Calculus Cheat Sheet Integrals - Lamar University

1 Calculus Cheat Sheet Visit for a complete set of Calculus notes. 2005 Paul Dawkins Integrals Definitions Definite Integral: Suppose ()fx is continuous on [ ],ab. Divide [ ],ab into n subintervals of width x a nd choose *ix from each interval. Then ()()*1liminbanif x dxf xx == . Anti-Derivative : A n anti-derivative of ()fx is a function, ( )Fx, such that ( ) ( )F x fx =. Indefinite Integral :( )( )f x dxF xc=+ where ( )Fx is an anti-derivative of ( )fx. Fundamental Theorem of Calculus Part I : If ( )fx is continuous on [],ab then ( )( )xag xf t dt= is also continuous on [ ],ab and ( )( )()xadg xf t dtf xdx == . Part II : ( )fxis continuous on[ ],ab, ( )Fx is an anti-derivative of( )fx( ( )( )F xf x dx= ) then( )( )()baf x dxF bF a= .Variants of Part I : ()( )()()uxadf t dtu x f u xdx = ( )( )( )( )bvxdf t dtv x f v xdx = ( )( )( )( )[ ]( )[ ]()()uxvxuxvxdftdt u xfv xfdx = Properties ( ) ( )( )( )f xg x dxf x dxg x dx = ( ) ( )( )()bbbaaaf xg x dxf x dxg x dx = ()0aaf x dx= ( )( )baabf x dxf x dx= ( )( )cf x dxc f x dx= , c is a constant ()()bbaacf x dxcf x dx= , c is a constant ()bac dxc b a= ( )( )bbaaf x dxf x dx ( )( )( )b cba acf x dxf x dxf x dx=+ for any value of c.

2 If () ( )f x gx ona xb then ( )( )bbaaf x dxg x dx If( )0fx on a xb then ( )0baf x dx If ( )m fx M on a xb then ()( )()bambafxdx Mba Common Integrals k dxk x c= + 111,1nnnx dxxc n++=+ 11lnxx dxdxxc == + 11lnaax bdxax bc+= ++ ( )lnlnu duuuu c= + uuduc= + ee cossinu duu c=+ sincosu duu c= + 2sectanu duu c=+ sec tansecuu duu c=+ csc cotcscuuduu c= + 2csccotu duu c= + tanln secu duuc=+ secln sectanu duuuc= ++ ( )11122tanuaaauduc +=+ ( )1221sinuaauduc =+ Calculus Cheat Sheet Visit for a complete set of Calculus notes. 2005 Paul Dawkins Standard Integration Techniques Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class. u Substitution : The substitution ( )u gx=will convert ( )()()( )( )()bgbagaf g x g x dxf u du = using ( )dug x dx =.

3 For indefinite Integrals drop the limits of integration. Ex. ( )23215cosxx dx 322133uxdux dxx dxdu= = = 33111::228xuxu= == = == ( )( )( )( )( )()232853118553315coscossinsin 8sin 1xx dxu duu=== Integration by Parts : u dv uvv du= and bbbaaau dv uvv du= . Choose u and dv from integral and compute du by differentiating u and compute v using vdv= . Ex. xxdx e xxuxdvdudx v = = = = ee xxxxxxdxxdxxc = + = + eeeee Ex. 53lnx dx 1lnxuxdvdxdudx vx= = = = ( )()( )( )55553333lnlnln5 ln 53 ln 32x dxxxdxxxx= = = Products and (some) Quotients of Trig Functions For sincosnmxx dx we have the following : 1. n odd. Strip 1 sine out and convert rest to cosines using 22sin1 cosxx= , then use the substitution cosux=. 2. m odd. Strip 1 cosine out and convert rest to sines using 22cos1 sinxx= , then use the substitution sinux=. 3. n and m both odd.

4 Use either 1. or 2. 4. n and m both even. U se double angle and/or half angle formulas to reduce the integral into a form that can be integrated. Fortansecnmxx dx we have the following : 1. n odd. Strip 1 tangent and 1 secant out and convert the rest to secants using 22tansec1xx= , then use the substitution secux=. 2. m even. Strip 2 secants out and convert rest to tangents using 22sec1 tanxx= +, then use the substitution tanux=. 3. n odd and m even. Use either 1. or 2. 4. n even and m odd. Each integral will be dealt with differently. Trig Formulas : ( )( ) ( )sin 22 sincosxxx=, ( )( )()212cos1 cos 2xx= +, ( )( )()212sin1 cos 2xx= Ex. 35tansecxx dx ()()()35242424751175tansectansectan secsec1 sectan sec1secsecsecxxdxxxxxdxxxxxdxuu duuxxxc== = == + Ex. 53sincosxxdx ()2211222254333223222433sin(sin)sinsinsi ncoscoscossin(1 cos)cos(1)12cossec2 ln coscosxxxxxxxxxxxuuuuudxdxdxdxuxduduxxxc +===== = =+ + Calculus Cheat Sheet Visit for a complete set of Calculus notes.

