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.
Limits and continuityUnit 1 Two special limits lim x → 0 sin x x = 1 \displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1 x → 0 lim x sin x = 1 · lim x → 0 1 − cos x x = 0 \displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 x → 0 lim x 1 − cos x = 0 Example: lim x → 0 sin ( 3 x ) x = 3 ⋅ lim x → 0 sin ( 3 x ) 3 x = 3 \displaystyle \lim_{x \to 0} \frac{\sin(3x)}{x} = 3 \cdot \lim_{x \to 0} \frac{\sin(3x)}{3x} = 3 x → 0 lim x sin ( 3 x ) = 3 ⋅ x → 0 lim 3 x sin ( 3 x ) = 3 Continuous at a f ( a ) \displaystyle f(a) f ( a ) exists, lim x → a f ( x ) \displaystyle \lim_{x \to a} f(x) x → a lim f ( x ) exists, and lim x → a f ( x ) = f ( a ) \displaystyle \lim_{x \to a} f(x) = f(a) x → a lim f ( x ) = f ( a ) Example: f ( x ) = x 2 − 1 x − 1 \displaystyle f(x) = \frac{x^2 - 1}{x - 1} f ( x ) = x − 1 x 2 − 1 is not continuous at x = 1 \displaystyle x = 1 x = 1 , because f ( 1 ) \displaystyle f(1) f ( 1 ) does not exist (even though the limit is 2). Intermediate Value Theorem If f \displaystyle f f is continuous on [ a , b ] \displaystyle [a, b] [ a , b ] and k \displaystyle k k is between f ( a ) \displaystyle f(a) f ( a ) and f ( b ) \displaystyle f(b) f ( b ) , then f ( c ) = k \displaystyle f(c) = k f ( c ) = k for some c \displaystyle c c in [ a , b ] \displaystyle [a, b] [ a , b ] . Example: f ( x ) = x 3 + x − 1 \displaystyle f(x) = x^3 + x - 1 f ( x ) = x 3 + x − 1 is continuous, f ( 0 ) = − 1 \displaystyle f(0) = -1 f ( 0 ) = − 1 and f ( 1 ) = 1 \displaystyle f(1) = 1 f ( 1 ) = 1 , so f ( c ) = 0 \displaystyle f(c) = 0 f ( c ) = 0 for some c \displaystyle c c in [ 0 , 1 ] \displaystyle [0, 1] [ 0 , 1 ] . Horizontal asymptote y = L \displaystyle y = L y = L when lim x → ∞ f ( x ) = L \displaystyle \lim_{x \to \infty} f(x) = L x → ∞ lim f ( x ) = L or lim x → − ∞ f ( x ) = L \displaystyle \lim_{x \to -\infty} f(x) = L x → − ∞ lim f ( x ) = L Example: lim x → ∞ 3 x 2 + 1 x 2 − 4 = 3 \displaystyle \lim_{x \to \infty} \frac{3x^2 + 1}{x^2 - 4} = 3 x → ∞ lim x 2 − 4 3 x 2 + 1 = 3 , so y = 3 \displaystyle y = 3 y = 3 . The derivativeUnit 2 Definition f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h \displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} f ′ ( x ) = h → 0 lim h f ( x + h ) − f ( x ) Example: for f ( x ) = x 2 \displaystyle f(x) = x^2 f ( x ) = x 2 , lim h → 0 ( x + h ) 2 − x 2 h = lim h → 0 ( 2 x + h ) = 2 x \displaystyle \lim_{h \to 0} \frac{(x + h)^2 - x^2}{h} = \lim_{h \to 0} (2x + h) = 2x h → 0 lim h ( x + h ) 2 − x 2 = h → 0 lim ( 2 x + h ) = 2 x At a point f ′ ( a ) = lim x → a f ( x ) − f ( a ) x − a \displaystyle f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} f ′ ( a ) = x → a lim x − a f ( x ) − f ( a ) Example: lim x → 2 x 3 − 8 x − 2 \displaystyle \lim_{x \to 2} \frac{x^3 - 8}{x - 2} x → 2 lim x − 2 x 3 − 8 is f ′ ( 2 ) \displaystyle f'(2) f ′ ( 2 ) for f ( x ) = x 3 \displaystyle f(x) = x^3 f ( x ) = x 3 , so it equals 3 ( 2 ) 2 = 12 \displaystyle 3(2)^2 = 12 3 ( 2 ) 2 = 12 . Average rate of change f ( b ) − f ( a ) b − a \displaystyle \frac{f(b) - f(a)}{b - a} b − a f ( b ) − f ( a ) Example: f ( x ) = x 2 \displaystyle f(x) = x^2 f ( x ) = x 2 on [ 1 , 3 ] \displaystyle [1, 3] [ 1 , 3 ] : 9 − 1 3 − 1 = 4 \displaystyle \frac{9 - 1}{3 - 1} = 4 3 − 1 9 − 1 = 4 Tangent line at x = a y = f ( a ) + f ′ ( a ) ( x − a ) \displaystyle y = f(a) + f'(a)(x - a) y = f ( a ) + f ′ ( a ) ( x − a ) Also the linear approximation of f near a. Example: f ( x ) = x 2 \displaystyle f(x) = x^2 f ( x ) = x 2 at x = 3 \displaystyle x = 3 x = 3 : y = 9 + 6 ( x − 3 ) \displaystyle y = 9 + 6(x - 3) y = 9 + 6 ( x − 3 ) , so f ( 3.1 ) ≈ 9.6 \displaystyle f(3.1) \approx 9.6 f ( 3.1 ) ≈ 9.6 Derivative rulesUnits 2–3 Product d d x [ f g ] = f ′ g + f g ′ \displaystyle \frac{d}{dx}\, [f g] = f' g + f g' d x d [ f g ] = f ′ g + f g ′ Example: d d x [ x 2 sin x ] = 2 x sin x + x 2 cos x \displaystyle \frac{d}{dx} [x^2 \sin x] = 2x \sin x + x^2 \cos x d x d [ x 2 sin x ] = 2 x sin x + x 2 cos x Quotient d d x [ f g ] = f ′ g − f g ′ g 2 \displaystyle \frac{d}{dx} \left[ \frac{f}{g} \right] = \frac{f' g - f g'}{g^2} d