Formulas for Calculus II Cheat Sheet, Cheat Sheet of Calculus

Download Formulas for Calculus II Cheat Sheet and more Cheat Sheet Calculus in PDF only on Docsity! / Calculus Cheat Sheet Limits Definitions Precise Definition : We say lim f ( x) = L if Limit at Infinity : We say lim f ( x) = L if we x➔a x➔oo for every s > O there is a 8 > O such that can make f ( x) as close to L as we want by whenever O < lx - ai < 8 then IJ ( x )- LI < B . taking x large enough and positive. "Working" Definition : We say lim f ( x) = L x➔a if we can make f ( x) as close to L as we want by taking x sufficiently close to a ( on either side of a) without letting x = a. Right hand limit : lim f ( x) = L. This has x➔a + the same definition as the limit except it requires x > a . Left �and limit : lim_ f ( x) = L . This has the 1 , x➔a same definition as the limit except it requires x<a. There is a similar definition for lim f ( x) = L x➔-oo except we require x large and negative. Infinite Limit : We say lim f ( x) = oo if we x➔a can make f ( x) arbitrarily large ( and positive) by taking x sufficiently close to a (on either side of a) without letting x =a. There is a similar definition for lim f ( x) = -oo x➔a except we make f ( x) arbitrarily large and negative. Relationship between the limit and one-sided limits limf(x)=L ⇒ lim f(x)= lim_f(x)=L lim f(x)= lim_f(x)=L ⇒ limf(x)=L x➔a x➔a + x➔a x➔a + x➔a x➔a lim f ( x) =1= lim_ f ( x) ⇒ lim f ( x) Does Not Exist x➔a + x➔a x➔a Properties Assume limf(x) and limg(x) both exist and e is any number then, .x➔a x➔a 1. lim[cf (x)] = climf(x)x➔a x➔a 3. lim [ f ( x) g ( x)] = lim f ( x) lim g ( x) x➔a x➔a x➔a 4. lim[f( (x ) )l )�In provided limg(x) * O x➔a g X hm g X x➔a x➔a 5. �i_Ta [f ( X )J = [� j (X) r 6. �� [ � j (X)]= n � f (X) Basic Limit Evaluations at ± oo Note : sgn (a)= 1 if a> O and sgn (a)= -1 if a< O . 1. limex=CXJ & lim ex =O 5. neven: lim xn= OO x➔oo x➔- oo x➔±<0 2. lim ln(x)=oo & lim ln(x)=-oo x➔oo x➔o- 3. If r > O then lim � = O x➔oo xr 4. If r > O and xr is real for negative x then lim �= O x➔-oo x' 6. n odd : lim xn = oo & lim xn = -oo x➔oo x➔-oo 7. n even : lim a xn + · · · + b x + e = sgn (a) oo x➔±oo 8. nodd: limaxn+ .. ·+bx+c= sgn(a)oo x➔oo 9. n odd : lim a xn + · · · + ex + d = -sgn (a) oo x➔-oo Calculus Cheat Sheet Evaluation Techniques Continuous Functions If / ( x) is continuous at a then lim f ( x) = f (a) x➔a Continuous Functions and Composition f ( x) is continuous at b and lim g ( x) = b then x➔a � f (g(x)) = !(�i�g( x)) = f (b) Factor and Cancel ]. x2 +4x-l2 1. (x-2)(x+6) 1m 2 = 1m x➔2 X -2x x➔2 X ( X - 2) = lim x + 6 = � = 4 x➔2 X 2 Rationalize Numerator/Denominator I. 3-✓x -1· 3-✓x 3+✓x,m i - ,m 2 , x ➔9 X -81 x➔9 X -81 3 + -y X . 9-x . -1 =hm------= hm------ x ➔9 (x2 -81)(3+✓x) x➔9 (x+9)(3+✓x ) -1 1 =---= (18) (6) 108 Combine Rational Expressions lim_!_(- 1 _ _!_)=lim_!_[ x-(x+h)) h➔O h X+ h X li➔O h X (X+ h) L'Hospital's Rule If lim f ( x) = .2. or lim f ( x) = ± 00 then x➔a g (X) O x➔a g (X) ± 00 ' lim / ( (x ) ) = lim f'(( x ) ) a is a number, oo or -oo x➔a g X x➔a g' X Polynomials at Infinity p ( x) and q ( x) are polynomials. To compute lim p ( ( x ) ) factor largest power of x in q ( x) out x➔±oo q X of both p ( x) and q ( x) then compute limit. 2 x2 (3 _..i._) 3 _..i._ lim 3 x - 4 = lim x 2 = lim __ x_i = _i x ➔-00 5x - 2x2 x➔-00 x2 ( f-2) x➔- 00 f- 2 2 Piecewise Function lim g(x) where g(x) ={x2 +5 x➔-2 l-3x Compute two one sided limits, lim g ( x) = lim x2 + 5 = 9 x➔-i- x➔-2- lim g ( x) = lim 1-3x = 7 x➔-2• x➔-2• ifx <-2 jf X:?: -2 One sided limits are different so lim g ( x) x➔-2 doesn't exist. Ifthe two one sided limits had been equa! then lim g ( x) would have existed x➔-2 and had the same value. Some Continuous Functions Partial 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 ( ) ( ) . division by zero. 8. tan x and sec x prov1ded 3. .ef; (n odd) for ali x. x * ... _ 37r _!!.._ !!