\int_0^1 e^u\,du = \frac{e - 1}{2}. Change the limits along with the variable.

47. Integration by parts

∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du

Example: ∫xex dx\int xe^x\,dx with u=xu = x, dv=exdxdv = e^xdx: xex−∫ex dx=(x−1)ex+Cxe^x - \int e^x\,dx = (x - 1)e^x + C. Also ∫ln⁡x dx=xln⁡x−x+C\int\ln x\,dx = x\ln x - x + C (take u=ln⁡xu = \ln x, dv=dxdv = dx).

48. Repeated integration by parts

When a polynomial multiplies an exponential or trig function, apply parts until the polynomial disappears.

Example: ∫x2ex dx=ex(x2−2x+2)+C\int x^2e^x\,dx = e^x(x^2 - 2x + 2) + C. Check by differentiating: ex(x2−2x+2)+ex(2x−2)=x2exe^x(x^2 - 2x + 2) + e^x(2x - 2) = x^2e^x.

49. Powers of sine and cosine

Use sin⁡2x=1−cos⁡2x2\sin^2 x = \frac{1 - \cos 2x}{2} and cos⁡2x=1+cos⁡2x2\cos^2 x = \frac{1 + \cos 2x}{2}.

Example: ∫0πsin⁡2x dx=[x2−sin⁡2x4]0π=π2\int_0^{\pi}\sin^2 x\,dx = \left[\frac{x}{2} - \frac{\sin 2x}{4}\right]_0^{\pi} = \frac{\pi}{2}.

50. Trigonometric substitution

For a2−x2\sqrt{a^2 - x^2} use x=asin⁡θx = a\sin\theta; for a2+x2\sqrt{a^2 + x^2}, x=atan⁡θx = a\tan\theta; for x2−a2\sqrt{x^2 - a^2}, x=asec⁡θx = a\sec\theta.

Example: With x=sin⁡θx = \sin\theta, ∫011−x2 dx=∫0π/2cos⁡2θ dθ=π4\int_0^1\sqrt{1 - x^2}\,dx = \int_0^{\pi/2}\cos^2\theta\,d\theta = \frac{\pi}{4}: a quarter of the unit disk.

51. Partial fractions

Split a rational function into simpler fractions before integrating.

Example: 1x2−1=12(1x−1−1x+1)\dfrac{1}{x^2 - 1} = \dfrac{1}{2}\left(\dfrac{1}{x - 1} - \dfrac{1}{x + 1}\right), so ∫dxx2−1=12ln⁡∣x−1x+1∣+C\displaystyle\int\frac{dx}{x^2 - 1} = \frac{1}{2}\ln\left|\frac{x - 1}{x + 1}\right| + C.

52. Convergent improper integrals

∫1∞dxxp=1p−1for p>1\int_1^\infty\frac{dx}{x^p} = \frac{1}{p - 1} \quad\text{for } p > 1

Example: ∫1∞dxx2=lim⁡b→∞(1−1b)=1\int_1^\infty\frac{dx}{x^2} = \lim_{b \to \infty}\left(1 - \frac{1}{b}\right) = 1. An infinitely long region can have finite area.

53. Divergent improper integrals and the Gaussian integral

For p≤1p \le 1 the integral above diverges: ∫1∞dxx=lim⁡b→∞ln⁡b=∞\int_1^\infty\frac{dx}{x} = \lim_{b \to \infty}\ln b = \infty. A famous convergent one:

∫−∞∞e−x2 dx=π\int_{-\infty}^{\infty}e^{-x^2}\,dx = \sqrt{\pi}

Example: This is why the normal distribution's density contains 12π\frac{1}{\sqrt{2\pi}} (see formula 83 for the proof).

54. Average value of a function

favg=1b−a∫abf(x) dxf_{\text{avg}} = \frac{1}{b - a}\int_a^b f(x)\,dx

Example: The average of sin⁡x\sin x over [0,π][0, \pi] is 1π∫0πsin⁡x dx=2π≈0.637\frac{1}{\pi}\int_0^\pi\sin x\,dx = \frac{2}{\pi} \approx 0.637.