x d [ g f ] = g 2 f ′ g − f g ′ Example: d d x [ x x + 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} d x d [ x + 1 x ] = ( x + 1 ) 2 ( x + 1 ) − x = ( x + 1 ) 2 1 Chain d d x f ( g ( x ) ) = f ′ ( g ( x ) ) g ′ ( x ) \displaystyle \frac{d}{dx}\, f(g(x)) = f'(g(x))\, g'(x) d x d f ( g ( x )) = f ′ ( g ( x )) g ′ ( x ) Example: d d x sin ( x 3 ) = cos ( x 3 ) ⋅ 3 x 2 \displaystyle \frac{d}{dx} \sin(x^3) = \cos(x^3) \cdot 3x^2 d x d sin ( x 3 ) = cos ( x 3 ) ⋅ 3 x 2 Inverse function ( f − 1 ) ′ ( a ) = 1 f ′ ( f − 1 ( a ) ) \displaystyle \left(f^{-1}\right)'(a) = \frac{1}{f'\left(f^{-1}(a)\right)} ( f − 1 ) ′ ( a ) = f ′ ( f − 1 ( a ) ) 1 Example: f ( x ) = x 3 + x \displaystyle f(x) = x^3 + x f ( x ) = x 3 + x has f ( 1 ) = 2 \displaystyle f(1) = 2 f ( 1 ) = 2 and f ′ ( 1 ) = 4 \displaystyle f'(1) = 4 f ′ ( 1 ) = 4 , so ( f − 1 ) ′ ( 2 ) = 1 4 \displaystyle \left(f^{-1}\right)'(2) = \frac{1}{4} ( f − 1 ) ′ ( 2 ) = 4 1 Implicit Differentiate both sides in x \displaystyle x x , writing d y d x \displaystyle \frac{dy}{dx} d x d y each time you differentiate y \displaystyle y y . Example: x 2 + y 2 = 25 \displaystyle x^2 + y^2 = 25 x 2 + y 2 = 25 gives 2 x + 2 y d y d x = 0 \displaystyle 2x + 2y \frac{dy}{dx} = 0 2 x + 2 y d x d y = 0 , so d y d x = − x y \displaystyle \frac{dy}{dx} = -\frac{x}{y} d x d y = − y x Derivatives to knowUnits 2–3 How to read these Each rule is written for any inside function f ( x ) \displaystyle f(x) f ( x ) , chain rule included. When the inside is just x \displaystyle x x , f ′ ( x ) = 1 \displaystyle f'(x) = 1 f ′ ( x ) = 1 and you get the basic rule. Power d d x [ 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) d x d [ f ( x ) ] n = n [ f ( x ) ] n − 1 ⋅ f ′ ( x ) Example: d d x ( 3 x 2 + 1 ) 5 = 5 ( 3 x 2 + 1 ) 4 ⋅ 6 x \displaystyle \frac{d}{dx} (3x^2 + 1)^5 = 5(3x^2 + 1)^4 \cdot 6x d x d ( 3 x 2 + 1 ) 5 = 5 ( 3 x 2 + 1 ) 4 ⋅ 6 x Exponential d d x e f ( x ) = e f ( x ) ⋅ f ′ ( x ) \displaystyle \frac{d}{dx}\, e^{f(x)} = e^{f(x)} \cdot f'(x) d x d e f ( x ) = e f ( x ) ⋅ f ′ ( x ) · d d x a f ( x ) = a f ( x ) ln a ⋅ f ′ ( x ) \displaystyle \frac{d}{dx}\, a^{f(x)} = a^{f(x)} \ln a \cdot f'(x) d x d a f ( x ) = a f ( x ) ln a ⋅ f ′ ( x ) Example: d d x e x 2 = e x 2 ⋅ 2 x \displaystyle \frac{d}{dx}\, e^{x^2} = e^{x^2} \cdot 2x d x d e x 2 = e x 2 ⋅ 2 x Log d d x ln ( f ( x ) ) = f ′ ( x ) f ( x ) \displaystyle \frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)} d x d ln ( f ( x )) = f ( x ) f ′ ( x ) · d d x log a ( f ( x ) ) = f ′ ( x ) f ( x ) ln a \displaystyle \frac{d}{dx} \log_a(f(x)) = \frac{f'(x)}{f(x) \ln a} d x d log a ( f ( x )) = f ( x ) ln a f ′ ( x ) Example: d d x ln ( x 2 + 4 ) = 2 x x 2 + 4 \displaystyle \frac{d}{dx} \ln(x^2 + 4) = \frac{2x}{x^2 + 4} d x d ln ( x 2 + 4 ) = x 2 + 4 2 x Sine and cosine d d x sin ( f ( x ) ) = cos ( f ( x ) ) ⋅ f ′ ( x ) \displaystyle \frac{d}{dx} \sin(f(x)) = \cos(f(x)) \cdot f'(x) d x d sin ( f ( x )) = cos ( f ( x )) ⋅ f ′ ( x ) · d d x cos ( f ( x ) ) = − sin ( f ( x ) ) ⋅ f ′ ( x ) \displaystyle \frac{d}{dx} \cos(f(x)) = -\sin(f(x)) \cdot f'(x) d x d cos ( f ( x )) = − sin ( f ( x )) ⋅ f ′ ( x ) Example: d d x cos ( 3 x ) = − 3 sin ( 3 x ) \displaystyle \frac{d}{dx} \cos(3x) = -3 \sin(3x) d x d cos ( 3 x ) = − 3 sin ( 3 x ) Tangent and cotangent d d x tan ( f ( x ) ) = sec 2 ( f ( x ) ) ⋅ f ′ ( x ) \displaystyle \frac{d}{dx} \tan(f(x)) = \sec^2(f(x)) \cdot f'(x) d x d tan ( f ( x )) = sec 2 ( f ( x )) ⋅ f ′ ( x ) · d d x cot ( f ( x ) ) = − csc 2 ( f ( x ) ) ⋅ f ′ ( x ) \displaystyle \frac{d}{dx} \cot(f(x)) = -\csc^2(f(x)) \cdot f'(x) d x d cot ( f ( x )) = − csc 2 ( f ( x )) ⋅ f ′ ( x ) Example: d d x tan ( x 2 ) = sec 2 ( x 2 ) ⋅ 2 x \displaystyle \frac{d}{dx} \tan(x^2) = \sec^2(x^2) \cdot 2x d x d tan ( x 2 ) = sec 2 ( x 2 ) ⋅ 2 x Secant and cosecant d d x sec ( 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) d x d sec ( f ( x )) = sec ( f ( x )) tan ( f ( x )) ⋅ f ′ ( x ) · d d x csc ( 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) d x d csc ( f ( x )) = − csc ( f ( x )) cot ( f ( x )) ⋅ f ′ ( x ) Example: d d x sec ( 4 x ) = 4 sec ( 4 x ) tan ( 4 x ) \displaystyle \frac{d}{dx} \sec(4x) = 4 \sec(4x) \tan(4x) d x d sec ( 4 x ) = 4 sec ( 4 x ) tan ( 4 x ) Inverse trig d d x arcsin ( f ( x ) ) = f ′ ( x ) 1 − [ f ( x ) ] 2 \displaystyle \frac{d}{dx} \arcsin(f(x)) = \frac{f'(x)}{\sqrt{1 - [f(x)]^2}} d x d arcsin ( f ( x )) = 1 − [ f ( x ) ] 2 f ′ ( x ) · d d x arccos ( f ( x ) ) = − f ′ ( x ) 1 − [ f ( x ) ] 2 \displaystyle \frac{d}{dx} \arccos(f(x)) = \frac{-f'(x)}{\sqrt{1 - [f(x)]^2}} d x d arccos ( f ( x )) = 1 − [ f ( x ) ] 2 − f ′ ( x ) · d d x arctan ( f ( x ) ) = f ′ ( x ) 1 + [ f ( x ) ] 2 \displaystyle \frac{d}{dx} \arctan(f(x)) = \frac{f'(x)}{1 + [f(x)]^2} d x d arctan ( f ( x )) = 1 + [ f ( x ) ] 2 f ′ ( x ) Example: d d x arctan ( 2 x ) = 2 1 + 4 x 2 \displaystyle \frac{d}{dx} \arctan(2x) = \frac{2}{1 + 4x^2} d x d arctan ( 2 x ) = 1 + 4 x 2 2 Derivatives in contextUnit 4 Motion v ( t ) = x ′ ( t ) \displaystyle v(t) = x'(t) v ( t ) = x ′ ( t ) · a ( t ) = v ′ ( t ) = x ′ ′ ( t ) \displaystyle a(t) = v'(t) = x''(t) a ( t ) = v ′ ( t ) = x ′′ ( t ) · speed = ∣ v ( t ) ∣ \displaystyle \text{speed} = |v(t)| speed = ∣ v ( t ) ∣ Example: x ( t ) = t 3 − 6 t \displaystyle x(t) = t^3 - 6t x ( t ) = t 3 − 6 t gives v ( t ) = 3 t 2 − 6 \displaystyle v(t) = 3t^2 - 6 v ( t ) = 3 t 2 − 6 and a ( t ) = 6 t \displaystyle a(t) = 6t a ( t ) = 6 t Speeding up When v ( t ) \displaystyle v(t) v ( t ) and a ( t ) \displaystyle a(t) 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 v ( 1 ) = − 3 and a ( 1 ) = 6 \displaystyle a(1) = 6 a ( 1 ) = 6 , so at t = 1 \displaystyle t = 1 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 t . Example: A = π r 2 \displaystyle A = \pi r^2 A = π r 2 gives d A d t = 2 π r d r d t \displaystyle \frac{dA}{dt} = 2\pi r \frac{dr}{dt} d t d A = 2 π r d t d r L'Hospital's Rule If lim f ( x ) g ( x ) \displaystyle \lim \frac{f(x)}{g(x)} lim g ( x ) f ( x ) has the form 0 0 \displaystyle \frac{0}{0} 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)} lim g ( x ) f ( x ) = lim g ′ ( x ) f ′ ( x ) . When the limit on the right exists. Check the form first. Example: lim x → 0 e x − 1 x \displaystyle \lim_{x \to 0} \frac{e^x - 1}{x} x → 0 lim x e x − 1 has the form 0 0 \displaystyle \frac{0}{0} 0 0 , so it equals lim x → 0 e x 1 = 1 \displaystyle \lim_{x \to 0} \frac{e^x}{1} = 1 x → 0 lim 1 e x = 1 Analyzing functionsUnit 5 Mean Value Theorem If f \displaystyle f f is continuous on [ a , b ] \displaystyle [a, b] [ a , b ] and differentiable on ( a , b ) \displaystyle (a, b) ( a , b ) , then f ′ ( c ) = f ( b ) − f ( a ) b − a \displaystyle f'(c) = \frac{f(b) - f(a)}{b - a} f ′ ( c ) = b − a f ( b ) − f ( a ) for some c \displaystyle c c in ( a , b ) \displaystyle (a, b) ( a , b ) . Example: f ( x ) = x 2 \displaystyle f(x) = x^2 f ( x ) = x 2 on [ 0 , 2 ] \displaystyle [0, 2] [ 0 , 2 ] : f ′ ( c ) = 4 − 0 2 − 0 = 2 \displaystyle f'(c) = \frac{4 - 0}{2 - 0} = 2 f ′ ( c ) = 2 − 0 4 − 0 = 2 , so 2 c = 2 \displaystyle 2c = 2 2 c = 2 and c = 1 \displaystyle c = 1 c = 1 . Extreme Value Theorem If f \displaystyle f f is continuous on [ a , b ] \displaystyle [a, b] [ a , b ] , then f \displaystyle f f has an absolute maximum and minimum on [ a , b ] \displaystyle [a, b] [ a , b ] . Example: f ( x ) = 1 x \displaystyle f(x) = \frac{1}{x} f ( x ) = x 1 on [ − 1 , 1 ] \displaystyle [-1, 1] [ − 1 , 1 ] is not continuous at x = 0 \displaystyle x = 0 x = 0 , so the theorem does not apply (and f \displaystyle f f has no maximum there). Critical point f ′ ( c ) = 0 \displaystyle f'(c) = 0 f ′ ( c ) = 0 or f ′ ( c ) \displaystyle f'(c) f ′ ( c ) does not existExample: f ( x ) = x 2 / 3 \displaystyle f(x) = x^{2/3} f ( x ) = x 2/3 has f ′ ( x ) = 2 3 x − 1 / 3 \displaystyle f'(x) = \frac{2}{3} x^{-1/3} f ′ ( x ) = 3 2 x − 1/3 , which does not exist at x = 0 \displaystyle x = 0 x = 0 , so x = 0 \displaystyle x = 0 x = 0 is a critical point. Increasing / decreasing f ′ > 0 \displaystyle f' > 0 f ′ > 0 : f \displaystyle f f increasing · f ′ < 0 \displaystyle f' < 0 f ′ < 0 : f \displaystyle f f decreasingExample: f ( x ) = x 2 − 4 x \displaystyle f(x) = x^2 - 4x f ( x ) = x 2 − 4 x has f ′ ( x ) = 2 x − 4 > 0 \displaystyle f'(x) = 2x - 4 > 0 f ′ ( x ) = 