Worksheets · AP Calculus AB

AP Calculus AB formula sheet

Every formula AP Calculus AB expects you to know, unit by unit, on one printable sheet. The exam gives you no formula sheet, so these are the ones to have by heart before May.

AP Calculus AB Formula SheetThe exam gives you none: these are the ones to know by heart.

Sigma Prep

Limits and continuityUnit 1

Two special limits
lim⁡x→0sin⁡xx=1\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1 · lim⁡x→01−cos⁡xx=0\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = 0Example: lim⁡x→0sin⁡(3x)x=3⋅lim⁡x→0sin⁡(3x)3x=3\displaystyle \lim_{x \to 0} \frac{\sin(3x)}{x} = 3 \cdot \lim_{x \to 0} \frac{\sin(3x)}{3x} = 3
Continuous at a
f(a)\displaystyle f(a) exists, lim⁡x→af(x)\displaystyle \lim_{x \to a} f(x) exists, and lim⁡x→af(x)=f(a)\displaystyle \lim_{x \to a} f(x) = f(a)Example: f(x)=x2−1x−1\displaystyle f(x) = \frac{x^2 - 1}{x - 1} is not continuous at x=1\displaystyle x = 1, because f(1)\displaystyle f(1) does not exist (even though the limit is 2).
Intermediate Value Theorem
If f\displaystyle f is continuous on [a,b]\displaystyle [a, b] and k\displaystyle k is between f(a)\displaystyle f(a) and f(b)\displaystyle f(b), then f(c)=k\displaystyle f(c) = k for some c\displaystyle c in [a,b]\displaystyle [a, b].Example: f(x)=x3+x−1\displaystyle f(x) = x^3 + x - 1 is continuous, f(0)=−1\displaystyle f(0) = -1 and f(1)=1\displaystyle f(1) = 1, so f(c)=0\displaystyle f(c) = 0 for some c\displaystyle c in [0,1]\displaystyle [0, 1].
Horizontal asymptote
y=L\displaystyle y = L when lim⁡x→∞f(x)=L\displaystyle \lim_{x \to \infty} f(x) = L or lim⁡x→−∞f(x)=L\displaystyle \lim_{x \to -\infty} f(x) = LExample: lim⁡x→∞3x2+1x2−4=3\displaystyle \lim_{x \to \infty} \frac{3x^2 + 1}{x^2 - 4} = 3, so y=3\displaystyle y = 3.

The derivativeUnit 2

Definition
f′(x)=lim⁡h→0f(x+h)−f(x)h\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}Example: for f(x)=x2\displaystyle f(x) = x^2, lim⁡h→0(x+h)2−x2h=lim⁡h→0(2x+h)=2x\displaystyle \lim_{h \to 0} \frac{(x + h)^2 - x^2}{h} = \lim_{h \to 0} (2x + h) = 2x
At a point
f′(a)=lim⁡x→af(x)−f(a)x−a\displaystyle f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}Example: lim⁡x→2x3−8x−2\displaystyle \lim_{x \to 2} \frac{x^3 - 8}{x - 2} is f′(2)\displaystyle f'(2) for f(x)=x3\displaystyle f(x) = x^3, so it equals 3(2)2=12\displaystyle 3(2)^2 = 12.
Average rate of change
f(b)−f(a)b−a\displaystyle \frac{f(b) - f(a)}{b - a}Example: f(x)=x2\displaystyle f(x) = x^2 on [1,3]\displaystyle [1, 3]: 9−13−1=4\displaystyle \frac{9 - 1}{3 - 1} = 4
Tangent line at x = a
y=f(a)+f′(a)(x−a)\displaystyle y = f(a) + f'(a)(x - a)Also the linear approximation of f near a. Example: f(x)=x2\displaystyle f(x) = x^2 at x=3\displaystyle x = 3: y=9+6(x−3)\displaystyle y = 9 + 6(x - 3), so f(3.1)≈9.6\displaystyle f(3.1) \approx 9.6