55. Area between curves

A=∫ab[f(x)−g(x)]dx,f≥gA = \int_a^b\left[f(x) - g(x)\right]dx, \quad f \ge g

Example: Between y=xy = x and y=x2y = x^2 on [0,1][0, 1]: ∫01(x−x2) dx=12−13=16\int_0^1(x - x^2)\,dx = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}.

56. Volume by disks

V=π∫ab[f(x)]2 dxV = \pi\int_a^b[f(x)]^2\,dx

Example: Rotating y=xy = \sqrt{x}, 0≤x≤40 \le x \le 4, about the xx-axis: V=π∫04x dx=8πV = \pi\int_0^4 x\,dx = 8\pi.

57. Volume by cylindrical shells

V=2π∫abx f(x) dx(rotation about the y-axis)V = 2\pi\int_a^b x\,f(x)\,dx \quad\text{(rotation about the } y\text{-axis)}

Example: Rotating y=x−x2y = x - x^2, 0≤x≤10 \le x \le 1, about the yy-axis: V=2π∫01(x2−x3) dx=2π(13−14)=π6V = 2\pi\int_0^1(x^2 - x^3)\,dx = 2\pi\left(\frac{1}{3} - \frac{1}{4}\right) = \frac{\pi}{6}.

58. Arc length

L=∫ab1+[f′(x)]2 dxL = \int_a^b\sqrt{1 + [f'(x)]^2}\,dx

Example: For y=23x3/2y = \frac{2}{3}x^{3/2} on [0,3][0, 3], f′(x)=xf'(x) = \sqrt{x}, so L=∫031+x dx=23(8−1)=143L = \int_0^3\sqrt{1 + x}\,dx = \frac{2}{3}\left(8 - 1\right) = \frac{14}{3}.

59. Work done by a variable force

W=∫abF(x) dxW = \int_a^b F(x)\,dx

Example: A spring with k=50k = 50 N/m obeys F=kxF = kx. Stretching it 0.2 m from rest takes W=∫00.250x dx=25(0.04)=1W = \int_0^{0.2}50x\,dx = 25(0.04) = 1 J.

60. Trapezoidal rule and Simpson's rule

Tn=h2[f0+2f1+⋯+2fn−1+fn],Sn=h3[f0+4f1+2f2+⋯+4fn−1+fn]T_n = \frac{h}{2}\left[f_0 + 2f_1 + \cdots + 2f_{n-1} + f_n\right], \qquad S_n = \frac{h}{3}\left[f_0 + 4f_1 + 2f_2 + \cdots + 4f_{n-1} + f_n\right]

Example: ∫0141+x2 dx=π\int_0^1\frac{4}{1 + x^2}\,dx = \pi. With n=2n = 2 (h=0.5h = 0.5, values 4,3.2,24, 3.2, 2): T2=3.1T_2 = 3.1 and S2=16(4+12.8+2)≈3.1333S_2 = \frac{1}{6}(4 + 12.8 + 2) \approx 3.1333, against π≈3.1416\pi \approx 3.1416. Simpson's rule is exact for cubics.

Sequences and series

61. Geometric series

∑n=0∞arn=a1−r,∣r∣<1\sum_{n=0}^{\infty}ar^n = \frac{a}{1 - r}, \quad |r| < 1

Example: 0.9‾=∑n=1∞9⋅10−n=0.91−0.1=10.\overline{9} = \sum_{n=1}^\infty 9 \cdot 10^{-n} = \frac{0.9}{1 - 0.1} = 1.

62. Divergence test and the harmonic series

If an↛0a_n \not\to 0, ∑an\sum a_n diverges. The converse fails: an→0a_n \to 0 does not guarantee convergence.

Example: The harmonic series ∑1n\sum\frac{1}{n} has terms going to 0 but diverges. Grouping terms, 13+14>12\frac{1}{3} + \frac{1}{4} > \frac{1}{2}, 15+⋯+18>12\frac{1}{5} + \cdots + \frac{1}{8} > \frac{1}{2}, and so on forever.

