### Transcription of 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=.

2 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. 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.

3 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 =. 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.

4 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. 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=.

5 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.

6 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.

7 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. 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.

8 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. 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.

9 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=.

10 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.