Derivative rulesUnits 2–3

Product
ddx [fg]=f′g+fg′\displaystyle \frac{d}{dx}\, [f g] = f' g + f g'Example: ddx[x2sin⁡x]=2xsin⁡x+x2cos⁡x\displaystyle \frac{d}{dx} [x^2 \sin x] = 2x \sin x + x^2 \cos x
Quotient
ddx[fg]=f′g−fg′g2\displaystyle \frac{d}{dx} \left[ \frac{f}{g} \right] = \frac{f' g - f g'}{g^2}Example: ddx[xx+1]=(x+1)−x(x+1)2=1(x+1)2\displaystyle \frac{d}{dx} \left[ \frac{x}{x + 1} \right] = \frac{(x + 1) - x}{(x + 1)^2} = \frac{1}{(x + 1)^2}
Chain
ddx f(g(x))=f′(g(x)) g′(x)\displaystyle \frac{d}{dx}\, f(g(x)) = f'(g(x))\, g'(x)Example: ddxsin⁡(x3)=cos⁡(x3)⋅3x2\displaystyle \frac{d}{dx} \sin(x^3) = \cos(x^3) \cdot 3x^2
Inverse function
(f−1)′(a)=1f′(f−1(a))\displaystyle \left(f^{-1}\right)'(a) = \frac{1}{f'\left(f^{-1}(a)\right)}Example: f(x)=x3+x\displaystyle f(x) = x^3 + x has f(1)=2\displaystyle f(1) = 2 and f′(1)=4\displaystyle f'(1) = 4, so (f−1)′(2)=14\displaystyle \left(f^{-1}\right)'(2) = \frac{1}{4}
Implicit
Differentiate both sides in x\displaystyle x, writing dydx\displaystyle \frac{dy}{dx} each time you differentiate y\displaystyle y.Example: x2+y2=25\displaystyle x^2 + y^2 = 25 gives 2x+2ydydx=0\displaystyle 2x + 2y \frac{dy}{dx} = 0, so dydx=−xy\displaystyle \frac{dy}{dx} = -\frac{x}{y}

Derivatives to knowUnits 2–3

How to read these
Each rule is written for any inside function f(x)\displaystyle f(x), chain rule included. When the inside is just x\displaystyle x, f′(x)=1\displaystyle f'(x) = 1 and you get the basic rule.
Power
ddx[f(x)]n=n[f(x)]n−1⋅f′(x)\displaystyle \frac{d}{dx} [f(x)]^n = n [f(x)]^{n - 1} \cdot f'(x)Example: ddx(3x2+1)5=5(3x2+1)4⋅6x\displaystyle \frac{d}{dx} (3x^2 + 1)^5 = 5(3x^2 + 1)^4 \cdot 6x
Exponential
ddx ef(x)=ef(x)⋅f′(x)\displaystyle \frac{d}{dx}\, e^{f(x)} = e^{f(x)} \cdot f'(x) · ddx af(x)=af(x)ln⁡a⋅f′(x)\displaystyle \frac{d}{dx}\, a^{f(x)} = a^{f(x)} \ln a \cdot f'(x)Example: ddx ex2=ex2⋅2x\displaystyle \frac{d}{dx}\, e^{x^2} = e^{x^2} \cdot 2x
Log
ddxln⁡(f(x))=f′(x)f(x)\displaystyle \frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)} · ddxlog⁡a(f(x))=f′(x)f(x)ln⁡a\displaystyle \frac{d}{dx} \log_a(f(x)) = \frac{f'(x)}{f(x) \ln a}Example: ddxln⁡(x2+4)=2xx2+4\displaystyle \frac{d}{dx} \ln(x^2 + 4) = \frac{2x}{x^2 + 4}
Sine and cosine
ddxsin⁡(f(x))=cos⁡(f(x))⋅f′(x)\displaystyle \frac{d}{dx} \sin(f(x)) = \cos(f(x)) \cdot f'(x) · ddxcos⁡(f(x))=−sin⁡(f(x))⋅f′(x)\displaystyle \frac{d}{dx} \cos(f(x)) = -\sin(f(x)) \cdot f'(x)Example: ddxcos⁡(3x)=−3sin⁡(3x)\displaystyle \frac{d}{dx} \cos(3x) = -3 \sin(3x)
Tangent and cotangent
ddxtan⁡(f(x))=sec⁡2(f(x))⋅f′(x)\displaystyle \frac{d}{dx} \tan(f(x)) = \sec^2(f(x)) \cdot f'(x) · ddxcot⁡(f(x))=−csc⁡2(f(x))⋅f′(x)\displaystyle \frac{d}{dx} \cot(f(x)) = -\csc^2(f(x)) \cdot f'(x)Example: ddxtan⁡(x2)=sec⁡2(x2)⋅2x\displaystyle \frac{d}{dx} \tan(x^2) = \sec^2(x^2) \cdot 2x
Secant and cosecant
ddxsec⁡(f(x))=sec⁡(f(x))tan⁡(f(x))⋅f′(x)\displaystyle \frac{d}{dx} \sec(f(x)) = \sec(f(x)) \tan(f(x)) \cdot f'(x) · ddxcsc⁡(f(x))=−csc⁡(f(x))cot⁡(f(x))⋅f′(x)\displaystyle \frac{d}{dx} \csc(f(x)) = -\csc(f(x)) \cot(f(x)) \cdot f'(x)Example: ddxsec⁡(4x)=4sec⁡(4x)tan⁡(4x)\displaystyle \frac{d}{dx} \sec(4x) = 4 \sec(4x) \tan(4x)
Inverse trig
ddxarcsin⁡(f(x))=f′(x)1−[f(x)]2\displaystyle \frac{d}{dx} \arcsin(f(x)) = \frac{f'(x)}{\sqrt{1 - [f(x)]^2}} · ddxarccos⁡(f(x))=−f′(x)1−[f(x)]2\displaystyle \frac{d}{dx} \arccos(f(x)) = \frac{-f'(x)}{\sqrt{1 - [f(x)]^2}} · ddxarctan⁡(f(x))=f′(x)1+[f(x)]2\displaystyle \frac{d}{dx} \arctan(f(x)) = \frac{f'(x)}{1 + [f(x)]^2}Example: ddxarctan⁡(2x)=21+4x2\displaystyle \frac{d}{dx} \arctan(2x) = \frac{2}{1 + 4x^2}

Derivatives in contextUnit 4

Motion
v(t)=x′(t)\displaystyle v(t) = x'(t) · a(t)=v′(t)=x′′(t)\displaystyle a(t) = v'(t) = x''(t) · speed=∣v(t)∣\displaystyle \text{speed} = |v(t)|Example: x(t)=t3−6t\displaystyle x(t) = t^3 - 6t gives v(t)=3t2−6\displaystyle v(t) = 3t^2 - 6 and a(t)=6t\displaystyle a(t) = 6t
Speeding up
When v(t)\displaystyle v(t) and a(t)\displaystyle a(t) have the same sign.Slowing down when they have opposite signs. Example: with the motion above, v(1)=−3\displaystyle v(1) = -3 and a(1)=6\displaystyle a(1) = 6, so at t=1\displaystyle t = 1 the particle is slowing down.
Related rates
Write an equation linking the quantities, then differentiate both sides with respect to t\displaystyle t.Example: A=πr2\displaystyle A = \pi r^2 gives dAdt=2πrdrdt\displaystyle \frac{dA}{dt} = 2\pi r \frac{dr}{dt}
L'Hospital's Rule
If lim⁡f(x)g(x)\displaystyle \lim \frac{f(x)}{g(x)} has the form 00\displaystyle \frac{0}{0} or ∞∞\displaystyle \frac{\infty}{\infty}, then lim⁡f(x)g(x)=lim⁡f′(x)g′(x)\displaystyle \lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}.When the limit on the right exists. Check the form first. Example: lim⁡x→0ex−1x\displaystyle \lim_{x \to 0} \frac{e^x - 1}{x} has the form 00\displaystyle \frac{0}{0}, so it equals lim⁡x→0ex1=1\displaystyle \lim_{x \to 0} \frac{e^x}{1} = 1

Analyzing functionsUnit 5

Mean Value Theorem
If f\displaystyle f is continuous on [a,b]\displaystyle [a, b] and differentiable on (a,b)\displaystyle (a, b), then f′(c)=f(b)−f(a)b−a\displaystyle f'(c) = \frac{f(b) - f(a)}{b - a} for some c\displaystyle c in (a,b)\displaystyle (a, b).Example: f(x)=x2\displaystyle f(x) = x^2 on [0,2]\displaystyle [0, 2]: f′(c)=4−02−0=2\displaystyle f'(c) = \frac{4 - 0}{2 - 0} = 2, so 2c=2\displaystyle 2c = 2 and c=1\displaystyle c = 1.
Extreme Value Theorem
If f\displaystyle f is continuous on [a,b]\displaystyle [a, b], then f\displaystyle f has an absolute maximum and minimum on [a,b]\displaystyle [a, b].Example: f(x)=1x\displaystyle f(x) = \frac{1}{x} on [−1,1]\displaystyle [-1, 1] is not continuous at x=0\displaystyle x = 0, so the theorem does not apply (and f\displaystyle f has no maximum there).
Critical point
f′(c)=0\displaystyle f'(c) = 0 or f′(c)\displaystyle f'(c) does not existExample: f(x)=x2/3\displaystyle f(x) = x^{2/3} has f′(x)=23x−1/3\displaystyle f'(x) = \frac{2}{3} x^{-1/3}, which does not exist at x=0\displaystyle x = 0, so x=0\displaystyle x = 0 is a critical point.
Increasing / decreasing
f′>0\displaystyle f' > 0: f\displaystyle f increasing · f′<0\displaystyle f' < 0: f\displaystyle f decreasingExample: f(x)=x2−4x\displaystyle f(x) = x^2 - 4x has f′(x)=2x−4>0\displaystyle f'(x) = 2x - 4 > 0 for x>2\displaystyle x > 2, so f\displaystyle f is increasing on (2,∞)\displaystyle (2, \infty).
First derivative test
f′\displaystyle f' changes from +\displaystyle + to −\displaystyle -: relative max · from −\displaystyle - to +\displaystyle +: relative minExample: f′(x)=(x−1)(x−3)\displaystyle f'(x) = (x - 1)(x - 3) goes +\displaystyle + to −\displaystyle - at x=1\displaystyle x = 1 (relative max) and −\displaystyle - to +\displaystyle + at x=3\displaystyle x = 3 (relative min).
Concavity
f′′>0\displaystyle f'' > 0 (or f′\displaystyle f' increasing): concave up · f′′<0\displaystyle f'' < 0 (or f′\displaystyle f' decreasing): concave downExample: f(x)=x3\displaystyle f(x) = x^3 has f′′(x)=6x>0\displaystyle f''(x) = 6x > 0 for x>0\displaystyle x > 0, so f\displaystyle f is concave up on (0,∞)\displaystyle (0, \infty).
Second derivative test
f′(c)=0\displaystyle f'(c) = 0 and f′′(c)<0\displaystyle f''(c) < 0: relative max · f′(c)=0\displaystyle f'(c) = 0 and f′′(c)>0\displaystyle f''(c) > 0: relative minExample: f(x)=x3−3x\displaystyle f(x) = x^3 - 3x has f′(1)=0\displaystyle f'(1) = 0 and f′′(1)=6>0\displaystyle f''(1) = 6 > 0, so f\displaystyle f has a relative min at x=1\displaystyle x = 1.
Point of inflection
Where f′′\displaystyle f'' changes sign (where f′\displaystyle f' changes from increasing to decreasing, or back)Example: f(x)=x4\displaystyle f(x) = x^4 has f′′(0)=0\displaystyle f''(0) = 0, but f′′(x)=12x2\displaystyle f''(x) = 12x^2 does not change sign, so there is no point of inflection at x=0\displaystyle x = 0.
Absolute extrema on [a, b]
Compare f\displaystyle f at the endpoints and at every critical point (the candidates test).Example: f(x)=x3−3x\displaystyle f(x) = x^3 - 3x on [0,2]\displaystyle [0, 2]: f(0)=0\displaystyle f(0) = 0, f(1)=−2\displaystyle f(1) = -2, f(2)=2\displaystyle f(2) = 2, so the absolute max is 2 and the absolute min is −2\displaystyle -2.