63. p-series and the Basel problem

∑1np\sum\frac{1}{n^p} converges exactly when p>1p > 1. Euler found the value for p=2p = 2:

∑n=1∞1n2=π26≈1.6449\sum_{n=1}^\infty\frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.6449

Example: ∑1n1.01\sum\frac{1}{n^{1.01}} converges, though very slowly; ∑1n\sum\frac{1}{\sqrt{n}} diverges.

64. Integral test

If ff is positive, continuous and decreasing with an=f(n)a_n = f(n), then ∑an\sum a_n and ∫1∞f(x) dx\int_1^\infty f(x)\,dx both converge or both diverge.

Example: ∑1nln⁡n\sum\frac{1}{n\ln n} (from n=2n = 2) diverges because ∫2∞dxxln⁡x=lim⁡b→∞ln⁡(ln⁡b)=∞\int_2^\infty\frac{dx}{x\ln x} = \lim_{b \to \infty}\ln(\ln b) = \infty.

65. Comparison test

If 0≤an≤bn0 \le a_n \le b_n and ∑bn\sum b_n converges, so does ∑an\sum a_n.

Example: 1n2+1<1n2\frac{1}{n^2 + 1} < \frac{1}{n^2}, so ∑1n2+1\sum\frac{1}{n^2 + 1} converges.

66. Ratio test

If lim⁡∣an+1an∣=L\lim\left|\frac{a_{n+1}}{a_n}\right| = L: L<1L < 1 converges absolutely, L>1L > 1 diverges, L=1L = 1 is inconclusive.

Example: For ∑n2n\sum\frac{n}{2^n}, the ratio is n+12n→12\frac{n + 1}{2n} \to \frac{1}{2}, so it converges (its sum is 2).

67. Alternating series test

If bnb_n decreases to 0, ∑(−1)n+1bn\sum(-1)^{n+1}b_n converges, and the error after NN terms is at most bN+1b_{N+1}.

Example: 1−12+13−⋯=ln⁡2≈0.69311 - \frac{1}{2} + \frac{1}{3} - \cdots = \ln 2 \approx 0.6931. After 100 terms the error is below 1101\frac{1}{101}.

68. Absolute and conditional convergence

A series converges absolutely if ∑∣an∣\sum|a_n| converges; it converges conditionally if it converges but not absolutely.

Example: The alternating harmonic series in formula 67 converges conditionally. By the Riemann rearrangement theorem its terms can be reordered to sum to any number you like, so the order of terms matters.

69. Radius of convergence of a power series

For ∑cn(x−a)n\sum c_n(x - a)^n, the ratio test usually gives R=lim⁡∣cncn+1∣R = \lim\left|\frac{c_n}{c_{n+1}}\right|. Check the endpoints separately.

Example: ∑xnn\sum\frac{x^n}{n} has R=1R = 1. At x=1x = 1 it is the harmonic series (diverges); at x=−1x = -1 it is alternating (converges). The interval is [−1,1)[-1, 1).

70. Taylor and Maclaurin series

f(x)=∑n=0∞f(n)(a)n!(x−a)nf(x) = \sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x - a)^n ex=∑xnn!,sin⁡x=∑(−1)nx2n+1(2n+1)!,cos⁡x=∑(−1)nx2n(2n)!e^x = \sum\frac{x^n}{n!}, \quad \sin x = \sum\frac{(-1)^nx^{2n+1}}{(2n + 1)!}, \quad \cos x = \sum\frac{(-1)^nx^{2n}}{(2n)!}

Example: sin⁡0.1≈0.1−0.0016=0.0998333\sin 0.1 \approx 0.1 - \frac{0.001}{6} = 0.0998333, correct to 7 decimal places.

The sine curve from minus 2 pi to 2 pi with its Taylor polynomials of degree 1, 3 and 5 around 0. Each higher-degree polynomial follows the sine curve over a wider interval before diverging.
Formula 70: each extra term of the series widens the interval where the polynomial matches sin x.

71. Taylor's theorem with remainder

The error of the degree-nn Taylor polynomial is

Rn(x)=f(n+1)(c)(n+1)!(x−a)n+1for some c between a and xR_n(x) = \frac{f^{(n+1)}(c)}{(n + 1)!}(x - a)^{n+1} \quad\text{for some } c \text{ between } a \text{ and } x

Example: To estimate ee with ∑k=081k!=2.7182788\sum_{k=0}^{8}\frac{1}{k!} = 2.7182788, the error is at most e9!<3362 880≈8.3×10−6\frac{e}{9!} < \frac{3}{362\,880} \approx 8.3 \times 10^{-6}. The true error is about 3.0×10−63.0 \times 10^{-6}.