_ 3n ... ' 2' 2' 2 ' 2 ' 4. .ef; (n even) for ali x:?: O . 5. ex for ali x. 9. cot(x) and csc (x) provided 6. ln x for x > O . x * •··,-27l',-7l',0,n,2n,- ·· Intermed.iate Value Theorem Suppose that f ( x) is continuous on [ a, b] and !et M be any number between / (a) and f ( b) . Then there exists a number e such that a < e < b and f (e) = M. \ Calculus Cheat Sheet Extrema Absolute Extrema 1. x = e is an absolute maximum of f ( x) if / (e) � f ( x) for all x in the domain. 2. x = e is an absolute minimum of f ( x) if / (e) ::; f ( x) for all x in the domain. Fermat's Theorem If f ( x) has a relative ( or locai) extrema at x =e, then x = e is a criticai point of f ( x) . Extreme Value Theorem If f ( x) is continuous on the closed interval [ a, b] then there exist numbers e and d so that, 1. a::; e, d::; b , 2. f (e) is the abs. max. in [a,b ]', 3. f ( d) is the abs. min. in [ a,b]. Finding Absolute Extrema To find the absolute extrema of the continuous function f ( x) on the interval [ a, b] use the following process. 1. Find all criticai points of f ( x) in [a, b] . 2. Evaluate f ( x) at ali points found in Step 1. 3. Evaluate f (a) and f (b). 4. Identify the abs. max. (largest function value) and the abs. min.(smallest function value) from the evaluations in Steps 2 & 3. Relative (locai) Extrema 1. x = e is a relative ( or locai) maximum of f ( x) if f (e)� f ( x) for ali x near c. 2. x = e is a relative (or locai) minimum of f ( x) if f (e)::; f ( x) for ali x near c. 1 st Derivative Test If x = e is a criticai point of f ( x) then x = e is 1. a rei. max. of f ( x) if f' ( x) > O to the left of x = e and f' ( x) < O to the right of x = e . 2. a rei. min. of f ( x) if f' ( x) < O to the left of x = e and f' ( x) > O to the right of x = e. 3. not a relative extrema of f ( x) if f' ( x) 1s the same sign on both sides of x = e . 2 nd Derivative Test If x = e is a criticai point of f ( x) such that f' (e) = O then x = e 1. is a relative maximum of f ( x) if f" (e)< O. 2. is a relative minimum of f ( x) if / 11 (e) > O . 3. may be a relative maximum, relative minimum, or neither if / 11 (e) = O . Finding Relative Extrema and/or Classify Criticai Points 1. Find ali criticai points of f ( x) . 2. Use the 1 st derivative test or the 2nd derivative test on each criticai point. Mean Value Theorem If f ( x) is continuous on the closed interval [ a, b] and differentiable on the open interval ( a, b) · '( ) f(b)-f(a) then there 1s a number a <e< b such that f e = --''---'-----'---"--. b-a Newton 's 1\'Iethod If x n is the nth guess for the root/solution of f ( x) = O then (n+ 1 f guess is x n+L = x n - 1,tn \ f Xn provided f' ( x n ) exists. Calcu/us Cheat Sheet Related Rates Sketch picture and identify known/unknown quantities. Write down equation relating quantities and differentiate with respect to t using implicit differentiation (i.e. add on a derivative every time ou differentiate a function of t . Plu in known quantities and solve for the unknown uanti . Ex. A 15 foot ladder is resting against a wall. Ex. Two people are 50 ft apart when one The bottom is initially 1 O ft away and is being starts walking north. The angle 0 changes at pushed towards the wall at ¼ft/sec. How fast 0.01 rad/min. At what rate is the distance is the top moving after 12 sec? between them changing when 0 = 0.5 rad? y' t y X -x'=-¼ x' is negative because x is decreasing. Using Pythagorean Theorem and differentiating, x2 + y2 =152 ⇒ 2xx'+2yy'=0 After 12 sec we have x = l O -12 ( ¼) = 7 and so y =;= ✓152 -72 = ✓176 . Plug in and solve I ' for y'. 7(-¼)+ ✓176 y' = O ⇒ y' = 4 � ft/sec t Moving P:on '---"5--'0_f"--t ____ .a.. Stationary Penon We have 0' = 0.01 rad/min. and want to find x'. We can use various trig fcns but easiest is, sec0 = � ⇒ sec0 tan 0 0' = i_ 50 50 We know0 = 0.05 so plug in 0' and solve. x' sec(0.5)tan(0.5)(0.01) = 50 x' = 0.3112 ft/sec Remember to have calculator in radians! Optimization Sketch picture if needed, write down equation to be optimized and constraint. Sol ve constraint for one of the two variables and plug into first equation. Find criticai points of equation in range of variables and verif that the are minimax as needed. ;,......-________________ _ Ex. We're enclosing a rectangular field with 500 ft of fence materiai and one side of the field is a building. Determine dimensions that will maximize the enclosed area. Building X Maximize A= xy subject to constraint of x + 2y = 500 . Sol ve constraint for x and plug into area. A= y(500-2y) x=500-2y ⇒ = 500y-2y2 Differentiate and find criticai point(s). A'= 500-4y ⇒ y = 125 By 2nd deriv. test this is a rei. max. and so is the answer we're after. Finally, find x. x=500-2(125)=250 The dimensions are then 250 x 125. Ex. Determine point(s) on y = x2 + 1 that are closest to (0,2). � ,:, (�:�) x,y) n Minimize f = d2 = (x-0}1 +(y-2)2 and the constraint is y = x2 + 1. Solve constraint for x2 and plug into the function. x2 = y-l ⇒ f = x2 +(y-2)2 = y-l+(y-2)2 = y2 -3y+3 Differentiate and finà criticai point(s). f' =2y-3 ⇒ y=f By the 2nd derivative test this is a rei. min. and so all we need to do is find x value(s). x2 = 1. -1 = .l ⇒ X = + _l_ 2 2 - ✓2 The 2 points are then ( }i , f) and ( -}i , ½) . "" Calculus Cheat Sheet lntegrals Definitions Definite Integrai: 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 /J. x and choose x; from each interval. Indefinite Integrai : f f ( x) dx= F ( x) + e Then J: J( x )dx= !�if ( x; )!:!i.x. where F ( x) is an anti-derivative of f ( x). Fundamental Theorem of Calculus Part I : If f ( x) is continuous on [ a, b] then Variants of Part I : d f u(x) ) [ )]g(x) = J: f(t)dt is also continuous on [a,b] dx a f(t)dt =u'(x f u(x and g'(x) = ! [ f(t)dt = f(x). ! f �x)f (t)dt =-v'(x)f[v(x)] . Part II: f ( x) is continuous on [ a,b], F( x) is .!!_f u(x) f (t)dt = u' ( x )f [ u(x)]-v' ( x) f [ v(x)]dx v(x) an an�i-derivative of f ( x) (i.e. F ( x) = f f ( x) dx) th�n· f: f ( x) dx·= F ( b) - F (a) . Properties f f(x)±g(x)dx= f f(x)dx± f g(x)dx f cf ( x) dx = e f f ( x) dx , e is a constant b f b f b LJ(x)±g(x)dx= af(x)dx± ag(x)dx f: cf ( x) dx = e f: f ( x) dx , e is a constant f: f(x)dx=O f:J(x)dx=-rf(x)dx f: f ( x) dx = f: f ( t) dt IJ: f(x)dxl � J:IJ(x) ldx If f (X) � g (X) on a � X � b then f: f (X) dx � f: g (X) dx If f ( x) � O on a � x � b then J: f ( x) dx � O If m � f ( x) � M on a � x � b then m ( b - a) � J: f ( x) dx � M ( b - a) f kdx=kx+c f Xn dx = _I Xn+l + C n "F- -1 n+l ' f x-1 dx= f fdx = lnlxl+c f axi+b dx=¾ ìnlax +bi+ e f ln u du = u In ( u )- u + e f eu du = eu +e Common Integrals f cos u du = sin u + e f sin u du = -cos u + e f sec 2 u du = tan u + e f sec u tan u du = sec u + e f csc u cot udu = - csc u + e f csc2 u du =-cotu+c f tan u du = lnlsecul +e f sec u du = ln isec u + tan ul + e f-1 -du = ..L tan-1 (!!..)+e a2+u2 a a f ,.!