Calculus_Cheat_Sheet_All

Calculus Cheat Sheet Limits DefinitionsPrecise Definition : We say lim f ( x) = L if Limit at Infinity : We say lim f ( x) = L if we x®a x®¥for every e > 0 there is a d > 0 such that can make f ( x) as close to L as we want bywhenever 0 < x - a < d then f ( x) - L < e . taking x large enough and positive.“Working” Definition : We say lim f ( x) = L There is a similar definition for lim f ( x) = L x®a x®-¥if we can make f ( x) as close to L as we want except we require x large and negative.by taking x sufficiently close to a (on either side Infinite Limit : We say lim f ( x) = ¥ if weof a) without letting x = a . x®aRight hand limit : lim f ( x) = L . This has can make f ( x) arbitrarily large (and positive) x®a+ by taking x sufficiently close to a (on either sidethe same definition as the limit except it of a) without letting x = a .requires x > a .Left hand limit : lim f ( x) = L . This has the There is a similar definition for lim f ( x) = -¥ x®a- x®asame definition as the limit except it requires except we make f ( x) arbitrarily large andx<a. negative. Relationship between the limit and one-sided limitslim f ( x) = L Þ lim f ( x) = lim f ( x) = L lim f ( x) = lim f ( x) = L Þ lim f ( x) = Lx®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®a+ x®a- x®a PropertiesAssume lim f ( x) and lim g ( x) both exist and c is any number then, x®a x®a1. lim éëcf ( x)ùû = c lim f (x) 4. lim é f (x) ù = lim f (x) provided lim g ( x) ¹ 0 x®a ê g (x) ú g (x) x®a x®a ë û x®a x®a2. lim f (x)± g ( x)ùû lim f ( x) ± lim g (x) lim x®a x®a x®a x®a ëé = 5. lim f ( x)ûùn ëélxi®ma f ( x ) ù n û x®a ëé =3. lim ëé f ( x ) g ( x )ùû = lim f (x) lim g ( x) 6. lim é n f ( x) ù = n lim f (x) ë û x®a x®a x®a x®a x®a Basic Limit Evaluations at ± ¥Note : sgn (a) = 1 if a > 0 and sgn (a) = -1 if a < 0 .1. lim ex = ¥ & lim ex = 0 5. n even : lim xn = ¥ x®¥ x®- ¥ x®± ¥2. lim ln ( x) = ¥ & lim ln ( x) = - ¥ 6. n odd : lim xn = ¥ & lim xn = -¥ x®¥ x®0 + x®¥ x®- ¥3. If r > 0 then lim b =0 7. n even : lim a xn +L + b x + c = sgn (a) ¥ x®¥ xr x®± ¥4. If r > 0 and xr is real for negative x 8. n odd : lim a xn +L + b x + c = sgn (a) ¥ x®¥ then lim b =0 9. n odd : lim a xn +L + c x + d = - sgn (a) ¥ x®-¥ xr x®-¥Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins

Calculus Cheat Sheet Evaluation TechniquesContinuous Functions L’Hospital’s RuleIf f ( x) is continuous at a then lim f ( x) = f (a) If lim f (x) = 0 or lim f (x) = ±¥ then, x®a g ( x) 0 g ( x) ±¥ x®a x®aContinuous Functions and Composition lim f (x) = lim f ¢(x) a is a number, ¥ or -¥ g ( x) x®a g¢(x)f ( x) is continuous at b and lim g ( x) = b then x®a x®a( )lim f ( g ( x)) = f lim g ( x) = f (b) Polynomials at Infinityx®a x®a p ( x) and q ( x) are polynomials. To computeFactor and Cancel p(x)lim x2 + 4x -12 = lim ( x - 2)(x + 6) lim q(x) factor largest power of x in q ( x) out x2 - 2x x(x - 2)x®2 x®2 x®±¥ x + 6 8 of both p ( x) and q ( x) then compute limit. x 2 = lim = = 4 ( )lim 3x2 x2 3 - 4 3- 4 