72. Binomial series

(1+x)k=1+kx+k(k−1)2!x2+⋯ ,∣x∣<1(1 + x)^k = 1 + kx + \frac{k(k - 1)}{2!}x^2 + \cdots, \quad |x| < 1

Example: 1.1=(1+0.1)1/2≈1+0.05−0.018=1.04875\sqrt{1.1} = (1 + 0.1)^{1/2} \approx 1 + 0.05 - \frac{0.01}{8} = 1.04875. The true value is 1.0488091.048809.

Multivariable calculus

73. Partial derivatives

Differentiate with respect to one variable while holding the others fixed.

Example: For f(x,y)=x2y+sin⁡(xy)f(x, y) = x^2y + \sin(xy): fx=2xy+ycos⁡(xy)f_x = 2xy + y\cos(xy) and fy=x2+xcos⁡(xy)f_y = x^2 + x\cos(xy).

74. Gradient and directional derivative

∇f=(fx,fy),Duf=∇f⋅u(∥u∥=1)\nabla f = (f_x, f_y), \qquad D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u} \quad (\lVert\mathbf{u}\rVert = 1)

Example: f(x,y)=x2yf(x, y) = x^2y at (1,2)(1, 2) has ∇f=(2xy,x2)=(4,1)\nabla f = (2xy, x^2) = (4, 1). In the direction u=(3,4)/5\mathbf{u} = (3, 4)/5: Duf=12+45=3.2D_{\mathbf{u}}f = \frac{12 + 4}{5} = 3.2.

75. The gradient points uphill

∇f\nabla f points in the direction of steepest increase, its length is the maximum rate of increase, and it is perpendicular to level curves.

Example: For ff in formula 74, the steepest rate at (1,2)(1, 2) is ∥(4,1)∥=17≈4.12\lVert(4, 1)\rVert = \sqrt{17} \approx 4.12. Gradient descent in machine learning steps in the direction −∇f-\nabla f.

76. Tangent plane

z=f(a,b)+fx(a,b)(x−a)+fy(a,b)(y−b)z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)

Example: For z=x2+y2z = x^2 + y^2 at (1,2)(1, 2): z=5+2(x−1)+4(y−2)z = 5 + 2(x - 1) + 4(y - 2).

77. Multivariable chain rule

dzdt=∂z∂xdxdt+∂z∂ydydt\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}

Example: z=x2yz = x^2y with x=cos⁡tx = \cos t, y=sin⁡ty = \sin t: dzdt=2xy(−sin⁡t)+x2cos⁡t=cos⁡t (1−3sin⁡2t)\frac{dz}{dt} = 2xy(-\sin t) + x^2\cos t = \cos t\,(1 - 3\sin^2 t). Backpropagation in neural networks is this rule applied through many layers.

78. Equality of mixed partials (Clairaut's theorem)

If fxyf_{xy} and fyxf_{yx} are continuous, they are equal.

Example: For f=x3y2f = x^3y^2: fxy=∂∂y(3x2y2)=6x2yf_{xy} = \frac{\partial}{\partial y}(3x^2y^2) = 6x^2y and fyx=∂∂x(2x3y)=6x2yf_{yx} = \frac{\partial}{\partial x}(2x^3y) = 6x^2y.

79. Second derivative test in two variables

At a critical point, let D=fxxfyy−fxy2D = f_{xx}f_{yy} - f_{xy}^2. D>0D > 0 and fxx>0f_{xx} > 0: minimum. D>0D > 0 and fxx<0f_{xx} < 0: maximum. D<0D < 0: saddle point.

Example: f(x,y)=x3−3x+y2f(x, y) = x^3 - 3x + y^2 has critical points (±1,0)(\pm 1, 0) and D=12xD = 12x. At (1,0)(1, 0), D=12>0D = 12 > 0 and fxx=6>0f_{xx} = 6 > 0: a local minimum. At (−1,0)(-1, 0), D=−12D = -12: a saddle.

