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Calculus Cheat Sheet

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  • 1. Calculus Cheat Sheet Calculus Cheat Sheet Evaluation Techniques Limits Continuous Functions L’Hospital’s Rule Definitions f ( x) 0 f ( x) ± ¥ If f ( x ) is continuous at a then lim f ( x ) = f ( a ) Precise Definition : We say lim f ( x ) = L if Limit at Infinity : We say lim f ( x ) = L if we = or lim = x ®a If lim then, x® a x ®¥ x® a g ( x ) x® a g ( x ) ±¥ 0 can make f ( x ) as close to L as we want by for every e > 0 there is a d > 0 such that f ( x) f ¢ ( x) whenever 0 < x - a < d then f ( x ) - L < e . Continuous Functions and Composition a is a number, ¥ or -¥ = lim taking x large enough and positive. f ( x ) is continuous at b and lim g ( x ) = b then lim x ®a g ( x ) x ®a g ¢ ( x ) x® a ( ) There is a similar definition for lim f ( x ) = L “Working” Definition : We say lim f ( x ) = L lim f ( g ( x ) ) = f lim g ( x ) = f ( b ) Polynomials at Infinity x ®-¥ p ( x ) and q ( x ) are polynomials. To compute x® a x® a x® a if we can make f ( x ) as close to L as we want except we require x large and negative. Factor and Cancel p ( x) ( x - 2)( x + 6) x2 + 4 x - 12 lim factor largest power of x out of both by taking x sufficiently close to a (on either side Infinite Limit : We say lim f ( x ) = ¥ if we x®±¥ q ( x ) = lim lim of a) without letting x = a . x ( x - 2) x - 2x 2 x® a x ®2 x ®2 p ( x ) and q ( x ) and then compute limit. can make f ( x ) arbitrarily large (and positive) x+6 8 Right hand limit : lim+ f ( x ) = L . This has ) ( = lim = =4 x 2 3 - 42 by taking x sufficiently close to a (on either side 3 - 42 x® a x ®2 x 2 3x2 - 4 3 of a) without letting x = a . x = lim 2 5 = lim 5 x = - the same definition as the limit except it lim ( ) Rationalize Numerator/Denominator x®-¥ 5 x - 2 x -2 -2 2 x®- ¥ x x®-¥ requires x > a . 2 3- x 3- x 3+ x x x There is a similar definition for lim f ( x ) = -¥ = lim 2 lim 2 Piecewise Function x®9 x - 81 x®9 x - 81 3 + x ®a Left hand limit : lim- f ( x ) = L . This has the x ì x 2 + 5 if x < -2 except we make f ( x ) arbitrarily large and lim g ( x ) where g ( x ) = í 9- x -1 x® a = lim = lim ) ( ) î1 - 3 x if x ³ -2 ( x 2 - 81)(3 + x x®9 ( x + 9 ) 3 + x x®-2 same definition as the limit except it requires x ®9 negative. x<a. Compute two one sided limits, -1 lim g ( x ) = lim- x 2 + 5 = 9 Relationship between the limit and one-sided limits 1 = =- lim f ( x ) = L Þ lim+ f ( x ) = lim- f ( x ) = L lim+ f ( x ) = lim- f ( x ) = L Þ lim f ( x ) = L (18 )( 6 ) 108 x®-2- x®-2 lim g ( x ) = lim+ 1 - 3 x = 7 x® a x® a x® a x ®a x ®a x ®a lim+ f ( x ) ¹ lim f ( x ) Þ lim f ( x ) Does Not Exist x®-2+ Combine Rational Expressions x®-2 One sided limits are different so lim g ( x ) - x® a 1 æ x - ( x + h) ö x ®a x ®a 1æ 1 1ö - ÷ = lim ç ÷ x®-2 lim ç x ø h ®0 h ç x ( x + h ) ÷ h ®0 h x + h è è ø doesn’t exist. If the two one sided limits had Properties been equal then lim g ( x ) would have existed Assume lim f ( x ) and lim g ( x ) both exist and c is any number then, 1 æ -h ö -1 1 x®-2 x ®a x® a = lim ç ÷ = lim =- 2 h ®0 h ç x ( x + h ) ÷ é f ( x ) ù lim f ( x ) h®0 x ( x + h ) 1. lim écf ( x ) ù = c lim f ( x ) and had the same value. x è ø x ®a ë û provided lim g ( x ) ¹ 0 ú= x® a x ®a 4. lim ê x ®a g ( x ) û lim g ( x ) x® a ë x® a 2. lim é f ( x ) ± g ( x ) ù = lim f ( x ) ± lim g ( x ) Some Continuous Functions x ®a ë û x ®a n 5. lim é f ( x )ù = é lim f ( x )ù n x ®a x ®a ë û ë x ®a û Partial list of continuous functions and the values of x for which they are continuous. 