----, du = sin -i (;)+e -.;a2 -u2 Calculus Cheat Sheet Applications of Integrals Net Area : J: f ( x) dx represents the net area between f ( x) and the x-axis with area above x-axis positive and area below x-axis negative. Area Between Curves : The generai formulas for the two main cases for each are, y = f (x) ⇒ A= J)upper function]-[!ower function]dx & X= f(y) ⇒ A= L d [rìght function]-[!eft function]dy If the curves intersect then the area of each portion must be found individually. Here are some sketches of a couple possible situations and formulas for a couple of possible cases. y y = f(x) Y y = f(x) 0 '- a y = g(x) a e A= J: f(x)-g(x)dx A= [f(x)-g(x)dx+ J: g(x)-f(x)dx Volumes of Revolution : The two main formulas are V= f A ( x) dx and V= f A (y) dy. Here is some generai information about each method of computing and some examples. Rings Cylinders À = n ( (outer radius)2 -(innor radius)2 ) Limits: x/y of right/bot ring to x/y of left/top ring Horz. Axis use f ( x), Vert. Axis use f (y), g ( x), A ( x) and dx. g (y), A (y) and dy. Ex. Axis : y = a > O y X Ex. Axis : y = a � O a ------- - ----- outer radius : a -f ( x) outer radius: lai+ g ( x) inner radius : a- g ( x) inner radius: lai+ f ( x) A= 2n (rac!ius)(width/height) Limits : x/y of inner cyl. to x/y of outer cyl. Horz. Axis use f (y), Vert. Axis use f (x), g(y),A(y) anddy. g(x),A(x) anddx. Ex. Axis : y = a > O radius : a-y width: f(y)-g(y) Ex. Axis : y = a � O y radius : lai+ 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 the y = a � O case with a = O . F or verti ca! axis of rotati on ( x = a > O and x = a � O) interchange x and y to get appropriate formulas. Calculus Cheat Sheet Work : Ifa force of F ( x) moves an object in a :=:;; x 5 b , the work done is W = f: F ( x) dx Average Function Value: The average value of f ( X ) on a 5 X 5 b is favg = b�a S: f ( X ) dx Are Length Surface Area : Note that this is often a Cale II topic. The three basic formulas are, L = f: ds SA = f: 2n y ds (rotate about x-axis) SA = f: 2n x ds (rotate about y-axis) where ds is dependent upon the form of the function being worked with as follows. ds=)1+(:f dx if y = f(x), a5x5b ds= (�) 2 +(�f d t if x = f(t),y = g(t), a5t5b ds=)1+(:r dy if x = f(y), a5y5b ds= r 2 +(:;r d 0 if r=/(0), a505b 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. lmproper Integrai · An improper integra! is an integra! with one or more infinite limits and/or discontinuous integrands. Integra! 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 Cale II topic. Infinite Limit 1. L 00 f (x)dx = !��(f (x)dx 2. roof (x)dx = ,�1!r f(x)dx 3. f _: f ( x) dx = f : 00 f ( x) dx+ f c00 f ( x) dx provided BOTH integrals are convergent. Discontinuous Integrand 1. Discont. at a: f b f ( x) dx = lim f b f ( x) dx a t➔a• l 2. Discont. at b : f b f ( x) dx = li� f I f ( x) dx a t➔ b a 3. Discontinuity at a < e < b : J: f ( x) dx = f: f ( x) dx+ f: f ( x) dx provided both are convergent. Comparison Test for lmproper Integrals : If f ( x) � g ( x) � O on [ a, oo) then, 1. If f: f(x)dx conv. then f: g(x)dx conv. 2. If f: g(x)dx divg. then L 00 f(x)dx divg. Useful fact : If a> O then f 00 � dx converges if p > 1 and diverges for p 51. a X Approximating Definite Integrals F or given integra! f: f ( x) dx and a n (must be even for Simpson' s Rule) define fu = b�a and divide [ a, b] into n subintervals [ x 0 , x 1 ] , [ x 1 , x 2 ] , ••• , [ x n -l, x n ] with x 0 = a and x n = b then, Midpoint Rule: J: f ( x) dx� !!u[f ( x;) + f ( x;)+ .. · + f ( x:)], x; is midpoint [ X;_i, X;] Trapezoid Rule: f: f (x) dx�� [f ( x0) + 2f ( x,) + +2/ (x2) + .. · + 2/ ( xn_,) + f ( xn )] Simpson's Rule: f: f ( x )dx�� [f (x0) + 4f ( X1) + 2/ (x2 )+ .. · + 2/ ( xn_2 ) + 4/ ( xn_,) + f ( xn)] ■