80. Lagrange multipliers

To optimize ff subject to g=cg = c, solve ∇f=λ∇g\nabla f = \lambda\nabla g together with g=cg = c.

Example: Maximize xyxy subject to x+y=10x + y = 10: (y,x)=λ(1,1)(y, x) = \lambda(1, 1) gives x=y=5x = y = 5 and a maximum product of 25. The optimization article works through the geometry with a diagram.

81. Double integrals over rectangles (Fubini's theorem)

For continuous ff on a rectangle, the double integral equals either iterated integral.

Example: ∬[0,1]×[0,2]xy dA=∫01x dx∫02y dy=12⋅2=1\iint_{[0,1]\times[0,2]}xy\,dA = \int_0^1 x\,dx\int_0^2 y\,dy = \frac{1}{2} \cdot 2 = 1.

82. Double integrals over general regions

Describe the region with variable limits.

Example: Over the region between y=x2y = x^2 and y=xy = x for 0≤x≤10 \le x \le 1: ∬x dA=∫01x(x−x2) dx=13−14=112\iint x\,dA = \int_0^1 x(x - x^2)\,dx = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}.

83. Polar coordinates

∬Rf dA=∬f(rcos⁡θ,rsin⁡θ) r dr dθ\iint_R f\,dA = \iint f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta

Example: The Gaussian integral: (∫e−x2dx)2=∬e−(x2+y2) dA=∫02π∫0∞e−r2r dr dθ=2π⋅12=π\left(\int e^{-x^2}dx\right)^2 = \iint e^{-(x^2+y^2)}\,dA = \int_0^{2\pi}\int_0^\infty e^{-r^2}r\,dr\,d\theta = 2\pi \cdot \frac{1}{2} = \pi, so ∫−∞∞e−x2dx=π\int_{-\infty}^\infty e^{-x^2}dx = \sqrt{\pi}.

84. Change of variables and the Jacobian

dx dy=∣∂(x,y)∂(u,v)∣du dvdx\,dy = \left|\frac{\partial(x, y)}{\partial(u, v)}\right|du\,dv

Example: For polar coordinates the Jacobian determinant is ∣cos⁡θ−rsin⁡θsin⁡θrcos⁡θ∣=r\begin{vmatrix}\cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta\end{vmatrix} = r. For spherical coordinates dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta.

85. Triple integrals

Example: The volume of a ball of radius RR in spherical coordinates:

∫02π∫0π∫0Rρ2sin⁡ϕ dρ dϕ dθ=2π⋅2⋅R33=43πR3\int_0^{2\pi}\int_0^{\pi}\int_0^R\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi \cdot 2 \cdot \frac{R^3}{3} = \frac{4}{3}\pi R^3

86. Center of mass

xˉ=1A∬Rx dA,yˉ=1A∬Ry dA\bar{x} = \frac{1}{A}\iint_R x\,dA, \qquad \bar{y} = \frac{1}{A}\iint_R y\,dA

Example: For a half-disk of radius rr above the xx-axis, ∬y dA=∫0π∫0rρ2sin⁡θ dρ dθ=2r33\iint y\,dA = \int_0^\pi\int_0^r\rho^2\sin\theta\,d\rho\,d\theta = \frac{2r^3}{3} and A=πr22A = \frac{\pi r^2}{2}, so yˉ=4r3π≈0.42r\bar{y} = \frac{4r}{3\pi} \approx 0.42r.

87. Line integrals and conservative fields

If F=∇ϕ\mathbf{F} = \nabla\phi, then ∫CF⋅dr=ϕ(end)−ϕ(start)\int_C\mathbf{F} \cdot d\mathbf{r} = \phi(\text{end}) - \phi(\text{start}) for any path.

Example: F=(y,x)=∇(xy)\mathbf{F} = (y, x) = \nabla(xy). The work from (0,0)(0, 0) to (1,2)(1, 2) along any path is 1⋅2−0=21 \cdot 2 - 0 = 2.