7. cos ( x ) and sin ( x ) for all x. 1. Polynomials for all x. 6. lim é n f ( x ) ù = n lim f ( x ) 3. lim é f ( x ) g ( x ) ù = lim f ( x ) lim g ( x ) x ®a ë û x ®a x ®a ë û 2. Rational function, except for x’s that give 8. tan ( x ) and sec ( x ) provided x ®a x ®a division by zero. 3p p p 3p Basic Limit Evaluations at ± ¥ x ¹ L, - ,L 3. n x (n odd) for all x. ,- , , Note : sgn ( a ) = 1 if a > 0 and sgn ( a ) = - 1 if a < 0 . 2 222 4. n x (n even) for all x ³ 0 . 9. cot ( x ) and csc ( x ) provided 5. e x for all x. 1. lim e x = ¥ & lim e x = 0 5. n even : lim x n = ¥ x ¹ L , -2p , -p , 0, p , 2p ,L 6. ln x for x > 0 . x®- ¥ x ®± ¥ x®¥ lim ln ( x ) = ¥ lim- ln ( x ) = - ¥ 6. n odd : lim x n = ¥ & lim x n = -¥ 2. & x ®¥ x ®¥ x ®- ¥ x ®0 Intermediate Value Theorem 7. n even : lim a x n + L + b x + c = sgn ( a ) ¥ b Suppose that f ( x ) is continuous on [a, b] and let M be any number between f ( a ) and f ( b ) . 3. If r > 0 then lim =0 x ®± ¥ x ®¥ x r 8. n odd : lim a xn + L + b x + c = sgn ( a ) ¥ Then there exists a number c such that a < c < b and f ( c ) = M . 4. If r > 0 and x r is real for negative x x ®¥ 9. n odd : lim a xn + L + c x + d = - sgn ( a ) ¥ b then lim r = 0 x ®-¥ x ®-¥ x © 2005 Paul Dawkins © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
  • 2. Calculus Cheat Sheet Calculus Cheat Sheet Chain Rule Variants Derivatives The chain rule applied to some specific functions. Definition and Notation ( ) f ( x + h) - f ( x) ( ) d d n-1 é f ( x )ù = n é f ( x ) ù f ¢ ( x ) cos é f ( x ) ù = - f ¢ ( x ) sin é f ( x ) ù n If y = f ( x ) then the derivative is defined to be f ¢ ( x ) = lim 1. 5. dx ë û ë û ë û ë û . dx h ®0 h ( ) ( ) d f ( x) d = f ¢( x)e tan é f ( x )ù = f ¢ ( x ) sec é f ( x ) ù f ( x) 2 2. e 6. ë û ë û If y = f ( x ) then all of the following are If y = f ( x ) all of the following are equivalent dx dx f ¢( x) ( sec [ f ( x)]) = f ¢( x ) sec [ f ( x)] tan [ f ( x )] ( ) d d ln é f ( x ) ù = notations for derivative evaluated at x = a . equivalent notations for the derivative. 7. 3. ë û f ( x) dx dx = ( f ( x) ) = Df ( x ) df dy d f ¢ ( x ) = y¢ = df dy f ¢ ( a ) = y ¢ x= a = = Df ( a ) = f ¢( x ) = ( ) d tan -1 é f ( x ) ù = ( ) d dx dx dx sin é f ( x ) ù = f ¢ ( x ) cos é f ( x ) ù dx x =a dx x =a 8. ë û 4. ë û ë û 1 + ë f ( x )û 2 é ù dx dx Interpretation of the Derivative If y = f ( x ) then, 2. f ¢ ( a ) is the instantaneous rate of Higher Order Derivatives The nth Derivative is denoted as The Second Derivative is denoted as 1. m = f ¢ ( a ) is the slope of the tangent change of f ( x ) at x = a . 