5 2005 Paul Dawkins Trig Substitutions : If the integral contains the following root use the given substitution and formula to convert into an integral involving trig functions. 222sinaba bxx = 22cos1 sin = 222secabbx ax = 22tansec1 = 222tanaba bxx + = 22sec1 tan = + Ex. 221649xxdx 2233sincosxdxd = = 2224 4 sin4 cos2 cos49x = = = Recall 2xx=. Because we have an indefinite integral we ll assume positive and drop absolute value bars. If we had a definite integral we d need to compute s and remove absolute value bars based on that and, if 0if 0xxxxx = < In this case we have 22 cos49x = . ()()23sin2 cos222491612sincos12 csc12 cotdddc == = + Use Right Triangle Trig to go back to x s. From substitution we have 32sinx = so, From this we see that 2493cotxx =. So, 2221644949xxxxdxc = + Partial Fractions : If integrating ( )( )PxQxdx where the degree of ( )Px is smaller than the degree of ( )Qx.

6 Factor denominator as completely as possible and find the partial fraction decomposition of the rational expression. Integrate the partial fraction decomposition ( ). For each factor in the denominator we get term(s) in the decomposition according to the following table. Factor in ( )Qx Term in Factor in ( )Qx Term in ax b+ Aax b+ ()kax b+ ()()122kkAAAax bax bax b++++++ 2axbx c++ 2Ax Baxbx c+++ ()2kaxbx c++ ()1122kkkAx BAx Baxbx caxbx c++++++++ Ex. 2()()214713xxxxdx ++ ()( )2222()()2132223164114431641447134 ln1ln48 tanxxxxxxxxxxxxdxdxdxxx + ++ +++= += ++= ++ + Here is partial fraction form and recombined. 22224) () ()()()()()21114414(713Bx CxxxxxxxAxBx CAxx+++ ++ +++=+= Set numerators equal and collect like terms. ()()227134xx A BxC Bx A C+ =+ + + Set coefficients equal to get a system and solve to get constants. 713404316A BC BACA BC+= = ==== An alternate method that sometimes works to find constants.

7 Start with setting numerators equal in previous example : ()() ()2271341xxA xBx Cx+ = ++ + . Chose nice values of x and plug in. For example if 1x= we get 20 5A= which gives 4A=. This won t always work easily. Calculus Cheat Sheet Visit for a complete set of Calculus notes. 2005 Paul Dawkins Applications of Integrals Net Area : ()baf x dx represents the net area between ( )fx and the x-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, ( )upper functionlower functionbayf xAdx = = & ( )right functionleft functiondcxf yAdy = = 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. ( )( )baAf xg x dx= ( ) ( )dcAf yg y dy= () ()( ) ()cbacAf xg x dxg xf x dx= + Volumes of Revolution : The two main formulas are ( )VA x dx= and ( )VA y dy=.

8 Here is some general information about each method of computing and some examples. Rings Cylinders ()()()22outer radiusinner radiusA = () ()radiuswidth / height2A = 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. Horz. Axis use( )fx, ()gx,( )Ax and dx. Vert. Axis use( )fy, ( )gy,( )Ay and dy. Horz. Axis use( )fy, ( )gy,( )Ay and dy. Vert. Axis use( )fx, ( )gx,( )Ax and dx. Ex. Axis : 0ya= > Ex. Axis : 0ya= Ex. Axis : 0ya= > Ex. Axis : 0ya= outer radius :( )a fx inner radius : ( )a gx outer radius:( )a gx+ inner radius:( )a fx+ radius :ay width : ( ) ( )f y gy radius :ay+ width : ( ) ( )f y gy These are only a few cases for horizontal axis of rotation. If axis of rotation is the x-axis use the 0ya= case with 0a=. For vertical axis of rotation (0xa= > and 0xa= ) interchange x and y to get appropriate formulas.

9 Calculus Cheat Sheet Visit for a complete set of Calculus notes. 2005 Paul Dawkins Work : If a force of( )Fxmoves an object ina xb , the work done is ( )baWF x dx= Average Function Value : The average value of ( )fx on a xb is( )1bavgabaff x dx = Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are, baLds= 2baSAy ds = (rotate about x-axis) 2baSAx ds = (rotate about y-axis) where ds is dependent upon the form of the function being worked with as follows. ( )( )21 if ,dydxdsdxyf xax b= += ()()21 if ,dxdydsdyxf yay b= += ( )()( )( )22 if ,,dydxdtdtdsdtxf tyg ta tb=+== ( )( )22 if ,drddsrdrfab = += 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.

10 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 1. ( )()limtaatf x dxf x dx = 2. ( )( )limbbttf x dxf x dx = 3. ( )( )( )ccf x dxf x dxf x dx =+ provided BOTH Integrals are convergent. Discontinuous Integrand 1. Discont. at a:( )( )limbbattaf x dxf x dx+ = 2. Discont. at b :()( )limbtaatbf x dxf x dx = 3. Discontinuity at acb<< : ()( )( )b cbaacf x dxf x dxf x dx=+ provided both are convergent. Comparison Test for Improper Integrals : If ( ) ( )0f x gx on [),a then, 1. If( )af x dx conv. then( )ag x dx conv. 2. If( )ag x dx divg. then( )af x dx divg. Useful fact : If 0a> then 1apxdx converges if 1p> and diverges for 1p . Approximating Definite Integrals For given integral ( )baf x dx and a n (must be even for Simpson s Rule) define banx = and divide [ ],ab into n subintervals []01,xx, []12,xx.]