2 x − 4 > 0 for x > 2 \displaystyle x > 2 x > 2 , so f \displaystyle f f is increasing on ( 2 , ∞ ) \displaystyle (2, \infty) ( 2 , ∞ ) . First derivative test f ′ \displaystyle f' 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) f ′ ( x ) = ( x − 1 ) ( x − 3 ) goes + \displaystyle + + to − \displaystyle - − at x = 1 \displaystyle x = 1 x = 1 (relative max) and − \displaystyle - − to + \displaystyle + + at x = 3 \displaystyle x = 3 x = 3 (relative min). Concavity f ′ ′ > 0 \displaystyle f'' > 0 f ′′ > 0 (or f ′ \displaystyle f' f ′ increasing): concave up · f ′ ′ < 0 \displaystyle f'' < 0 f ′′ < 0 (or f ′ \displaystyle f' f ′ decreasing): concave downExample: f ( x ) = x 3 \displaystyle f(x) = x^3 f ( x ) = x 3 has f ′ ′ ( x ) = 6 x > 0 \displaystyle f''(x) = 6x > 0 f ′′ ( x ) = 6 x > 0 for x > 0 \displaystyle x > 0 x > 0 , so f \displaystyle f f is concave up on ( 0 , ∞ ) \displaystyle (0, \infty) ( 0 , ∞ ) . Second derivative test f ′ ( c ) = 0 \displaystyle f'(c) = 0 f ′ ( c ) = 0 and f ′ ′ ( c ) < 0 \displaystyle f''(c) < 0 f ′′ ( c ) < 0 : relative max · f ′ ( c ) = 0 \displaystyle f'(c) = 0 f ′ ( c ) = 0 and f ′ ′ ( c ) > 0 \displaystyle f''(c) > 0 f ′′ ( c ) > 0 : relative minExample: f ( x ) = x 3 − 3 x \displaystyle f(x) = x^3 - 3x f ( x ) = x 3 − 3 x has f ′ ( 1 ) = 0 \displaystyle f'(1) = 0 f ′ ( 1 ) = 0 and f ′ ′ ( 1 ) = 6 > 0 \displaystyle f''(1) = 6 > 0 f ′′ ( 1 ) = 6 > 0 , so f \displaystyle f f has a relative min at x = 1 \displaystyle x = 1 x = 1 . Point of inflection Where f ′ ′ \displaystyle f'' f ′′ changes sign (where f ′ \displaystyle f' f ′ changes from increasing to decreasing, or back) Example: f ( x ) = x 4 \displaystyle f(x) = x^4 f ( x ) = x 4 has f ′ ′ ( 0 ) = 0 \displaystyle f''(0) = 0 f ′′ ( 0 ) = 0 , but f ′ ′ ( x ) = 12 x 2 \displaystyle f''(x) = 12x^2 f ′′ ( x ) = 12 x 2 does not change sign, so there is no point of inflection at x = 0 \displaystyle x = 0 x = 0 . Absolute extrema on [a, b] Compare f \displaystyle f f at the endpoints and at every critical point (the candidates test). Example: f ( x ) = x 3 − 3 x \displaystyle f(x) = x^3 - 3x f ( x ) = x 3 − 3 x on [ 0 , 2 ] \displaystyle [0, 2] [ 0 , 2 ] : f ( 0 ) = 0 \displaystyle f(0) = 0 f ( 0 ) = 0 , f ( 1 ) = − 2 \displaystyle f(1) = -2 f ( 1 ) = − 2 , f ( 2 ) = 2 \displaystyle f(2) = 2 f ( 2 ) = 2 , so the absolute max is 2 and the absolute min is − 2 \displaystyle -2 − 2 . Integrals to knowUnit 6 How to read these Most exam integrals have an inside of the form k x \displaystyle kx k x : integrate as if it were x \displaystyle x x , then divide by k \displaystyle k k . For any other inside function, use u \displaystyle u u -substitution (below). Power ∫ x n d x = x n + 1 n + 1 + C \displaystyle \int x^n \, dx = \frac{x^{n + 1}}{n + 1} + C ∫ x n d x = n + 1 x n + 1 + C for n ≠ − 1 \displaystyle n \ne -1 n = − 1 Example: ∫ ( 2 x + 1 ) 3 d x = ( 2 x + 1 ) 4 8 + C \displaystyle \int (2x + 1)^3 \, dx = \frac{(2x + 1)^4}{8} + C ∫ ( 2 x + 1 ) 3 d x = 8 ( 2 x + 1 ) 4 + C Log ∫ 1 x d x = ln ∣ x ∣ + C \displaystyle \int \frac{1}{x} \, dx = \ln |x| + C ∫ x 1 d x = ln ∣ x ∣ + C · ∫ f ′ ( x ) f ( x ) d x = ln ∣ f ( x ) ∣ + C \displaystyle \int \frac{f'(x)}{f(x)} \, dx = \ln |f(x)| + C ∫ f ( x ) f ′ ( x ) d x = ln ∣ f ( x ) ∣ + C Example: ∫ 2 x x 2 + 5 d x = ln ( x 2 + 5 ) + C \displaystyle \int \frac{2x}{x^2 + 5} \, dx = \ln(x^2 + 5) + C ∫ x 2 + 5 2 x d x = ln ( x 2 + 5 ) + C Exponential ∫ e k x d x = 1 k e k x + C \displaystyle \int e^{kx} \, dx = \frac{1}{k} e^{kx} + C ∫ e k x d x = k 1 e k x + C · ∫ a k x d x = a k x k ln a + C \displaystyle \int a^{kx} \, dx = \frac{a^{kx}}{k \ln a} + C ∫ a k x d x = k ln a a k x + C Example: ∫ e 3 x d x = 1 3 e 3 x + C \displaystyle \int e^{3x} \, dx = \frac{1}{3} e^{3x} + C ∫ e 3 x d x = 3 1 e 3 x + C Sine and cosine ∫ cos ( k x ) d x = 1 k sin ( k x ) + C \displaystyle \int \cos(kx) \, dx = \frac{1}{k} \sin(kx) + C ∫ cos ( k x ) d x = k 1 sin ( k x ) + C · ∫ sin ( k x ) d x = − 1 k cos ( k x ) + C \displaystyle \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C ∫ sin ( k x ) d x = − k 1 cos ( k x ) + C Example: ∫ sin ( 2 x ) d x = − 1 2 cos ( 2 x ) + C \displaystyle \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C ∫ sin ( 2 x ) d x = − 2 1 cos ( 2 x ) + C Secant² and cosecant² ∫ sec 2 ( k x ) d x = 1 k tan ( k x ) + C \displaystyle \int \sec^2(kx) \, dx = \frac{1}{k} \tan(kx) + C ∫ sec 2 ( k x ) d x = k 1 tan ( k x ) + C · ∫ csc 2 ( k x ) d x = − 1 k cot ( k x ) + C \displaystyle \int \csc^2(kx) \, dx = -\frac{1}{k} \cot(kx) + C ∫ csc 2 ( k x ) d x = − k 1 cot ( k x ) + C Example: ∫ sec 2 ( 5 x ) d x = 1 5 tan ( 5 x ) + C \displaystyle \int \sec^2(5x) \, dx = \frac{1}{5} \tan(5x) + C ∫ sec 2 ( 5 x ) d x = 5 1 tan ( 5 x ) + C Secant‑tangent and cosecant‑cotangent ∫ sec ( k x ) tan ( k x ) d x = 1 k sec ( k x ) + C \displaystyle \int \sec(kx) \tan(kx) \, dx = \frac{1}{k} \sec(kx) + C ∫ sec ( k x ) tan ( k x ) d x = k 1 sec ( k x ) + C · ∫ csc ( k x ) cot ( k x ) d x = − 1 k csc ( k x ) + C \displaystyle \int \csc(kx) \cot(kx) \, dx = -\frac{1}{k} \csc(kx) + C ∫ csc ( k x ) cot ( k x ) d x = − k 1 csc ( k x ) + C Example: ∫ sec ( 3 x ) tan ( 3 x ) d x = 1 3 sec ( 3 x ) + C \displaystyle \int \sec(3x) \tan(3x) \, dx = \frac{1}{3} \sec(3x) + C ∫ sec ( 3 x ) tan ( 3 x ) d x = 3 1 sec ( 3 x ) + C Tangent and cotangent ∫ tan x d x = − ln ∣ cos x ∣ + C \displaystyle \int \tan x \, dx = -\ln |\cos x| + C ∫ tan x d x = − ln ∣ cos x ∣ + C · ∫ cot x d x = ln ∣ sin x ∣ + C \displaystyle \int \cot x \, dx = \ln |\sin x| + C ∫ cot x d x = ln ∣ sin x ∣ + C Both come from the log rule: tan x = sin x cos x \displaystyle \tan x = \frac{\sin x}{\cos x} tan x = cos x sin x . Inverse trig ∫ 1 1 − x 2 d x = arcsin x + C \displaystyle \int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin x + C ∫ 1 − x 2 1 d x = arcsin x + C · ∫ 1 1 + x 2 d x = arctan x + C \displaystyle \int \frac{1}{1 + x^2} \, dx = \arctan x + C ∫ 1 + x 2 1 d x = arctan x + C Example: ∫ 0 1 1 1 + x 2 d x = arctan 1 − arctan 0 = π 4 \displaystyle \int_0^1 \frac{1}{1 + x^2} \, dx = \arctan 1 - \arctan 0 = \frac{\pi}{4} ∫ 0 1 1 + x 2 1 d x = arctan 1 − arctan 0 = 4 π u-substitution ∫ g ( f ( x ) ) f ′ ( x ) d x = ∫ g ( u ) d u \displaystyle \int g(f(x)) \, f'(x) \, dx = \int g(u) \, du ∫ g ( f ( x )) f ′ ( x ) d x = ∫ g ( u ) d u with u = f ( x ) \displaystyle u = f(x) u = f ( x ) , d u = f ′ ( x ) d x \displaystyle du = f'(x) \, dx d u = f ′ ( x ) d x Example: ∫ 2 x cos ( x 2 ) d x = sin ( x 2 ) + C \displaystyle \int 2x \cos(x^2) \, dx = \sin(x^2) + C ∫ 2 x cos ( x 2 ) d x = sin ( x 2 ) + C , with u = x 2 \displaystyle u = x^2 u = x 2 Definite integrals With u = f ( x ) \displaystyle u = f(x) u = f ( x ) , change the limits too: ∫ a b g ( f ( x ) ) f ′ ( x ) d x = ∫ f ( a ) f ( b ) g ( u ) d u \displaystyle \int_a^b g(f(x)) \, f'(x) \, dx = \int_{f(a)}^{f(b)} g(u) \, du ∫ a b g ( f ( x )) f ′ ( x ) d x = ∫ f ( a ) f ( b ) g ( u ) d u Example: ∫ 0 2 2 x e x 2 d x = ∫ 0 4 e u d u = e 4 − 1 \displaystyle \int_0^2 2x \, e^{x^2} \, dx = \int_0^4 e^u \, du = e^4 - 1 ∫ 0 2 2 x e x 2 d x = ∫ 0 4 e u d u = e 4 − 1 , with u = x 2 \displaystyle u = x^2 u = x 2 Accumulation and the Fundamental TheoremUnit 6 Fundamental Theorem: evaluating ∫ a b f ′ ( x ) d x = f ( b ) − f ( a ) \displaystyle \int_a^b f'(x) \, dx = f(b) - f(a) ∫ a b f ′ ( x ) d x = f ( b ) − f ( a ) Example: ∫ 0 π sin x d x = − cos π − ( − cos 0 ) = 2 \displaystyle \int_0^{\pi} \sin x \, dx = -\cos \pi - (-\cos 0) = 2 ∫ 0 π sin x d x = − cos π − ( − cos 0 ) = 2 Net change f ( b ) = f ( a ) + ∫ a b f ′ ( x ) d x \displaystyle f(b) = f(a) + \int_a^b f'(x) \, dx f ( b ) = f ( a ) + ∫ a b f ′ ( x ) d x Example: if f ( 2 ) = 5 \displaystyle f(2) = 5 f ( 2 ) = 5 and f ′ ( x ) = 3 x 2 \displaystyle f'(x) = 3x^2 f ′ ( x ) = 3 x 2 , then f ( 4 ) = 5 + ∫ 2 4 3 x 2 d x = 5 + 56 = 61 \displaystyle f(4) = 5 + \int_2^4 3x^2 \, dx = 5 + 56 = 61 f ( 4 ) = 5 + ∫ 2 4 3 x 2 d x = 5 + 56 = 61 Fundamental Theorem: derivative of an integral d d x ∫ a x f ( t ) d t = f ( x ) \displaystyle \frac{d}{dx} \int_a^x f(t) \, dt = f(x) d x d ∫ a x f ( t ) d t = f ( x ) · d d x ∫ a g ( x ) f ( t ) d t = f ( g ( x ) ) ⋅ g ′ ( x ) \displaystyle \frac{d}{dx} \int_a^{g(x)} f(t) \, dt = f(g(x)) \cdot g'(x) d x d ∫ a g ( x ) f ( t ) d t = f ( g ( x )) ⋅ g ′ ( x ) Example: if G ( x ) = ∫ 0 x 2 cos t d t \displaystyle G(x) = \int_0^{x^2} \cos t \, dt G ( x ) = ∫ 0 x 2 cos t d t , then G ′ ( x ) = cos ( x 2 ) ⋅ 2 x \displaystyle G'(x) = \cos(x^2) \cdot 2x G ′ ( x ) = cos ( x 2 ) ⋅ 2 x . Plug the top limit into f \displaystyle f f , then multiply by its derivative. Both limits move d d x ∫ h ( x ) g ( x ) f ( t ) d t = 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) d x d ∫ h ( x ) g ( x ) f ( t ) d t = f ( g ( x )) ⋅ g ′ ( x ) − f ( h ( x )) ⋅ h ′ ( x ) Example: d d x ∫ x 3 x e t 2 d t = 3 e 9 x 2 − e x 2 \displaystyle \frac{d}{dx} \int_{x}^{3x} e^{t^2} \, dt = 3e^{9x^2} - e^{x^2} d x d ∫ x 3 x e t 2 d t = 3 e 9 x 2 − e x 2 Properties ∫ a a f ( x ) d x = 0 \displaystyle \int_a^a f(x) \, dx = 0 ∫ a a f ( x ) d x = 0 · ∫ b a f ( x ) d x = − ∫ a b f ( x ) d x \displaystyle \int_b^a f(x) \, dx = -\int_a^b f(x) \, dx ∫ b a f ( x ) d x = − ∫ a b f ( x ) d x · ∫ a b f ( x ) d x + ∫ b c f ( x ) d x = ∫ a c f ( x ) d x \displaystyle \int_a^b f(x) \, dx + \int_b^c f(x) \, dx = \int_a^c f(x) \, dx ∫ a b f ( x ) d x + ∫ b c f ( x ) d x = ∫ a c f ( x ) d x Example: if ∫ 1 3 f ( x ) d x = 4 \displaystyle \int_1^3 f(x) \, dx = 4 ∫ 1 3 f ( x ) d x = 4 and ∫ 3 5 f ( x ) d x = 2 \displaystyle \int_3^5 f(x) \, dx = 2 ∫ 3 5 f ( x ) d x = 2 , then ∫ 1 5 f ( x ) d x = 6 \displaystyle \int_1^5 f(x) \, dx = 6 ∫ 1 5 f ( x ) d x = 6 and ∫ 3 1 f ( x ) d x = − 4 \displaystyle \int_3^1 f(x) \, dx = -4 ∫ 3 1 f ( x ) d x = − 4 Approximating an integralUnit 6 Left and right sums Left: Δ x [ f ( x 0 ) + f ( x 1 ) + ⋯ + f ( x n − 1 ) ] \displaystyle \Delta x \, [f(x_0) + f(x_1) + \dots + f(x_{n-1})] Δ x [ f ( x 0 ) + f ( x 1 ) + ⋯ + f ( x n − 1 )] · Right: Δ x [ f ( x 1 ) + f ( x 2 ) + ⋯ + f ( x n ) ] \displaystyle \Delta x \, [f(x_1) + f(x_2) + \dots + f(x_n)] Δ x [ f ( x 1 ) + f ( x 2 ) + ⋯ + f ( x n )] With unequal widths (a table), multiply each height by its own width. Example: f ( x ) = x 2 \displaystyle f(x) = x^2 f ( x ) = x 2 on [ 0 , 2 ] \displaystyle [0, 2] [ 0 , 2 ] with Δ x = 1 \displaystyle \Delta x = 1 Δ x = 1 : left = 1 ⋅ ( 0 + 1 ) = 1 \displaystyle = 1 \cdot (0 + 1) = 1 = 1 ⋅ ( 0 + 1 ) = 1 , right = 1 ⋅ ( 1 + 4 ) = 5 \displaystyle = 1 \cdot (1 + 4) = 5 = 1 ⋅ ( 1 + 4 ) = 5 Midpoint sum Δ x [ f ( m 1 ) + f ( m 2 ) + ⋯ + f ( m n ) ] \displaystyle \Delta x \, [f(m_1) + f(m_2) + \dots + f(m_n)] Δ x [ f ( m 1 ) + f ( m 2 ) + ⋯ + f ( m n )] , each m i \displaystyle m_i m i the midpoint of its intervalExample: same f \displaystyle f 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 1 ⋅ [ f ( 0.5 ) + f ( 1.5 )] = 0.25 + 2.25 = 2.5 Trapezoidal sum Δ x 2 [ f ( x 0 ) + 2 f ( x 1 ) + ⋯ + 2 f ( x n − 1 ) + f ( x n ) ] \displaystyle \frac{\Delta x}{2} \, [f(x_0) + 2f(x_1) + \dots + 2f(x_{n-1}) + f(x_n)] 2 Δ x [ f ( x 0 ) + 2 f ( x 1 ) + ⋯ + 2 f ( x n − 1 ) + f ( x n )] Example: same f \displaystyle f f and intervals: 1 2 [ 0 + 2 ( 1 ) + 4 ] = 3 \displaystyle \frac{1}{2} [0 + 2(1) + 4] = 3 2 1 [ 0 + 2 ( 1 ) + 4 ] = 3 Over or under A left sum underestimates when f \displaystyle f f is increasing and overestimates when it is decreasing; a trapezoidal sum underestimates when f \displaystyle f f is concave down. Example: x 2 \displaystyle x^2 x 2 on [ 0 , 2 ] \displaystyle [0, 2] [ 0 , 2 ] is increasing and concave up, so left = 1 \displaystyle = 1 = 1 is under and trapezoid = 3 \displaystyle = 3 = 3 is over the exact 8 3 \displaystyle \frac{8}{3} 3 8 . Limit of a sum ∫ a b f ( x ) d x = lim n → ∞ ∑ k = 1 n f ( x k ) Δ x \displaystyle \int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{k = 1}^{n} f(x_k) \, \Delta x ∫ a b f ( x ) d x = n → ∞ lim k = 1 ∑ n f ( x k ) Δ x with Δ x = b − a n \displaystyle \Delta x = \frac{b - a}{n} Δ x = n b − a Example: lim n → ∞ ∑ k = 1 n ( k n ) 2 1 n = ∫ 0 1 x 2 d x = 1 3 \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} n → ∞ lim k = 1 ∑ n ( n k ) 2 n 1 = ∫ 0 1 x 2 d x = 3 1 Differential equationsUnit 7 Separation of variables d y d x = g ( x ) h ( y ) \displaystyle \frac{dy}{dx} = g(x)\, h(y) d x d y = g ( x ) h ( y ) gives ∫ d y h ( y ) = ∫ g ( x ) d x \displaystyle \int \frac{dy}{h(y)} = \int g(x) \, dx ∫ h ( y ) d y = ∫ g ( x ) d x Add + C