5x - x2 x2 x2 x®2 ( )x®-¥ - 4 3 2 x2 2Rationalize Numerator/Denominator = lim 5 - 2 = lim 5 - 2 = - x x 3- x 3- x 3+ x x®-¥ x®-¥ x2 - 81 x2 - 81 3+ xlim = lim Piecewise Functionx®9 x®9= lim 9-x = lim -1 lim g ( x) where g ( x ) = ìx2 +5 if x < -2 íî1 - 3x if x ³ -2 ( )( ) (x®9 x2 - 81 3 + x x®9 ( x + 9) 3 + x) x®-2 -1 1 Compute two one sided limits, 108= (18) ( = - lim g ( x) = lim x2 + 5 = 9 6) x®-2- x®-2-Combine Rational Expressions lim g ( x) = lim 1- 3x = 7 x®-2+ x®-2+lim 1 æ x 1 h - 1 ö = lim 1 æ x -(x + h) ö One sided limits are different so lim g ( x) h èç + x ø÷ h çèç x(x+ h) ÷ø÷ x®-2h®0 h®0 doesn’t exist. If the two one sided limits had = lim 1 æ x( -h h) ö = lim x( -1 h) = - 1 been equal then lim g ( x) would have existed h èçç x+ ÷÷ø x+ x2 x®-2 h®0 h®0 and had the same value. Some Continuous FunctionsPartial list of continuous functions and the values of x for which they are continuous.1. Polynomials for all x. 7. cos ( x) and sin ( x) for all x.2. Rational function, except for x’s that give 8. tan ( x) and sec ( x) provided division by zero. 3p 3p 2 23. n x (n odd) for all x. x ¹ L, - ,- p , p , ,L 2 24. n x (n even) for all x ³ 0 .5. ex for all x. 9. cot ( x) and csc ( x) provided6. ln x for x > 0 . x ¹ L, -2p , -p , 0,p , 2p ,L Intermediate Value TheoremSuppose that f ( x) is continuous on [a, b] and let M be any number between f (a) and f (b) .Then there exists a number c such that a < c < b and f (c) = M .Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins

Calculus Cheat Sheet Derivatives Definition and NotationIf y= f (x) then the derivative is defined to be f ¢( x) = lim f (x + h) - f (x) . h®0 hIf y = f ( x) then all of the following are If y = f ( x) all of the following are equivalentequivalent notations for the derivative. notations for derivative evaluated at x = a .f ¢( x) = y¢ = df = dy = d ( f ( x)) = Df ( x) f ¢(a) = y¢ x=a = df = dy = Df (a) dx dx dx dx dx x=a x=aIf y = f ( x) then, Interpretation of the Derivative 2. f ¢(a) is the instantaneous rate of 1. m = f ¢(a) is the slope of the tangent change of f ( x) at x = a . line to y = f ( x) at x = a and the 3. If f ( x) is the position of an object at equation of the tangent line at x = a is time x then f ¢(a) is the velocity of given by y = f (a) + f ¢(a)( x - a) . the object at x = a . Basic Properties and FormulasIf f ( x) and g ( x) are differentiable functions (the derivative exists), c and n are any real numbers,1. (c f )¢ = c f ¢( x) 5. d ( c) = 0 dx2. ( f ± g )¢ = f ¢( x) ± g¢( x) ( )6.d xn = n xn-1 – Power Rule dx3. ( f g )¢ = f ¢ g + f g¢ – Product Rule 7. d ( f (g (x))) = f ¢( g ( x)) g¢( x) ö¢ dx4. æf ÷ = f¢g- f g¢ – Quotient Rule ç g2 This is the Chain Rule è g ø Common Derivativesd ( x) = 1 d ( csc x) = - csc x cot x d ( a x ) = a x ln ( a )dx dx dxd (sin x ) = cos x d (cot x) = - csc2 x ( )d ex = exdx dx dxd ( )d 1 d ( x)) 1dx 1- x2 dx x dx ( cos x) = - sin x sin-1 x = ln ( = , x>0d ( tan x) = sec2 x ( )d cos-1 x =- 1 d ( ln x ) = 1 , x¹0dx 1- x2 dx x dxd d 1dx (sec x) = sec x tan x ( )d 1 dx ( log ( x )) = ln , x>0 = 1+ x2 dx tan-1 x a x aVisit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins

Calculus Cheat Sheet Chain Rule VariantsThe chain rule applied to some specific functions.( )1.d d dx ëé f ( x)ûùn = n éë f ( x)ûùn-1 f ¢( x) ( )5. dx = - f ¢( x)sin éë f ( x)ùû cos ëé f ( x)ùû( )2.d d dx e f (x) = f ¢( x)e f (x) ( )6. dx = f ¢( x)sec2 éë f ( x)ùû tan ëé f ( x)ùû3. (d ln ëé f ( x)ûù) = f ¢(x) 7. d (sec [ f ( x)]) = f ¢(x) sec[ f (x)] tan [ f (x)] f (x) dx dx f (x) d ( )8. d ëé ¢ ( x)ùû( )4.dx dx tan-1 éë f ( x)ùû = f sin ëé f ( x)ûù = f ¢( x) cos éë f ( x)ûù 1 2 + Higher Order Derivatives The nth Derivative is denoted asThe Second Derivative is denoted asf ¢¢( x) = f (2) ( x) = d2 f and is defined as f (n) (x) = dn f and is defined as dx2 dxnf ¢¢( x) = ( f ¢( x))¢ , i.e. the derivative of the ( )f (n) ( x) = f (n-1) ( x) ¢ , i.e. the derivative offirst derivative, f ¢( x) . the (n-1)st derivative, f (n-1) ( x) . Implicit DifferentiationFind y¢ if e2x-9 y + x3 y2 = sin ( y) +11x . Remember y = y ( x) here, so products/quotients of x and ywill use the product/quotient rule and derivatives of y will use the chain rule. The “trick” is todifferentiate as normal and every time you differentiate a y you tack on a y¢ (from the chain rule).After differentiating solve for y¢ . e2x-9 y (2 - 9 y¢) + 3x2 y2 + 2x3 y y¢ = cos ( y ) y¢ +11 Þ y¢ = 11- 2e2x-9 y - 3x2 y2 2e2x-9 y - 9 y¢e2x-9 y + 3x2 y2 + 2x3 y y¢ = cos ( y) y¢ +11 2x3 y - 9e2x-9 y - cos ( y ) ( )2x3 y - 9e2x-9 y - cos ( y) y¢ = 11- 2e2x-9y - 3x2 y2 Increasing/Decreasing – Concave Up/Concave DownCritical Pointsx = c is a critical point of f ( x) provided either Concave Up/Concave Down 1. If f ¢¢( x) > 0 for all x in an interval I then1. f ¢(c) = 0 or 2. f ¢(c) doesn’t exist. f ( x) is concave up on the interval I.Increasing/Decreasing 2. If f ¢¢( x) < 0 for all x in an interval I then f ( x) is concave down on the interval I.1. If f ¢( x) > 0 for all x in an interval I then f ( x) is increasing on the interval I.2. If f ¢( x) < 0 for all x in an interval I then Inflection Points f ( x) is decreasing on the interval I. x = c is a inflection point of f ( x) if the3. If f ¢( x) = 0 for all x in an interval I then concavity changes at x = c . f ( x) is constant on the interval I.Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins

Calculus Cheat SheetAbsolute Extrema Extrema Relative (local) Extrema1. x = c is an absolute maximum of f ( x) 1. x = c is a relative (or local) maximum of if f (c) ³ f ( x) for all x in the domain. f ( x) if f (c) ³ f ( x) for all x near c.2. x = c is an absolute minimum of f ( x) if f (c) £ f ( x) for all x in the domain. 2. x = c is a relative (or local) minimum of f ( x) if f (c) £ f ( x) for all x near c.Fermat’s Theorem 1st Derivative TestIf f ( x) has a relative (or local) extrema at If x = c is a critical point of f ( x) then x = c isx = c , then x = c is a critical point of f ( x) . 