2 dn f df f ¢¢ ( x ) = f ( 2) ( x ) = 2 and is defined as f ( n) ( x ) = n and is defined as line to y = f ( x ) at x = a and the 3. If f ( x ) is the position of an object at dx dx time x then f ¢ ( a ) is the velocity of equation of the tangent line at x = a is ( ) ¢ , i.e. the derivative of the ¢ f ¢¢ ( x ) = ( f ¢ ( x ) ) f ( x ) = f ( ) ( x ) , i.e. the derivative of ( n) n-1 given by y = f ( a ) + f ¢ ( a )( x - a ) . the object at x = a . first derivative, f ¢ ( x ) . () the (n-1)st derivative, f ( n -1) x . Basic Properties and Formulas If f ( x ) and g ( x ) are differentiable functions (the derivative exists), c and n are any real numbers, Implicit Differentiation Find y¢ if e2 x -9 y + x3 y 2 = sin ( y ) + 11x . Remember y = y ( x ) here, so products/quotients of x and y ( c f )¢ = c f ¢ ( x ) d (c) = 0 1. 5. will use the product/quotient rule and derivatives of y will use the chain rule. The “trick” is to dx ± g )¢ = f ¢ ( x ) ± g ¢ ( x ) (f differentiate as normal and every time you differentiate a y you tack on a y¢ (from the chain rule). ( x ) = n x n-1 – Power Rule dn 2. 6. After differentiating solve for y¢ . dx ( f g )¢ = f ¢ g + f g ¢ – Product Rule 3. ( ) f ( g ( x) ) = f ¢ ( g ( x) ) g¢ ( x) d e2 x-9 y ( 2 - 9 y¢ ) + 3x 2 y 2 + 2 x 3 y y¢ = cos ( y ) y¢ + 11 7. æ f ö¢ f ¢ g - f g ¢ dx 11 - 2e 2 x -9 y - 3 x 2 y 2 4. ç ÷ = 2e 2 x -9 y - 9 y¢e 2x - 9 y + 3x 2 y 2 + 2 x 3 y y¢ = cos ( y ) y¢ + 11 – Quotient Rule y¢ = Þ This is the Chain Rule g2 2 x3 y - 9e2 x -9 y - cos ( y ) ègø ( 2 x y - 9e x - cos ( y ) ) y ¢ = 11 - 2e2 x - 9 y - 3 x 2 y 2 2 -9 y 3 Common Derivatives ( a ) = a x ln ( a ) d d dx ( x) = 1 ( csc x ) = - csc x cot x Increasing/Decreasing – Concave Up/Concave Down dx dx dx Critical Points ( e ) = ex x = c is a critical point of f ( x ) provided either d d dx ( sin x ) = cos x ( cot x ) = - csc2 x Concave Up/Concave Down 1. If f ¢¢ ( x ) > 0 for all x in an interval I then dx dx dx 1. f ¢ ( c ) = 0 or 2. f ¢ ( c ) doesn’t exist. ( sin -1 x ) = 1 2 ( ln ( x ) ) = 1 , x > 0 d d d ( cos x ) = - sin x f ( x ) is concave up on the interval I. 1- x dx dx dx x 2. If f ¢¢ ( x ) < 0 for all x in an interval I then Increasing/Decreasing ( ln x ) = 1 , x ¹ 0 d d ( tan x ) = sec 2 x ( cos x ) = - 1 2 d 1. If f ¢ ( x ) > 0 for all x in an interval I then -1 f ( x ) is concave down on the interval I. dx dx x 1- x dx f ( x ) is increasing on the interval I. ( log a ( x )) = x ln a , x > 0 d d 1 ( sec x ) = sec x tan x ( tan -1 x ) = 1 +1x2 d 2. If f ¢ ( x ) < 0 for all x in an interval I then dx dx Inflection Points dx x = c is a inflection point of f ( x ) if the f ( x ) is decreasing on the interval I. concavity changes at x = c . 3. If f ¢ ( x ) = 0 for all x in an interval I then f ( x ) is constant on the interval I. © 2005 Paul Dawkins © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
  • 3. Calculus Cheat Sheet Calculus Cheat Sheet Extrema Related Rates Absolute Extrema Relative (local) Extrema Sketch picture and identify known/unknown quantities. Write down equation relating quantities 1. x = c is an absolute maximum of f ( x ) 1. x = c is a relative (or local) maximum of and differentiate with respect to t using implicit differentiation (i.e. add on a derivative every time f ( x ) if f ( c ) ³ f ( x ) for all x near c. you differentiate a function of t). Plug in known quantities and solve for the unknown quantity. if f ( c ) ³ f ( x ) for all x in the domain. Ex. A 15 foot ladder is resting against a wall. Ex. Two people are 50 ft apart when one 2. x = c is a relative (or local) minimum of 2. x = c is an absolute minimum of f ( x ) The bottom is initially 10 ft away and is being starts walking north. The angleq changes at f ( x ) if f ( c ) £ f ( x ) for all x near c. if f ( c ) £ f ( x ) for all x in the domain. pushed towards the wall at 1 ft/sec. How fast 0.01 rad/min. At what rate is the distance 4 between them changing when q = 0.5 rad? is the top moving after 12 sec? 1st Derivative Test If x = c is a critical point of f ( x ) then x = c is Fermat’s Theorem If f ( x ) has a relative (or local) extrema at 1. a rel. max. of f ( x ) if f ¢ ( x ) > 0 to the left x = c , then x = c is a critical point of f ( x ) . of x = c and f ¢ ( x ) < 0 to the right of x = c . We have q ¢ = 0.01 rad/min. and want to find 2. a rel. min. of f ( x ) if f ¢ ( x ) < 0 to the left x¢ is negative because x is decreasing. Using Extreme Value Theorem x¢ . We can use various trig fcns but easiest is, If f ( x ) is continuous on the closed interval of x = c and f ¢ ( x ) > 0 to the right of x = c . Pythagorean Theorem and differentiating, x¢ x Þ sec q tan q q ¢ = sec q = x 2 + y 2 = 15 2 Þ 2 x x¢ + 2 y y¢ = 0 [a, b] then there exist numbers c and d so that, 3. not a relative extrema of f ( x ) if f ¢ ( x ) is 50 50 After 12 sec we have x = 10 - 12 ( 1 ) = 7 and We know q = 0.05 so plug in q ¢ and solve. 1. a £ c, d £ b , 2. f ( c ) is the abs. max. in the same sign on both sides of x = c . 4 x¢ sec ( 0.5 ) tan ( 0.5 )( 0.01) = so y = 152 - 7 2 = 176 . Plug in and solve [a, b] , 3. f ( d ) is the abs. min. in [a, b] . 50 2nd Derivative Test for y¢ . x¢ = 0.3112 ft/sec If x = c is a critical point of f ( x ) such that 7 7 ( - 1 ) + 176 y¢ = 0 Þ y¢ = ft/sec Finding Absolute Extrema Remember to have calculator in radians! f ¢ ( c ) = 0 then x = c 4 4 176 To find the absolute extrema of the continuous function f ( x ) on the interval [ a , b ] use the 1. is a relative maximum of f ( x ) if f ¢¢ ( c ) < 0 . Optimization 2. is a relative minimum of f ( x ) if f ¢¢ ( c ) > 0 . following process. Sketch picture if needed, write down equation to be optimized and constraint. Solve constraint for 1. Find all critical points of f ( x ) in [ a, b ] . one of the two variables and plug into first equation. Find critical points of equation in range of 3. may be a relative maximum, relative 2. Evaluate f ( x ) at all points found in Step 1. minimum, or neither if f ¢¢ ( c ) = 0 . variables and verify that they are min/max as needed. Ex. Determine point(s) on y = x 2 + 1 that are 3. Evaluate f ( a ) and f ( b ) . Ex. We’re enclosing a rectangular field with 500 ft of fence material and one side of the closest to (0,2). Finding Relative Extrema and/or 4. Identify the abs. max. (largest function field is a building. Determine dimensions that Classify Critical Points value) and the abs. min.(smallest function will maximize the enclosed area. 1. Find all critical points of f ( x ) . value) from the evaluations in Steps 2 & 3. 2. Use the 1st derivative test or the 2nd Minimize f = d 2 = ( x - 0 ) + ( y - 2 ) and the derivative test on each critical point. 