before solving for y, then use the initial condition. Example: d y d x = x y \displaystyle \frac{dy}{dx} = xy d x d y = x y with y ( 0 ) = 2 \displaystyle y(0) = 2 y ( 0 ) = 2 : ln ∣ y ∣ = x 2 2 + C \displaystyle \ln |y| = \frac{x^2}{2} + C ln ∣ y ∣ = 2 x 2 + C , so y = 2 e x 2 / 2 \displaystyle y = 2e^{x^2/2} y = 2 e x 2 /2 Exponential growth and decay d y d t = k y \displaystyle \frac{dy}{dt} = k y d t d y = k y gives y = y 0 e k t \displaystyle y = y_0 \, e^{k t} y = y 0 e k t Example: d P d t = 0.03 P \displaystyle \frac{dP}{dt} = 0.03P d t d P = 0.03 P with P ( 0 ) = 100 \displaystyle P(0) = 100 P ( 0 ) = 100 gives P = 100 e 0.03 t \displaystyle P = 100e^{0.03t} P = 100 e 0.03 t Slope field At each point ( x , y ) \displaystyle (x, y) ( x , y ) , draw a short segment with slope d y d x \displaystyle \frac{dy}{dx} d x d y . Example: for d y d x = x − y \displaystyle \frac{dy}{dx} = x - y d x d y = x − y , the segment at ( 2 , 1 ) \displaystyle (2, 1) ( 2 , 1 ) has slope 2 − 1 = 1 \displaystyle 2 - 1 = 1 2 − 1 = 1 . Applications of integrationUnit 8 Average value f avg = 1 b − a ∫ a b f ( x ) d x \displaystyle f_{\text{avg}} = \frac{1}{b - a} \int_a^b f(x) \, dx f avg = b − a 1 ∫ a b f ( x ) d x Example: f ( x ) = x 2 \displaystyle f(x) = x^2 f ( x ) = x 2 on [ 0 , 3 ] \displaystyle [0, 3] [ 0 , 3 ] : 1 3 ∫ 0 3 x 2 d x = 1 3 ⋅ 9 = 3 \displaystyle \frac{1}{3} \int_0^3 x^2 \, dx = \frac{1}{3} \cdot 9 = 3 3 1 ∫ 0 3 x 2 d x = 3 1 ⋅ 9 = 3 Displacement ∫ a b v ( t ) d t \displaystyle \int_a^b v(t) \, dt ∫ a b v ( t ) d t Example: v ( t ) = t − 2 \displaystyle v(t) = t - 2 v ( t ) = t − 2 on [ 0 , 4 ] \displaystyle [0, 4] [ 0 , 4 ] : ∫ 0 4 ( t − 2 ) d t = 0 \displaystyle \int_0^4 (t - 2) \, dt = 0 ∫ 0 4 ( t − 2 ) d t = 0 , so the particle ends where it started. Total distance ∫ a b ∣ v ( t ) ∣ d t \displaystyle \int_a^b |v(t)| \, dt ∫ a b ∣ v ( t ) ∣ d t Example: same v ( t ) \displaystyle v(t) v ( t ) : ∫ 0 4 ∣ t − 2 ∣ d t = 2 + 2 = 4 \displaystyle \int_0^4 |t - 2| \, dt = 2 + 2 = 4 ∫ 0 4 ∣ t − 2∣ d t = 2 + 2 = 4 Position x ( t ) = x ( 0 ) + ∫ 0 t v ( s ) d s \displaystyle x(t) = x(0) + \int_0^t v(s) \, ds x ( t ) = x ( 0 ) + ∫ 0 t v ( s ) d s Example: x ( 0 ) = 3 \displaystyle x(0) = 3 x ( 0 ) = 3 and v ( t ) = 2 t \displaystyle v(t) = 2t v ( t ) = 2 t : x ( 3 ) = 3 + ∫ 0 3 2 s d s = 3 + 9 = 12 \displaystyle x(3) = 3 + \int_0^3 2s \, ds = 3 + 9 = 12 x ( 3 ) = 3 + ∫ 0 3 2 s d s = 3 + 9 = 12 Area between curves ∫ a b [ top − bottom ] d x \displaystyle \int_a^b [\text{top} - \text{bottom}] \, dx ∫ a b [ top − bottom ] d x · ∫ c d [ right − left ] d y \displaystyle \int_c^d [\text{right} - \text{left}] \, dy ∫ c d [ right − left ] d y Example: between y = x \displaystyle y = x y = x and y = x 2 \displaystyle y = x^2 y = x 2 on [ 0 , 1 ] \displaystyle [0, 1] [ 0 , 1 ] : ∫ 0 1 ( x − x 2 ) d x = 1 6 \displaystyle \int_0^1 (x - x^2) \, dx = \frac{1}{6} ∫ 0 1 ( x − x 2 ) d x = 6 1 Disc V = π ∫ a b [ R ( x ) ] 2 d x \displaystyle V = \pi \int_a^b [R(x)]^2 \, dx V = π ∫ a b [ R ( x ) ] 2 d x Example: y = x \displaystyle y = \sqrt{x} y = x on [ 0 , 4 ] \displaystyle [0, 4] [ 0 , 4 ] about the x \displaystyle x x -axis: π ∫ 0 4 x d x = 8 π \displaystyle \pi \int_0^4 x \, dx = 8\pi π ∫ 0 4 x d x = 8 π Washer V = π ∫ a b ( [ R ( x ) ] 2 − [ r ( x ) ] 2 ) d x \displaystyle V = \pi \int_a^b \left( [R(x)]^2 - [r(x)]^2 \right) dx V = π ∫ a b ( [ R ( x ) ] 2 − [ r ( x ) ] 2 ) d x R and r are measured from the axis of rotation. Example: the region between y = x \displaystyle y = x y = x and y = x 2 \displaystyle y = x^2 y = x 2 about the x \displaystyle x x -axis: π ∫ 0 1 ( x 2 − x 4 ) d x = 2 π 15 \displaystyle \pi \int_0^1 (x^2 - x^4) \, dx = \frac{2\pi}{15} π ∫ 0 1 ( x 2 − x 4 ) d x = 15 2 π Cross sections V = ∫ a b A ( x ) d x \displaystyle V = \int_a^b A(x) \, dx V = ∫ a b A ( x ) d x · squares A = s 2 \displaystyle A = s^2 A = s 2 · semicircles A = 1 2 π ( s 2 ) 2 \displaystyle A = \frac{1}{2}\pi \left(\frac{s}{2}\right)^2 A = 2 1 π ( 2 s ) 2 · equilateral triangles A = 3 4 s 2 \displaystyle A = \frac{\sqrt{3}}{4} s^2 A = 4 3 s 2 Example: base under y = x \displaystyle y = \sqrt{x} y = x on [ 0 , 4 ] \displaystyle [0, 4] [ 0 , 4 ] , square cross sections perpendicular to the x \displaystyle x x -axis: ∫ 0 4 ( x ) 2 d x = 8 \displaystyle \int_0^4 (\sqrt{x})^2 \, dx = 8 ∫ 0 4 ( x ) 2 d x = 8 Trig and geometry you still needAll units Identities sin 2 x + cos 2 x = 1 \displaystyle \sin^2 x + \cos^2 x = 1 sin 2 x + cos 2 x = 1 · 1 + tan 2 x = sec 2 x \displaystyle 1 + \tan^2 x = \sec^2 x 1 + tan 2 x = sec 2 x Example: 1 − cos 2 ( 3 x ) = sin 2 ( 3 x ) \displaystyle 1 - \cos^2(3x) = \sin^2(3x) 1 − cos 2 ( 3 x ) = sin 2 ( 3 x ) Double angle sin 2 x = 2 sin x cos x \displaystyle \sin 2x = 2 \sin x \cos x sin 2 x = 2 sin x cos x · cos 2 x = cos 2 x − sin 2 x = 1 − 2 sin 2 x \displaystyle \cos 2x = \cos^2 x - \sin^2 x = 1 - 2\sin^2 x cos 2 x = cos 2 x − sin 2 x = 1 − 2 sin 2 x Example: sin x cos x = 1 2 sin ( 2 x ) \displaystyle \sin x \cos x = \frac{1}{2} \sin(2x) sin x cos x = 2 1 sin ( 2 x ) , so ∫ sin x cos x d x = − 1 4 cos ( 2 x ) + C \displaystyle \int \sin x \cos x \, dx = -\frac{1}{4} \cos(2x) + C ∫ sin x cos x d x = − 4 1 cos ( 2 x ) + C Unit circle sin π 6 = 1 2 \displaystyle \sin \frac{\pi}{6} = \frac{1}{2} sin 6 π = 2 1 · sin π 4 = 2 2 \displaystyle \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} sin 4 π = 2 2 · sin π 3 = 3 2 \displaystyle \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} sin 3 π = 2 3 · cos π 6 = 3 2 \displaystyle \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} cos 6 π = 2 3 · cos π 4 = 2 2 \displaystyle \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} cos 4 π = 2 2 · cos π 3 = 1 2 \displaystyle \cos \frac{\pi}{3} = \frac{1}{2} cos 3 π = 2 1 Example: cos 2 π 3 = − 1 2 \displaystyle \cos \frac{2\pi}{3} = -\frac{1}{2} cos 3 2 π = − 2 1 : reference angle π 3 \displaystyle \frac{\pi}{3} 3 π , and cosine is negative in quadrant II. Logs ln ( a b ) = ln a + ln b \displaystyle \ln(ab) = \ln a + \ln b ln ( ab ) = ln a + ln b · ln a b = ln a − ln b \displaystyle \ln \frac{a}{b} = \ln a - \ln b ln b a = ln a − ln b · ln a k = k ln a \displaystyle \ln a^k = k \ln a ln a k = k ln a · ln e = 1 \displaystyle \ln e = 1 ln e = 1 · ln 1 = 0 \displaystyle \ln 1 = 0 ln 1 = 0 Example: ln x 2 y = 2 ln x − ln y \displaystyle \ln \frac{x^2}{y} = 2 \ln x - \ln y ln y x 2 = 2 ln x − ln y Shapes (related rates) Circle A = π r 2 \displaystyle A = \pi r^2 A = π r 2 , C = 2 π r \displaystyle C = 2\pi r C = 2 π r · sphere V = 4 3 π r 3 \displaystyle V = \frac{4}{3}\pi r^3 V = 3 4 π r 3 · cone V = 1 3 π r 2 h \displaystyle V = \frac{1}{3}\pi r^2 h V = 3 1 π r 2 h · cylinder V = π r 2 h \displaystyle V = \pi r^2 h V = π r 2 h Example: sphere V = 4 3 π r 3 \displaystyle V = \frac{4}{3}\pi r^3 V = 3 4 π r 3 gives d V d t = 4 π r 2 d r d t \displaystyle \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} d t d V = 4 π r 2 d t d r Practice every formula on this sheet. Free AP Calculus AB practice tests for all 8 units, with answer keys and FRQ scoring guides: sigmaprep.io/worksheets/ap-calculus-ab 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.
The calculator does not replace them. In the calculator sections, Desmos (built into Bluebook) can evaluate a derivative at a point or a definite integral, but you still write the setup, and more than half of the exam time allows no calculator at all.Free response grades the setup. A volume, an average value or a net change earns its first point for the integral you write, before any number.Theorems need their conditions. An IVT, EVT or MVT justification earns its point only when it checks continuity (and, for the MVT, differentiability).Questions about the AP Calculus AB formula sheet Is there a formula sheet on the AP Calculus AB exam? + 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.
Which AP Calculus AB formulas should I memorize first? + 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.
Do I need to know the Mean Value Theorem and the Intermediate Value Theorem word for word? + 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.
Does this sheet cover AP Calculus BC? + 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.
Is the formula sheet free to print? + 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.