1. a rel. max. of f ( x) if f ¢( x) > 0 to the left of x = c and f ¢( x) < 0 to the right of x = c .Extreme Value Theorem 2. a rel. min. of f ( x) if f ¢( x) < 0 to the left of x = c and f ¢( x) > 0 to the right of x = c .If f ( x) is continuous on the closed interval[a,b] then there exist numbers c and d so that, 3. not a relative extrema of f ( x) if f ¢( x) is1. a £ c, d £ b , 2. f (c) is the abs. max. in the same sign on both sides of x = c .[a,b] , 3. f (d ) is the abs. min. in [a,b] . 2nd Derivative TestFinding Absolute Extrema If x = c is a critical point of f ( x) such thatTo find the absolute extrema of the continuous f ¢(c) = 0 then x = cfunction f ( x) on the interval [a,b] use the 1. is a relative maximum of f ( x) if f ¢¢(c) < 0 .following process. 2. is a relative minimum of f ( x) if f ¢¢(c) > 0 .1. Find all critical points of f ( x) in [a,b] .2. Evaluate f ( x) at all points found in Step 1. 3. may be a relative maximum, relative minimum, or neither if f ¢¢(c) = 0 .3. Evaluate f (a) and f (b) . Finding Relative Extrema and/or Classify Critical Points4. Identify the abs. max. (largest function value) and the abs. min.(smallest function 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 derivative test on each critical point. Mean Value TheoremIf f ( x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b)then there is a number a<c<b such that f ¢(c) = f (b) - f (a) . - a b Newton’s MethodIf xn is the nth guess for the root/solution of f (x) = 0 then (n+1)st guess is xn+1 = xn - f ( xn ) f ¢( xn )provided f ¢( xn ) exists.Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins

Calculus Cheat Sheet Related RatesSketch picture and identify known/unknown quantities. Write down equation relating quantitiesand differentiate with respect to t using implicit differentiation (i.e. add on a derivative every timeyou differentiate a function of t). Plug in known quantities and solve for the unknown quantity.Ex. A 15 foot ladder is resting against a wall. Ex. Two people are 50 ft apart when one starts walking north. The angleq changes atThe bottom is initially 10 ft away and is being 0.01 rad/min. At what rate is the distance between them changing when q = 0.5 rad?pushed towards the wall at 1 ft/sec. How fast 4is the top moving after 12 sec?x¢ is negative because x is decreasing. Using We have q ¢ = 0.01 rad/min. and want to findPythagorean Theorem and differentiating, x¢ . We can use various trig fcns but easiest is,x2 + y2 = 152 Þ 2x x¢ + 2 y y¢ = 0 secq = x Þ secq tan q q ¢ = x¢ 50 50After 12 sec we have x = 10 - 12 ( 1 ) = 7 and 4 We knowq = 0.5 so plug in q ¢ and solve.so y = 152 - 72 = 176 . Plug in and solve sec (0.5) tan (0.5) ( 0.01) = x¢for y¢ . 