Integrals to knowUnit 6

How to read these
Most exam integrals have an inside of the form kx\displaystyle kx: integrate as if it were x\displaystyle x, then divide by k\displaystyle k. For any other inside function, use u\displaystyle u-substitution (below).
Power
∫xn dx=xn+1n+1+C\displaystyle \int x^n \, dx = \frac{x^{n + 1}}{n + 1} + C for n≠−1\displaystyle n \ne -1Example: ∫(2x+1)3 dx=(2x+1)48+C\displaystyle \int (2x + 1)^3 \, dx = \frac{(2x + 1)^4}{8} + C
Log
∫1x dx=ln⁡∣x∣+C\displaystyle \int \frac{1}{x} \, dx = \ln |x| + C · ∫f′(x)f(x) dx=ln⁡∣f(x)∣+C\displaystyle \int \frac{f'(x)}{f(x)} \, dx = \ln |f(x)| + CExample: ∫2xx2+5 dx=ln⁡(x2+5)+C\displaystyle \int \frac{2x}{x^2 + 5} \, dx = \ln(x^2 + 5) + C
Exponential
∫ekx dx=1kekx+C\displaystyle \int e^{kx} \, dx = \frac{1}{k} e^{kx} + C · ∫akx dx=akxkln⁡a+C\displaystyle \int a^{kx} \, dx = \frac{a^{kx}}{k \ln a} + CExample: ∫e3x dx=13e3x+C\displaystyle \int e^{3x} \, dx = \frac{1}{3} e^{3x} + C
Sine and cosine
∫cos⁡(kx) dx=1ksin⁡(kx)+C\displaystyle \int \cos(kx) \, dx = \frac{1}{k} \sin(kx) + C · ∫sin⁡(kx) dx=−1kcos⁡(kx)+C\displaystyle \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + CExample: ∫sin⁡(2x) dx=−12cos⁡(2x)+C\displaystyle \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C
Secant² and cosecant²
∫sec⁡2(kx) dx=1ktan⁡(kx)+C\displaystyle \int \sec^2(kx) \, dx = \frac{1}{k} \tan(kx) + C · ∫csc⁡2(kx) dx=−1kcot⁡(kx)+C\displaystyle \int \csc^2(kx) \, dx = -\frac{1}{k} \cot(kx) + CExample: ∫sec⁡2(5x) dx=15tan⁡(5x)+C\displaystyle \int \sec^2(5x) \, dx = \frac{1}{5} \tan(5x) + C
Secant‑tangent and cosecant‑cotangent
∫sec⁡(kx)tan⁡(kx) dx=1ksec⁡(kx)+C\displaystyle \int \sec(kx) \tan(kx) \, dx = \frac{1}{k} \sec(kx) + C · ∫csc⁡(kx)cot⁡(kx) dx=−1kcsc⁡(kx)+C\displaystyle \int \csc(kx) \cot(kx) \, dx = -\frac{1}{k} \csc(kx) + CExample: ∫sec⁡(3x)tan⁡(3x) dx=13sec⁡(3x)+C\displaystyle \int \sec(3x) \tan(3x) \, dx = \frac{1}{3} \sec(3x) + C
Tangent and cotangent
∫tan⁡x dx=−ln⁡∣cos⁡x∣+C\displaystyle \int \tan x \, dx = -\ln |\cos x| + C · ∫cot⁡x dx=ln⁡∣sin⁡x∣+C\displaystyle \int \cot x \, dx = \ln |\sin x| + CBoth come from the log rule: tan⁡x=sin⁡xcos⁡x\displaystyle \tan x = \frac{\sin x}{\cos x}.
Inverse trig
∫11−x2 dx=arcsin⁡x+C\displaystyle \int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin x + C · ∫11+x2 dx=arctan⁡x+C\displaystyle \int \frac{1}{1 + x^2} \, dx = \arctan x + CExample: ∫0111+x2 dx=arctan⁡1−arctan⁡0=π4\displaystyle \int_0^1 \frac{1}{1 + x^2} \, dx = \arctan 1 - \arctan 0 = \frac{\pi}{4}
u-substitution
∫g(f(x)) f′(x) dx=∫g(u) du\displaystyle \int g(f(x)) \, f'(x) \, dx = \int g(u) \, du with u=f(x)\displaystyle u = f(x), du=f′(x) dx\displaystyle du = f'(x) \, dxExample: ∫2xcos⁡(x2) dx=sin⁡(x2)+C\displaystyle \int 2x \cos(x^2) \, dx = \sin(x^2) + C, with u=x2\displaystyle u = x^2
Definite integrals
With u=f(x)\displaystyle u = f(x), change the limits too: ∫abg(f(x)) f′(x) dx=∫f(a)f(b)g(u) du\displaystyle \int_a^b g(f(x)) \, f'(x) \, dx = \int_{f(a)}^{f(b)} g(u) \, duExample: ∫022x ex2 dx=∫04eu du=e4−1\displaystyle \int_0^2 2x \, e^{x^2} \, dx = \int_0^4 e^u \, du = e^4 - 1, with u=x2\displaystyle u = x^2

Accumulation and the Fundamental TheoremUnit 6

Fundamental Theorem: evaluating
∫abf′(x) dx=f(b)−f(a)\displaystyle \int_a^b f'(x) \, dx = f(b) - f(a)Example: ∫0πsin⁡x dx=−cos⁡π−(−cos⁡0)=2\displaystyle \int_0^{\pi} \sin x \, dx = -\cos \pi - (-\cos 0) = 2
Net change
f(b)=f(a)+∫abf′(x) dx\displaystyle f(b) = f(a) + \int_a^b f'(x) \, dxExample: if f(2)=5\displaystyle f(2) = 5 and f′(x)=3x2\displaystyle f'(x) = 3x^2, then f(4)=5+∫243x2 dx=5+56=61\displaystyle f(4) = 5 + \int_2^4 3x^2 \, dx = 5 + 56 = 61
Fundamental Theorem: derivative of an integral
ddx∫axf(t) dt=f(x)\displaystyle \frac{d}{dx} \int_a^x f(t) \, dt = f(x) · ddx∫ag(x)f(t) dt=f(g(x))⋅g′(x)\displaystyle \frac{d}{dx} \int_a^{g(x)} f(t) \, dt = f(g(x)) \cdot g'(x)Example: if G(x)=∫0x2cos⁡t dt\displaystyle G(x) = \int_0^{x^2} \cos t \, dt, then G′(x)=cos⁡(x2)⋅2x\displaystyle G'(x) = \cos(x^2) \cdot 2x. Plug the top limit into f\displaystyle f, then multiply by its derivative.
Both limits move
ddx∫h(x)g(x)f(t) dt=f(g(x))⋅g′(x)−f(h(x))⋅h′(x)\displaystyle \frac{d}{dx} \int_{h(x)}^{g(x)} f(t) \, dt = f(g(x)) \cdot g'(x) - f(h(x)) \cdot h'(x)Example: ddx∫x3xet2 dt=3e9x2−ex2\displaystyle \frac{d}{dx} \int_{x}^{3x} e^{t^2} \, dt = 3e^{9x^2} - e^{x^2}
Properties
∫aaf(x) dx=0\displaystyle \int_a^a f(x) \, dx = 0 · ∫baf(x) dx=−∫abf(x) dx\displaystyle \int_b^a f(x) \, dx = -\int_a^b f(x) \, dx · ∫abf(x) dx+∫bcf(x) dx=∫acf(x) dx\displaystyle \int_a^b f(x) \, dx + \int_b^c f(x) \, dx = \int_a^c f(x) \, dxExample: if ∫13f(x) dx=4\displaystyle \int_1^3 f(x) \, dx = 4 and ∫35f(x) dx=2\displaystyle \int_3^5 f(x) \, dx = 2, then ∫15f(x) dx=6\displaystyle \int_1^5 f(x) \, dx = 6 and ∫31f(x) dx=−4\displaystyle \int_3^1 f(x) \, dx = -4

Approximating an integralUnit 6

Left and right sums
Left: Δx [f(x0)+f(x1)+⋯+f(xn−1)]\displaystyle \Delta x \, [f(x_0) + f(x_1) + \dots + f(x_{n-1})] · Right: Δx [f(x1)+f(x2)+⋯+f(xn)]\displaystyle \Delta x \, [f(x_1) + f(x_2) + \dots + f(x_n)]With unequal widths (a table), multiply each height by its own width. Example: f(x)=x2\displaystyle f(x) = x^2 on [0,2]\displaystyle [0, 2] with Δx=1\displaystyle \Delta x = 1: left =1⋅(0+1)=1\displaystyle = 1 \cdot (0 + 1) = 1, right =1⋅(1+4)=5\displaystyle = 1 \cdot (1 + 4) = 5
Midpoint sum
Δx [f(m1)+f(m2)+⋯+f(mn)]\displaystyle \Delta x \, [f(m_1) + f(m_2) + \dots + f(m_n)], each mi\displaystyle m_i the midpoint of its intervalExample: same f\displaystyle f and intervals: 1⋅[f(0.5)+f(1.5)]=0.25+2.25=2.5\displaystyle 1 \cdot [f(0.5) + f(1.5)] = 0.25 + 2.25 = 2.5
Trapezoidal sum
Δx2 [f(x0)+2f(x1)+⋯+2f(xn−1)+f(xn)]\displaystyle \frac{\Delta x}{2} \, [f(x_0) + 2f(x_1) + \dots + 2f(x_{n-1}) + f(x_n)]Example: same f\displaystyle f and intervals: 12[0+2(1)+4]=3\displaystyle \frac{1}{2} [0 + 2(1) + 4] = 3
Over or under
A left sum underestimates when f\displaystyle f is increasing and overestimates when it is decreasing; a trapezoidal sum underestimates when f\displaystyle f is concave down.Example: x2\displaystyle x^2 on [0,2]\displaystyle [0, 2] is increasing and concave up, so left =1\displaystyle = 1 is under and trapezoid =3\displaystyle = 3 is over the exact 83\displaystyle \frac{8}{3}.
Limit of a sum
∫abf(x) dx=lim⁡n→∞∑k=1nf(xk) Δx\displaystyle \int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{k = 1}^{n} f(x_k) \, \Delta x with Δx=b−an\displaystyle \Delta x = \frac{b - a}{n}Example: lim⁡n→∞∑k=1n(kn)21n=∫01x2 dx=13\displaystyle \lim_{n \to \infty} \sum_{k = 1}^{n} \left(\frac{k}{n}\right)^2 \frac{1}{n} = \int_0^1 x^2 \, dx = \frac{1}{3}

Differential equationsUnit 7

Separation of variables
dydx=g(x) h(y)\displaystyle \frac{dy}{dx} = g(x)\, h(y) gives ∫dyh(y)=∫g(x) dx\displaystyle \int \frac{dy}{h(y)} = \int g(x) \, dxAdd + C before solving for y, then use the initial condition. Example: dydx=xy\displaystyle \frac{dy}{dx} = xy with y(0)=2\displaystyle y(0) = 2: ln⁡∣y∣=x22+C\displaystyle \ln |y| = \frac{x^2}{2} + C, so y=2ex2/2\displaystyle y = 2e^{x^2/2}
Exponential growth and decay
dydt=ky\displaystyle \frac{dy}{dt} = k y gives y=y0 ekt\displaystyle y = y_0 \, e^{k t}Example: dPdt=0.03P\displaystyle \frac{dP}{dt} = 0.03P with P(0)=100\displaystyle P(0) = 100 gives P=100e0.03t\displaystyle P = 100e^{0.03t}
Slope field
At each point (x,y)\displaystyle (x, y), draw a short segment with slope dydx\displaystyle \frac{dy}{dx}.Example: for dydx=x−y\displaystyle \frac{dy}{dx} = x - y, the segment at (2,1)\displaystyle (2, 1) has slope 2−1=1\displaystyle 2 - 1 = 1.

Applications of integrationUnit 8

Average value
favg=1b−a∫abf(x) dx\displaystyle f_{\text{avg}} = \frac{1}{b - a} \int_a^b f(x) \, dxExample: f(x)=x2\displaystyle f(x) = x^2 on [0,3]\displaystyle [0, 3]: 13∫03x2 dx=13⋅9=3\displaystyle \frac{1}{3} \int_0^3 x^2 \, dx = \frac{1}{3} \cdot 9 = 3
Displacement
∫abv(t) dt\displaystyle \int_a^b v(t) \, dtExample: v(t)=t−2\displaystyle v(t) = t - 2 on [0,4]\displaystyle [0, 4]: ∫04(t−2) dt=0\displaystyle \int_0^4 (t - 2) \, dt = 0, so the particle ends where it started.
Total distance
∫ab∣v(t)∣ dt\displaystyle \int_a^b |v(t)| \, dtExample: same v(t)\displaystyle v(t): ∫04∣t−2∣ dt=2+2=4\displaystyle \int_0^4 |t - 2| \, dt = 2 + 2 = 4
Position
x(t)=x(0)+∫0tv(s) ds\displaystyle x(t) = x(0) + \int_0^t v(s) \, dsExample: x(0)=3\displaystyle x(0) = 3 and v(t)=2t\displaystyle v(t) = 2t: x(3)=3+∫032s ds=3+9=12\displaystyle x(3) = 3 + \int_0^3 2s \, ds = 3 + 9 = 12
Area between curves
∫ab[top−bottom] dx\displaystyle \int_a^b [\text{top} - \text{bottom}] \, dx · ∫cd[right−left] dy\displaystyle \int_c^d [\text{right} - \text{left}] \, dyExample: between y=x\displaystyle y = x and y=x2\displaystyle y = x^2 on [0,1]\displaystyle [0, 1]: ∫01(x−x2) dx=16\displaystyle \int_0^1 (x - x^2) \, dx = \frac{1}{6}
Disc
V=π∫ab[R(x)]2 dx\displaystyle V = \pi \int_a^b [R(x)]^2 \, dxExample: y=x\displaystyle y = \sqrt{x} on [0,4]\displaystyle [0, 4] about the x\displaystyle x-axis: π∫04x dx=8π\displaystyle \pi \int_0^4 x \, dx = 8\pi
Washer
V=π∫ab([R(x)]2−[r(x)]2)dx\displaystyle V = \pi \int_a^b \left( [R(x)]^2 - [r(x)]^2 \right) dxR and r are measured from the axis of rotation. Example: the region between y=x\displaystyle y = x and y=x2\displaystyle y = x^2 about the x\displaystyle x-axis: π∫01(x2−x4) dx=2π15\displaystyle \pi \int_0^1 (x^2 - x^4) \, dx = \frac{2\pi}{15}
Cross sections
V=∫abA(x) dx\displaystyle V = \int_a^b A(x) \, dx · squares A=s2\displaystyle A = s^2 · semicircles A=12π(s2)2\displaystyle A = \frac{1}{2}\pi \left(\frac{s}{2}\right)^2 · equilateral triangles A=34s2\displaystyle A = \frac{\sqrt{3}}{4} s^2Example: base under y=x\displaystyle y = \sqrt{x} on [0,4]\displaystyle [0, 4], square cross sections perpendicular to the x\displaystyle x-axis: ∫04(x)2 dx=8\displaystyle \int_0^4 (\sqrt{x})^2 \, dx = 8

Trig and geometry you still needAll units

Identities
sin⁡2x+cos⁡2x=1\displaystyle \sin^2 x + \cos^2 x = 1 · 1+tan⁡2x=sec⁡2x\displaystyle 1 + \tan^2 x = \sec^2 xExample: 1−cos⁡2(3x)=sin⁡2(3x)\displaystyle 1 - \cos^2(3x) = \sin^2(3x)
Double angle
sin⁡2x=2sin⁡xcos⁡x\displaystyle \sin 2x = 2 \sin x \cos x · cos⁡2x=cos⁡2x−sin⁡2x=1−2sin⁡2x\displaystyle \cos 2x = \cos^2 x - \sin^2 x = 1 - 2\sin^2 xExample: sin⁡xcos⁡x=12sin⁡(2x)\displaystyle \sin x \cos x = \frac{1}{2} \sin(2x), so ∫sin⁡xcos⁡x dx=−14cos⁡(2x)+C\displaystyle \int \sin x \cos x \, dx = -\frac{1}{4} \cos(2x) + C
Unit circle
sin⁡π6=12\displaystyle \sin \frac{\pi}{6} = \frac{1}{2} · sin⁡π4=22\displaystyle \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} · sin⁡π3=32\displaystyle \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} · cos⁡π6=32\displaystyle \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} · cos⁡π4=22\displaystyle \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} · cos⁡π3=12\displaystyle \cos \frac{\pi}{3} = \frac{1}{2}Example: cos⁡2π3=−12\displaystyle \cos \frac{2\pi}{3} = -\frac{1}{2}: reference angle π3\displaystyle \frac{\pi}{3}, and cosine is negative in quadrant II.
Logs
ln⁡(ab)=ln⁡a+ln⁡b\displaystyle \ln(ab) = \ln a + \ln b · ln⁡ab=ln⁡a−ln⁡b\displaystyle \ln \frac{a}{b} = \ln a - \ln b · ln⁡ak=kln⁡a\displaystyle \ln a^k = k \ln a · ln⁡e=1\displaystyle \ln e = 1 · ln⁡1=0\displaystyle \ln 1 = 0Example: ln⁡x2y=2ln⁡x−ln⁡y\displaystyle \ln \frac{x^2}{y} = 2 \ln x - \ln y
Shapes (related rates)
Circle A=πr2\displaystyle A = \pi r^2, C=2πr\displaystyle C = 2\pi r · sphere V=43πr3\displaystyle V = \frac{4}{3}\pi r^3 · cone V=13πr2h\displaystyle V = \frac{1}{3}\pi r^2 h · cylinder V=πr2h\displaystyle V = \pi r^2 hExample: sphere V=43πr3\displaystyle V = \frac{4}{3}\pi r^3 gives dVdt=4πr2drdt\displaystyle \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}
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Is there a formula sheet on the AP Calculus AB exam?

No. AP Calculus AB gives you no formula or reference sheet, so every formula on this page has to be memorized.

Questions about the AP Calculus AB formula sheet

No. AP Calculus AB comes with no formula or reference sheet, unlike AP Physics or AP Chemistry, so every formula has to be memorized. In the calculator sections, the graphing calculator (Desmos is built into the Bluebook app) can find a derivative at a point or a definite integral numerically, but it does not give you the formulas or the setup.

The derivative and integral rules (power, product, quotient, chain, and the basic trig, exponential and log derivatives and integrals), the Fundamental Theorem of Calculus with its net-change form, and the Unit 8 setups: average value, area between curves, and volume by discs, washers and cross sections. Free response leans hardest on Unit 8 and on reading a rate from a table or graph, so those setups pay off most.

You need their conditions. On free response, a justification earns its point only when it checks the hypotheses: continuity on the closed interval for the IVT and the EVT, plus differentiability on the open interval for the MVT. Both statements are on the sheet with their conditions.

It covers everything in AB, which is also most of BC. BC adds integration by parts and partial fractions, logistic growth, Euler’s method, arc length, parametric, polar and vector functions, and series, none of which are on this sheet.

Yes. Download the PDF or print it straight from the page. There is no account and no sign-up, and you can hand out as many copies as you like.

AP is a registered trademark of the College Board, which was not involved in making this sheet and does not endorse it. Written by Sigma Prep.