88. Green's theorem

∮CP dx+Q dy=∬R(∂Q∂x−∂P∂y)dA\oint_C P\,dx + Q\,dy = \iint_R\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dA

Example: With P=−y/2P = -y/2 and Q=x/2Q = x/2 the integrand is 1, so 12∮(x dy−y dx)\frac{1}{2}\oint(x\,dy - y\,dx) is the area enclosed. For a circle x=rcos⁡tx = r\cos t, y=rsin⁡ty = r\sin t, it gives 12∫02πr2 dt=πr2\frac{1}{2}\int_0^{2\pi}r^2\,dt = \pi r^2. A planimeter measures area this way.

Vector calculus

89. Divergence and curl

∇⋅F=∂F1∂x+∂F2∂y+∂F3∂z,∇×F=∣ijk∂x∂y∂zF1F2F3∣\nabla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}, \qquad \nabla \times \mathbf{F} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ F_1 & F_2 & F_3\end{vmatrix}

Example: For F=(x2,xy,z)\mathbf{F} = (x^2, xy, z): ∇⋅F=2x+x+1=3x+1\nabla \cdot \mathbf{F} = 2x + x + 1 = 3x + 1 and ∇×F=(0,0,y)\nabla \times \mathbf{F} = (0, 0, y). Divergence measures outflow from a point; curl measures rotation.

90. Divergence theorem

∬SF⋅n dS=∭V∇⋅F dV\iint_S\mathbf{F} \cdot \mathbf{n}\,dS = \iiint_V\nabla \cdot \mathbf{F}\,dV

Example: For F=(x,y,z)\mathbf{F} = (x, y, z) and the unit ball, ∇⋅F=3\nabla \cdot \mathbf{F} = 3, so the flux is 3⋅43π=4π3 \cdot \frac{4}{3}\pi = 4\pi. Directly: on the unit sphere F⋅n=1\mathbf{F} \cdot \mathbf{n} = 1, and the area is 4π4\pi. Gauss's law for electric fields is this theorem.

91. Stokes' theorem

∮CF⋅dr=∬S(∇×F)⋅n dS\oint_C\mathbf{F} \cdot d\mathbf{r} = \iint_S(\nabla \times \mathbf{F}) \cdot \mathbf{n}\,dS

Example: F=(−y,x,0)\mathbf{F} = (-y, x, 0) has curl (0,0,2)(0, 0, 2). Over the unit disk, ∬2 dA=2π\iint 2\,dA = 2\pi, which equals the circulation ∮(−y dx+x dy)=∫02π1 dt=2π\oint(-y\,dx + x\,dy) = \int_0^{2\pi}1\,dt = 2\pi around the unit circle.

92. Test for a conservative field

On a simply connected region, a smooth field is conservative exactly when its curl is zero (in 2D: ∂Q/∂x=∂P/∂y\partial Q/\partial x = \partial P/\partial y).

Example: F=(2xy,x2)\mathbf{F} = (2xy, x^2): ∂x(x2)=2x=∂y(2xy)\partial_x(x^2) = 2x = \partial_y(2xy), so F=∇(x2y)\mathbf{F} = \nabla(x^2y). Gravity and electrostatic fields are conservative, which is why potential energy is well defined.

Differential equations

93. Exponential growth and decay

dydt=ky⟹y=y0ekt\frac{dy}{dt} = ky \quad\Longrightarrow\quad y = y_0e^{kt}

Example: Carbon-14 has a half-life of about 5,730 years, so k=−ln⁡25730k = -\frac{\ln 2}{5730}. A sample with 25% of its original carbon-14 is two half-lives old, about 11,460 years.

94. Newton's law of cooling

dTdt=−k(T−Troom)⟹T=Troom+(T0−Troom)e−kt\frac{dT}{dt} = -k(T - T_{\text{room}}) \quad\Longrightarrow\quad T = T_{\text{room}} + (T_0 - T_{\text{room}})e^{-kt}

Example: Coffee at 90 °C in a 20 °C room cools to 60 °C in 10 minutes, so e−10k=4070e^{-10k} = \frac{40}{70} and k≈0.0560k \approx 0.0560 per minute. It reaches 40 °C when e−kt=2070e^{-kt} = \frac{20}{70}, at t=ln⁡3.5k≈22.4t = \frac{\ln 3.5}{k} \approx 22.4 minutes.