507 ( - 1 ) + 176 y¢ = 0 Þ y¢ = 7 ft/sec x¢ = 0.3112 ft/sec 4 4 176 Remember to have calculator in radians! OptimizationSketch picture if needed, write down equation to be optimized and constraint. Solve constraint forone of the two variables and plug into first equation. Find critical points of equation in range ofvariables and verify that they are min/max as needed.Ex. We’re enclosing a rectangular field with Ex. Determine point(s) on y = x2 +1 that are500 ft of fence material and one side of the closest to (0,2).field is a building. Determine dimensions thatwill maximize the enclosed area. Minimize f = d 2 = ( x - 0)2 + ( y - 2)2 and theMaximize A = xy subject to constraint of constraint is y = x2 +1. Solve constraint forx + 2 y = 500 . Solve constraint for x and plug x2 and plug into the function.into area. x2 = y -1 Þ f = x2 + ( y - 2)2 x = 500 - 2 y Þ A = y (500 - 2 y) = y -1+ ( y - 2)2 = y2 - 3y + 3 = 500 y - 2 y2 Differentiate and find critical point(s).Differentiate and find critical point(s). f ¢= 2y-3 Þ y = 3 A¢ = 500 - 4 y Þ y = 125 2By 2nd deriv. test this is a rel. max. and so is By the 2nd derivative test this is a rel. min. andthe answer we’re after. Finally, find x. so all we need to do is find x value(s). x = 500 - 2(125) = 250 x2 = 3 -1 = 1 Þ x=± 1The dimensions are then 250 x 125. 2 2 2 ( ) ( )The 2 points are then ,1 3 and - ,1 3 . 22 22Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins

Calculus Cheat Sheet Integrals DefinitionsDefinite Integral: Suppose f ( x) is continuous Anti-Derivative : An anti-derivative of f ( x)on [a,b] . Divide [a,b] into n subintervals of is a function, F ( x) , such that F¢( x) = f ( x) .width D x and choose xi* from each interval. Indefinite Integral : ò f ( x) dx = F ( x) + c b f ( x) dx = lim ¥ f where F ( x) is an anti-derivative of f ( x) . a n®¥ i=1ò å ( )Then xi* D x . Fundamental Theorem of CalculusPart I : If f ( x) is continuous on [a,b] then Variants of Part I :g ( x ) = x f (t ) dt is also continuous on [a,b] d ò u(x) f (t ) dt = u¢ ( x) f ëéu ( x)ûù dx a òaand g¢(x) = d x f (t ) dt = f ( x) . d b f (t ) dt = -v¢( x) f éëv ( x)ùû dx dx òa òv( x)Part II : f ( x) is continuous on[a,b] , F ( x) is d ò u(x) f (t ) dt = u¢( x) f [u(x)] - v¢( x) f [v(x)] dx v( x)an anti-derivative of f ( x) (i.e. F ( x) = ò f ( x) dx )then b f ( x) dx = F (b) - F (a) . òa Propertiesò f ( x) ± g ( x) dx = ò f ( x) dx ± ò g ( x) dx ò cf ( x) dx = cò f ( x) dx , c is a constant b f ( x) ± g ( x) dx = b f ( x) dx ± b g ( x) dx b cf ( x) dx = òc b f ( x) dx , c is a constant aòa òa òa òa a f ( x) dx = 0 b c dx = c (b - a )òa òa b f ( x) dx = a f ( x) dx b f ( x) dx b f (x) dxòa -òb òa £ òa b f ( x) dx = c f ( x ) dx + b f ( x) dx for any value of c.òa òa òcIf f (x) ³ g(x) on a £ x £ b then b f ( x) dx ³ b g ( x) dx òa òaIf f (x) ³ 0 on a£ x£b then b f ( x) dx ³ 0 òaIf m£ f (x) £ M on a£ x£b then m(b - a) £ b f ( x) dx £ M (b - a) òaò k dx = k x + c Common Integrals ò tan u du = ln sec u + cò xn dx = 1 x n +1 + c, n ¹ -1 ò cos u du = sin u + c ò sec u du = ln sec u + tan u + c n+1 ò sin u du = - cos u + c ò sec2 u du = tan u + cò òx-1 dx = tan -1 u 1 dx = ln x +c ò ( )1 u2 du = 1 a +c a2+ x aò1 dx 1 ln ax b c ò sec u tan u du = sec u + c ò ( )1 u ax +b ò csc u cot udu = - csc u + c a2- u2 a = a + + ò csc2 u du = - cot u + c du = sin -1 +cò ln u du = u ln (u) - u + cò eu du = eu + cVisit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins

Calculus Cheat Sheet Standard Integration TechniquesNote that at many schools all but the Substitution Rule tend to be taught in a Calculus II class.u Substitution : The substitution u = g ( x) will convert b f ( g ( x )) g¢ ( x ) dx = ò g(b) f (u) du using g(a) òadu = g¢( x) dx . For indefinite integrals drop the limits of integration.( )òEx. 2 5x2 cos x3 dx ò ( ) ò2 5x2 cos x3 8 5 cos ( ) 1 1 1 3 dx = u duu = x3 Þ du = 3x2dx Þ x2dx = 1 du = 5 sin (u) 8 = 5 (sin (8) - sin (1)) 3 3 1 3x = 1 Þ u = 13 = 1 :: x = 2 Þ u = 23 = 8 b u dv = uv b b v du . a a aò ò ò òIntegration by Parts : Choose u and dv from u dv = uv - v du and -integral and compute du by differentiating u and compute v using v = ò dv .òEx. xe-x dx òEx. 5 ln x dx 3 u = x dv = e-x Þ du = dx v = -e-x u = ln x dv = dx Þ du = 1 dx v=xò òxe-x dx = -xe-x + e-x dx = -xe-x - e-x + c x 5 dx = 3 ò ò ( )5 5 5 3 3 3 ln x dx = x ln x - x ln ( x) - x = 5ln (5) - 3ln (3) - 2Products and (some) Quotients of Trig FunctionsòFor sinn x cosm x dx we have the following : òFor tann x secm x dx we have the following :1. n odd. Strip 1 sine out and convert rest to 1. n odd. Strip 1 tangent and 1 secant out and cosines using sin2 x = 1- cos2 x , then use convert the rest to secants using the substitution u = cos x . tan2 x = sec2 x -1, then use the substitution2. m odd. Strip 1 cosine out and convert rest u = sec x . to sines using cos2 x = 1- sin2 x , then use 2. m even. Strip 2 secants out and convert rest the substitution u = sin x . to tangents using sec2 x = 1+ tan2 x , then3. n and m both odd. Use either 1. or 2. use the substitution u = tan x .4. n and m both even. Use double angle 3. n odd and m even. Use either 1. or 2. and/or half angle formulas to reduce the 4. n even and m odd. Each integral will be integral into a form that can be integrated. dealt with differently.Trig Formulas : sin (2x) = 2sin ( x) cos ( x) , cos2 (x) = 1 (1 + cos ( 2 x ) ) , sin 2 (x) = 1 (1- cos(2x)) 2 2òEx. tan3 x sec5 x dx òEx. sin5 x dx cos3 xò òtan3 x sec5 xdx = tan2 x sec4 x tan x sec xdx ò ò òsin5 (sin2 x)2 sin cos3 cos3 x x dx = sin4 xsin x dx = x dx x cos3 x ( )ò= sec2 x -1 sec4 x tan x sec xdx ò= (1-cos2 x)2 sin x dx (u = cos x) cos3 x = ò (u2 -1)u4du (u = sec x) (1-u2 )2 du 1-2u2 +u4 du u3 u3 sec7 sec5 ò ò= - = - = 1 x - 1 x + c = 1 sec2 x + 2 ln cos x - 1 cos2 x + c 7 5 2 2Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins

Calculus Cheat SheetTrig Substitutions : If the integral contains the following root use the given substitution andformula to convert into an integral involving trig functions. a2 - b2x2 Þ x = a sin q b2x2 - a2 Þ x = a secq a2 + b2x2 Þ x = a tanq b b b cos2 q = 1- sin2 q tan2 q = sec2 q -1 sec2 q = 1+ tan2 qòEx. 16 dx ( ) òó 16 cos 12 dq x2 4-9 x2 õ4 sin2 2 q dq = sin2q 9 q (2 cosq ) 3x = 2 sin q Þ dx = 2 cosq dq = ò12 csc2 dq = -12 cotq + c 3 3 4 - 9x2 = 4 - 4sin2 q = 4 cos2 q = 2 cosq