2 2 Maximize A = xy subject to constraint of constraint is y = x 2 + 1 . Solve constraint for Mean Value Theorem x + 2 y = 500 . Solve constraint for x and plug If f ( x ) is continuous on the closed interval [ a , b ] and differentiable on the open interval ( a , b ) x 2 and plug into the function. into area. x2 = y -1 Þ f = x2 + ( y - 2 ) f (b) - f ( a) 2 A = y ( 500 - 2 y ) then there is a number a < c < b such that f ¢ ( c ) = . x = 500 - 2 y Þ b-a = y - 1+ ( y - 2) = y2 - 3 y + 3 2 = 500 y - 2 y 2 Differentiate and find critical point(s). Differentiate and find critical point(s). Newton’s Method f ¢ = 2y - 3 Þ y= 3 A¢ = 500 - 4 y Þ y = 125 f ( xn ) 2 If xn is the n guess for the root/solution of f ( x ) = 0 then (n+1)st guess is xn+1 = xn - th nd By 2nd deriv. test this is a rel. max. and so is f ¢ ( xn ) By the 2 derivative test this is a rel. min. and so all we need to do is find x value(s). the answer where after. Finally, find x. provided f ¢ ( xn ) exists. x = 500 - 2 (125 ) = 250 x 2 = 3 - 1 = 1 Þ x = ± 12 2 2 ( ) ( ) The dimensions are then 250 x 125. , 3 and - 1 1 ,3 . The 2 points are then 2 2 2 2 © 2005 Paul Dawkins © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
  • 4. Calculus Cheat Sheet Calculus Cheat Sheet Standard Integration Techniques Integrals Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class. Definitions Definite Integral: Suppose f ( x ) is continuous Anti-Derivative : An anti-derivative of f ( x ) () ò a f ( g ( x )) g ¢ ( x ) dx = ò g (a ) f ( u ) du u Substitution : The substitution u = g ( x ) will convert b gb on [ a , b ] . Divide [ a , b ] into n subintervals of is a function, F ( x ) , such that F ¢ ( x ) = f ( x ) . using Indefinite Integral : ò f ( x ) dx = F ( x ) + c du = g ¢ ( x ) dx . For indefinite integrals drop the limits of integration. width D x and choose x from each interval. * i ¥ where F ( x ) is an anti-derivative of f ( x ) . ò a f ( x ) dx = n å f ( x ) D x . cos ( x3 ) dx cos ( x3 ) dx = ò cos ( u ) dx 2 8 b 2 ò 1 5x ò 1 5x 5 2 * 2 Then lim Ex. 13 i ®¥ =1 i = 5 sin ( u ) 1 = 5 ( sin (8 ) - sin (1) ) u= x Þ du = 3 x dx Þ x dx = du 3 2 2 8 1 3 3 3 Fundamental Theorem of Calculus x = 1 Þ u = 13 = 1 :: x = 2 Þ u = 2 3 = 8 Part I : If f ( x ) is continuous on [ a, b ] then Variants of Part I : d u(x) f ( t ) dt = u ¢ ( x ) f éu ( x ) ù dx ò a g ( x ) = ò f ( t ) dt is also continuous on [ a , b ] x b b Integration by Parts : ò u dv = uv - ò v du and ò a u dv = uv - ò v du . Choose u and dv from ë û b a a a db integral and compute du by differentiating u and compute v using v = ò dv . f ( t ) dt = - v¢ ( x ) f é v ( x )ù dx ò v( x) d and g ¢ ( x ) = f ( t ) dt = f ( x ) . x dx ò a ë û ò xe -x 5 ò3 ln x dx Part II : f ( x ) is continuous on [ a , b ] , F ( x ) is d u( x) dx Ex. f ( t ) dt = u ¢ ( x ) f [ u ( x) ] - v¢ ( x ) f [ v( x ) ] dx ò v( x) Ex. -x -x u=x dv = e Þ du = dx v = -e an anti-derivative of f ( x ) (i.e. F ( x ) = ò f ( x ) dx ) u = ln x dv = dx Þ du = 1 dx v = x x ò xe dx = - xe + ò e dx = - xe - e + c -x -x -x -x -x ò3 ln x dx = x ln x 3 - ò3 dx = ( x ln ( x ) - x ) 3 5 5 5 5 then ò f ( x ) dx = F ( b ) - F ( a ) . b a = 5ln ( 5 ) - 3ln ( 3) - 2 Properties ò f ( x) ± g ( x) dx = ò f ( x) dx ± ò g ( x) dx ò cf ( x) dx = c ò f ( x) dx , c is a constant Products and (some) Quotients of Trig Functions For ò sin n x cos m x dx we have the following : For ò tan n x sec m x dx we have the following : ò a f ( x ) ± g ( x ) dx = ò a f ( x ) dx ± ò a g ( x ) dx ò a cf ( x ) dx = c ò a f ( x ) dx , c is a constant b b b b b 1. n odd. Strip 1 sine out and convert rest to 1. n odd. Strip 1 tangent and 1 secant out and ò a f ( x ) dx = 0 ò a f ( x ) dx = ò a f ( t ) dt a b b cosines using sin 2 x = 1 - cos 2 x , then use convert the rest to secants using tan 2 x = sec 2 x - 1 , then use the substitution the substitution u = cos x . ò a f ( x) dx = - òb f ( x) dx b a ò f ( x ) dx £ ò f ( x ) dx b b u = sec x . 2. m odd. Strip 1 cosine out and convert rest a a to sines using cos2 x = 1 - sin 2 x , then use 2. m even. Strip 2 secants out and convert rest ò f ( x) dx ³ ò g ( x ) dx If f ( x ) ³ g ( x ) on a £ x £ b then b a to tangents using sec 2 x = 1 + tan 2 x , then the substitution u = sin x . a b use the substitution u = tan x . 3. n and m both odd. Use either 1. or 2. If f ( x ) ³ 0 on a £ x £ b then f ( x ) dx ³ 0 ò b 4. n and m both even. Use double angle 3. n odd and m even. Use either 1. or 2. a and/or half angle formulas to reduce the 4. n even and m odd. Each integral will be If m £ f ( x ) £ M on a £ x £ b then m ( b - a ) £ ò f ( x ) dx £ M ( b - a ) b integral into a form that can be integrated. dealt with differently. 2 (1 + cos ( 2 x ) ) , sin ( x ) = 2 (1 - cos ( 2 x ) ) Trig Formulas : sin ( 2 x ) = 2sin ( x ) cos ( x ) , cos 2 ( x ) = a 2 1 1 Common Integrals ò tan ò k dx = k x + c ò cos u du = sin u + c ò tan u du = ln sec u + c ò cos x dx sin 5 x 3 x sec5 x dx Ex. Ex. 3 ò x dx = n+1 x + c, n ¹ -1 ò sin u du = - cos u + c ò sec u du = ln sec u + tan u + c ò tan x sec xdx = ò tan x sec x tan x sec xdx +1 n n 1 3 5 2 4 ò cos x dx = ò cos x dx = ò cos x dx 2 2 5 4 (sin x) sin x sin x sin x sin x 3 3 3 ò a + u du = a tan ( a ) + c = ò ( sec2 x - 1) sec 4 x tan x sec xdx ò x dx = ò x dx = ln x + c ò sec u du = tan u + c -1 -1 u 2 1 1 1 (1- cos x ) sin x ( u = cos x ) =ò 2 2 2 2 dx ò a x + b dx = a ln ax + b + c ò sec u tan u du = secu + c 3 ò a - u du = sin ( a ) + c = ò ( u 2 - 1) u 4 du cos x -1 ( u = sec x ) 1 u 1 1 (1-u ) = -ò du = - ò 22 1- 2u + u du 2 4 2 2 ò csc u cot udu = - csc u + c ò ln u du = u ln ( u ) - u + c 3 3 u u = 1 sec 7 x - 1 sec5 x + c = 1 sec 2 x + 2 ln cos x - 1 cos2 x + c 7 5 ò csc u du = - cot u + c ò e du = e + c 2 2 2 u u © 2005 Paul Dawkins © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
  • 5. Calculus Cheat Sheet Calculus Cheat Sheet Trig Substitutions : If the integral contains the following root use the given substitution and Applications of Integrals ò a f ( x ) dx represents the net area between f ( x ) and the b formula to convert into an integral involving trig functions. Net Area : a 2 - b 2 x 2 Þ x = a sin q b 2 x2 - a 2 Þ x = a sec q a 2 + b 2 x 2 Þ x = a tan q b b b x-axis with area above x-axis positive and area below x-axis negative. cos 2 q = 1 - sin 2 q tan 2 q = sec 2 q - 1 sec 2 q = 1 + tan 2 q òx ( 2 cos q ) dq = ò sin2 q dq ó 16 12 16 Ex. dx õ Area Between Curves : The general formulas for the two main cases for each are, 2 4- 9 x2 3 4 sin 2 q ( 2cosq ) 9 x = sin q Þ dx = cos q d q y = f ( x) Þ A = ò - élower function ù dx & x = f ( y ) Þ A = ò b d 2 2 - éleft function ù dy = ò 12 csc dq = -12 cot q + c é ù é right function ù 2 ë û ë û ë upper function û ë û 3 3 a c 4 - 9x = 4 - 4 sin q = 4 cos q = 2 cosq 2 2 2 If the curves intersect then the area of each portion must be found individually. Here are some Use Right Triangle Trig to go back to x’s. From x 2 = x . Because we have an indefinite substitution we have sin q = 2 so, sketches of a couple possible situations and formulas for a couple of possible cases. 3x Recall integral we’ll assume positive and drop absolute value bars. If we had a definite integral we’d need to compute q ’s and remove absolute value bars based on that and, ì x if x ³ 0 4 -9 x2 From this we see that cot q = x =í . So, î- x if x < 0 3x òx 4 4- 9 x 2 dx = - +c 16 4 - 9x 2 = 2 cos q . A = ò f ( y ) - g ( y ) dy d A = ò f ( x ) - g ( x ) dx b A = ò f ( x ) - g ( x ) dx + ò g ( x ) - f ( x ) dx c b In this case we have x 2 4 -9 x2 c a a c P( x ) ò Q( x) dx where the degree of P ( x ) is smaller than the degree of Partial Fractions : If integrating Volumes of Revolution : The two main formulas are V = ò A ( x ) dx and V = ò A ( y ) dy . Here is Q ( x ) . Factor denominator as completely as possible and find the partial fraction decomposition of some general information about each method of computing and some examples. the rational expression. Integrate the partial fraction decomposition (P.F.D.). For each factor in the Rings Cylinders ( ) denominator we get term(s) in the decomposition according to the following table. A = 2p ( radius ) ( width / height ) A = p ( outer radius ) 2 - ( inner radius ) 2 Factor in Q ( x ) Term in P.F.D Factor in Q ( x ) Limits: x/y of right/bot ring to x/y of left/top ring Limits : x/y of inner cyl. to x/y of outer cyl. Term in P.F.D Horz. Axis use f ( x ) , Vert. Axis use f ( y ) , Horz. Axis use f ( y ) , Vert. Axis use f ( x ) , A1 A2 Ak +L+ A + ( ax + b ) g ( x ) , A ( x ) and dx. g ( y ) , A ( y ) and dy. g ( y ) , A ( y ) and dy. g ( x ) , A ( x ) and dx. ax + b k ax + b ( ax + b )2 ( ax + b ) k ax + b Ak x + Bk A1 x + B1 Ax + B +L + ( ax + bx + c ) Ex. Axis : y = a > 0 Ex. Axis : y = a £ 0 Ex. Axis : y = a > 0 Ex. Axis : y = a £ 0 k ax 2 + bx + c 2 ( ax 2 + bx + c ) ax 2 + bx + c k ax + bx + c 2 ò 7 x2 +13 x + Bx+C = A( x 2 +4)+ ( Bx +C ) ( x -1) 7 x 2 +13 x = A Ex. dx x -1 ( x -1)( x 2 + 4 ) ( x -1)( x 2 + 4 ) x2 + 4 ( x -1)( x 2 + 4 ) ò dx = ò 7 x2 +13 x 3 x +16 Set numerators equal and collect like terms. + 4 dx x -1 7 x 2 + 13 x = ( A + B ) x 2 + ( C - B ) x + 4 A - C ( x -1)( x 2 + 4 ) x2 + 4 =ò + + 4 3x 16 dx x -1 Set coefficients equal to get a system and solve x2 +4 x2 +4 = 4 ln x - 1 + 3 ln ( x 2 + 4 ) + 8 tan -1 ( x ) to get constants. outer radius : a - f ( x ) outer radius: a + g ( x ) radius : a - y radius : a + y A+ B = 7 C - B = 13 4A- C = 0 2 2 width : f ( y ) - g ( y ) Here is partial fraction form and recombined. inner radius : a - g ( x ) inner radius: a + f ( x ) width : f ( y ) - g ( y ) A=4 B=3 C = 16 An alternate method that sometimes works to find constants. Start with setting numerators equal in These are only a few cases for horizontal axis of rotation. If axis of rotation is the x-axis use the previous example : 7 x 2 + 13 x = A ( x 2 + 4 ) + ( Bx + C ) ( x - 1) . Chose nice values of x and plug in. y = a £ 0 case with a = 0 . For vertical axis of rotation ( x = a > 0 and x = a £ 0 ) interchange x and For example if x = 1 we get 20 = 5A which gives A = 4 . This won’t always work easily. y to get appropriate formulas. © 2005 Paul Dawkins © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
  • 6. Calculus Cheat Sheet Work : If a force of F ( x ) moves an object Average Function Value : The average value of f ( x ) on a £ x £ b is f avg = ò f ( x ) dx b 1 in a £ x £ b , the work done is W = ò F ( x ) dx b b-a a a Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are, b b b L = ò ds SA = ò 2p y ds (rotate about x-axis) SA = ò 2p x ds (rotate about y-axis) a a a where ds is dependent upon the form of the function being worked with as follows. () () ( dx ) 2 2 dx if y = f ( x ) , a £ x £ b dt if x = f ( t ) , y = g ( t ) , a £ t £ b 2 dy dy ds = 1 + ds = + dt dx dt 1+ ( ) ds = r 2 + ( dq ) d q if r = f (q ) , a £ q £ b 2 2 dy if x = f ( y ) , a £ y £ b dr ds = dx dy With surface area you may have to substitute in for the x or y depending on your choice of ds to match the differential in the ds. With parametric and polar you will always need to substitute. Improper Integral An improper integral is an integral with one or more infinite limits and/or discontinuous integrands. Integral is called convergent if the limit exists and has a finite value and divergent if the limit doesn’t exist or has infinite value. This is typically a Calc II topic. Infinite Limit ¥ f ( x ) dx = lim ò f ( x ) dx ò ¥ f ( x ) dx = lim ò f ( x ) dx t b b ò 1. 2. - t ®¥ t ®-¥ a a t ¥ ¥ ò ¥ f ( x ) dx = ò ¥ f ( x ) dx + ò f ( x ) dx provided BOTH integrals are convergent. c 3. - - c Discontinuous Integrand 1. Discont. at a: ò f ( x ) dx = lim ò f ( x ) dx 2. Discont. at b : ò f ( x ) dx = lim ò f ( x ) dx b b b t + - t ®a t ®b a t a a ò f ( x ) dx = ò f ( x ) dx + ò f ( x ) dx provided both are convergent. b c b 3. Discontinuity at a < c < b : a a c Comparison Test for Improper Integrals : If f ( x ) ³ g ( x ) ³ 0 on [ a, ¥ ) then, ¥ ¥ ¥ ¥ f ( x ) dx conv. then ò g ( x ) dx conv. 2. If ò g ( x ) dx divg. then ò f ( x ) dx divg. 1. If ò a a a a ¥ òa Useful fact : If a > 0 then dx converges if p > 1 and diverges for p £ 1 . 1 p x Approximating Definite Integrals f ( x ) dx and a n (must be even for Simpson’s Rule) define Dx = b -a and b òa For given integral n divide [ a , b ] into n subintervals [ x0 , x1 ] , [ x1 , x2 ] , … , [ xn -1 , xn ] with x0 = a and xn = b then, ò f ( x ) dx » Dx é f ( x ) + f ( x ) + L + f ( x ) ù , xi is midpoint [ xi -1 , xi ] b * * * * Midpoint Rule : ë û 1 2 n a Dx ò f ( x ) dx » 2 é f ( x ) + 2 f ( x ) + +2 f ( x ) + L + 2 f ( x ) + f ( x ) ù b Trapezoid Rule : ë û n -1 0 1 2 n a Dx ò f ( x ) dx » 3 é f ( x ) + 4 f ( x ) + 2 f ( x ) +L + 2 f ( x ) + 4 f ( x ) + f ( x )ù b Simpson’s Rule : ë û n -2 n -1 0 1 2 n a © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.