95. Logistic growth

dPdt=rP(1−PK)⟹P(t)=K1+K−P0P0e−rt\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) \quad\Longrightarrow\quad P(t) = \frac{K}{1 + \frac{K - P_0}{P_0}e^{-rt}}

Example: Growth is fastest when P=K/2P = K/2, where the curve has its inflection point. Populations, technology adoption and the spread of rumors often follow this S-shaped curve.

96. First-order linear equations (integrating factor)

For y′+p(x)y=q(x)y' + p(x)y = q(x), multiply by μ=e∫p dx\mu = e^{\int p\,dx}; then (μy)′=μq(\mu y)' = \mu q.

Example: y′+y=xy' + y = x: μ=ex\mu = e^x, (exy)′=xex(e^xy)' = xe^x, so exy=(x−1)ex+Ce^xy = (x - 1)e^x + C and y=x−1+Ce−xy = x - 1 + Ce^{-x}.

97. Second-order linear equations with constant coefficients

For ay′′+by′+cy=0ay'' + by' + cy = 0, try y=erxy = e^{rx} and solve ar2+br+c=0ar^2 + br + c = 0.

Example: y′′−3y′+2y=0y'' - 3y' + 2y = 0 gives r=1,2r = 1, 2, so y=c1ex+c2e2xy = c_1e^x + c_2e^{2x}.

98. Simple harmonic motion

my′′+ky=0⟹y=Acos⁡(ωt)+Bsin⁡(ωt),ω=k/mmy'' + ky = 0 \quad\Longrightarrow\quad y = A\cos(\omega t) + B\sin(\omega t), \quad \omega = \sqrt{k/m}

Example: A 0.5 kg mass on a spring with k=8k = 8 N/m oscillates with ω=4\omega = 4 rad/s and period 2π4=π2≈1.57\frac{2\pi}{4} = \frac{\pi}{2} \approx 1.57 s.

99. Damped oscillations

my′′+by′+ky=0my'' + by' + ky = 0 has characteristic roots r=−b±b2−4mk2mr = \frac{-b \pm \sqrt{b^2 - 4mk}}{2m}. Underdamped (b2<4mkb^2 < 4mk) systems oscillate while decaying; critically damped (b2=4mkb^2 = 4mk) systems return fastest without oscillating.

Example: For the spring in formula 98, critical damping needs b=4⋅0.5⋅8=4b = \sqrt{4 \cdot 0.5 \cdot 8} = 4 N·s/m. Car shock absorbers are designed to be close to critically damped.

100. Euler's method

yn+1=yn+h f(xn,yn)y_{n+1} = y_n + h\,f(x_n, y_n)

Example: For y′=yy' = y, y(0)=1y(0) = 1 and step h=0.25h = 0.25, four steps give y(1)≈1.254=2.441y(1) \approx 1.25^4 = 2.441, against the exact e≈2.718e \approx 2.718. Halving hh roughly halves the error, which is why practical solvers use higher-order methods such as Runge–Kutta.

Common mistakes

  • Using a theorem without its hypotheses. The mean value theorem needs continuity on the closed interval and differentiability inside; L'Hôpital's rule only applies to 0/00/0 or ∞/∞\infty/\infty forms.
  • Dropping the inner derivative in the chain rule. ddxsin⁡(x2)=2xcos⁡(x2)\frac{d}{dx}\sin(x^2) = 2x\cos(x^2), not cos⁡(x2)\cos(x^2).
  • Forgetting to change the limits in a substitution. Either change them to uu-values or substitute back before evaluating.
  • Assuming an→0a_n \to 0 means the series converges. The harmonic series is the standard counterexample (formula 62).
  • Forgetting the Jacobian. In polar coordinates dA=r dr dθdA = r\,dr\,d\theta, not dr dθdr\,d\theta.
  • Treating an improper integral as ordinary. ∫−11dxx2\int_{-1}^1\frac{dx}{x^2} is not −2-2; the integrand blows up at 0 and the integral diverges.
  • Using degrees. Every derivative and series of a trigonometric function in this article assumes radians.

Next steps

References