Use Right Triangle Trig to go back to x’s. FromRecall x2 = x . Because we have an indefinite substitution we have sin q = 3x so, 2integral we’ll assume positive and drop absolutevalue bars. If we had a definite integral we’dneed to compute q ’s and remove absolute valuebars based on that and, x = ìx if x ³ 0 From this we see that cotq = 4-9 x2 . So, îí- x if x < 0 3xIn this case we have 4 - 9x2 = 2 cosq . ò x2 16 dx = - 4 4-9 x2 +c x 4-9 x2òPartial Fractions : If integrating P(x) dx where the degree of P(x) is smaller than the degree of Q( x)Q ( x) . Factor denominator as completely as possible and find the partial fraction decomposition ofthe rational expression. Integrate the partial fraction decomposition (P.F.D.). For each factor in thedenominator we get term(s) in the decomposition according to the following table. Factor in Q ( x) Term in P.F.D Factor in Q ( x) Term in P.F.D (ax + b)k ax + b A A1 + A2 +L+ Ak ax + b ( )ax2 + bx + c k ax + b ax2 + bx + c (ax + b)2 (ax + b)k Ax + B ax2 + bx + c + B1 Ak x + Bk bx + ax2 + bx + c k ( )A1x c + L + ax2 + 7 x2 +13x = + =7x2 +13x A( x2 +4)+( Bx+C ) ( x-1) ( x-1)( x2 +4) ( x-1)( x2 +4) ( x-1)( x2 +4)òEx. dx A Bx+C x-1 x2 +4ò ò7x2+13x dx = 4 + 3 x +16 dx Set numerators equal and collect like terms. ( x-1)( x2 +4) x-1 x2 +4 7x2 +13x = ( A + B) x2 + (C - B) x + 4A - C ò= 4 + 3x + 16 dx x-1 x2 +4 x2 +4 Set coefficients equal to get a system and solve to get constants. ( )= -1 x 4 ln x -1 + 3 ln x2 + 4 + 8 tan ( ) A + B = 7 C - B = 13 4 A - C = 0 2 2Here is partial fraction form and recombined. A=4 B=3 C = 16An alternate method that sometimes works to find constants. Start with setting numerators equal in( )previous example : 7x2 +13x = A x2 + 4 + ( Bx + C ) ( x -1) . Chose nice values of x and plug in.For example if x = 1 we get 20 = 5A which gives A = 4 . This won’t always work easily.Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins

Calculus Cheat Sheet Applications of IntegralsNet Area : b f ( x) dx represents the net area between f (x) and the òax-axis with area above x-axis positive and area below x-axis negative.Area Between Curves : The general formulas for the two main cases for each are, b d c -a ëéupper functionûù éëlower( ) ò ( ) òy = fx A= x= f y A= functionûù dy Þ functionùû dx & Þ ëéright -functionùû ëéleftIf the curves intersect then the area of each portion must be found individually. Here are somesketches of a couple possible situations and formulas for a couple of possible cases.A = b f ( x) - g ( x) dx A = d f ( y) - g( y) dy A = c f (x) - g ( x) dx + b g (x) - f ( x) dx òa òc òa òcVolumes of Revolution : The two main formulas are V = ò A( x) dx and V = ò A( y) dy . Here issome general information about each method of computing and some examples. Rings Cylinders ( )pA = -( ) ( )2 2 A = 2p ( )radius (width / )height outer radius inner radiusLimits: 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.Horz. Axis use f ( x) , Vert. Axis use f ( y) , Horz. Axis use f ( y) , Vert. Axis use f ( x) ,g ( x) , A( x) and dx. g ( y) , A( y) and dy. g ( y) , A( y) and dy. g ( x) , A( x) and dx.Ex. Axis : y = a > 0 Ex. Axis : y = a £ 0 Ex. Axis : y = a > 0 Ex. Axis : y = a £ 0outer radius : a - f ( x) outer radius: a + g ( x) radius : a - y radius : a + yinner radius : a - g ( x) inner radius: a + f ( x) width : f ( y) - g ( y) width : f ( y) - g ( y)These are only a few cases for horizontal axis of rotation. If axis of rotation is the x-axis use they = a £ 0 case with a = 0 . For vertical axis of rotation ( x = a > 0 and x = a £ 0 ) interchange x andy to get appropriate formulas.Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins

Calculus Cheat SheetWork : If a force of F ( x) moves an object Average Function Value : The average value f (x) on 1 b ( x) òof is favg b-a a f dxin a £ x £ b , the work done is W = b F ( x) dx a£ x£b = òaArc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are,òL = b ds òSA = b 2p y ds (rotate about x-axis) òSA = b 2p x ds (rotate about y-axis) a a awhere ds is dependent upon the form of the function being worked with as follows. dy 2 2 dy 2 dx dt dx + dt( ) ( ) ( )ds = dx if x = f (t ), y = g (t ), a £ t £ b 1+ if y = f ( x), a £ x £ b ds = dt 2 dr 2 dq dq dy( ) ( )ds = dx r2 + if r = f (q ), a £ q £ b 1+ dy if x = f ( y), a £ y £ b ds =With surface area you may have to substitute in for the x or y depending on your choice of ds tomatch the differential in the ds. With parametric and polar you will always need to substitute. Improper IntegralAn 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 limitdoesn’t exist or has infinite value. This is typically a Calc II topic.Infinite Limitò ò1. ¥ f ( x) dx = lim t f ( x) dx ò ò2. b f ( x) dx = lim b f ( x) dx a t®¥ a -¥ t®-¥ t3. ¥ f ( x) dx = c f ( x ) dx + ¥ f ( x) dx provided BOTH integrals are convergent. ò-¥ ò-¥ òcDiscontinuous Integrandò ò1. Discont. at a: b f ( x) dx = lim b f ( x) dx ò ò2. Discont. at b : b f ( x) dx = lim t f ( x) dx a t®a+ t a t®b- a3. Discontinuity at a<c<b : b f ( x) dx = c f ( x ) dx + b f ( x) dx provided both are convergent. òa òa òcComparison Test for Improper Integrals : If f ( x) ³ g ( x) ³ 0 on [a,¥) then,1. òIf ¥ f ( x) dx conv. then ò ¥ g ( x ) dx conv. 2. If ò ¥ g ( x ) dx divg. òthen ¥ f ( x) dx divg. a a a aòUseful fact : If a > 0 then ¥ 1 dx converges if p > 1 and diverges for p £1. a xp Approximating Definite IntegralsòFor given integral b f ( x) dx and a n (must be even for Simpson’s Rule) define Dx = b-a and a ndivide [a,b] into n subintervals [ x0, x1] , [ x1, x2 ] , … , [ xn-1, xn ] with x0 = a and xn = b then,ò ( ) ( ) ( )Midpoint Rule : b [ ]xn* ùû , xi* is midpoint xi-1, xi a f ( x) dx » Dx ëé f x1* + f x2* +L + fTrapezoid Rule : b f ( x) dx » Dx éë f ( x0 ) + 2 f ( x1 ) + +2 f ( x2 ) +L + 2 f ( xn-1 ) + f ( xn )ùû 2 òaòSimpson’s Rule : b f ( x) dx » Dx ëé f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) +L + 2 f ( xn-2 ) + 4 f ( xn-1 ) + f ( xn )ùû a 3Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins