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Calculus Cheat Sheet Calculus Cheat Sheet
Integrals Standard Integration Techniques
Definitions Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class.
Definite Integral: Suppose fx is continuous Anti-Derivative : An anti-derivative of fx
( ) ( ) bgb
¢ ( )
u Substitution : The substitution u= gxwill convert fgxgxdx= fudu using
( ) () () ()
¢ ( )
on ab, . Divide ab, into n subintervals of is a function, Fx, such that Fx= fx.
[ ] [ ] ( ) ( ) ( ) òò
aga
()
width Dx and choose x* from each interval. Indefinite Integral : fxdx=+Fxc ¢
( ) ( ) du=gxdx. For indefinite integrals drop the limits of integration.
i ò ( )
b ¥ 2 28
* where Fx is an anti-derivative of fx. 23 235
Then . ( ) ( ) Ex. 5xcos xdx 5xcosxdx= cos udu
fxdx=Dlim fxx ()
() å () ( ) ( )
i 3
ò ò òò
a n®¥ i=1 1 11
322 8
1 55
u=xÞdu=3xdxÞ=xdxdu
3 =sinu =-sin8sin1
() () ()
( )
33
Fundamental Theorem of Calculus 331
Part I : If fx is continuous on ab, then Variants of Part I : x=1Þu=1=1::xu=2Þ==28
( ) [ ]
ux
x d () bb
¢ b
ftdt=uxféùux
() () ()
gx= ftdt is also continuous on ab, ò ëû Integration by Parts : udv=-uvvdu and udv=-uvvdu. Choose u and dv from
()ò () [ ] dx a òòòò
a a
aa
dx d b ¢ integral and compute du by differentiating u and compute v using v= dv .
ftdt=-vxféùvx
¢ () () ()
and gx==ftdtfx. ò
() () () ò ëû
vx
ò dx()
dx a -x 5
ux Ex. xe dx
Part II : fxis continuous on ab, , Fx is d () ò Ex. lnxdx
( ) [ ] ( ) ¢¢ ò
ftdt=-uxfu(x)vxfvx()
() ()[ ] ()[ ] 3
ò
vx
dx () --xx
u=xdv=eeÞdu=dxv=- 1
an anti-derivative of fx(i.e. Fx= fxdx) u=lnxdv=dxÞdu==dxvx
( ) ( ) ò ( ) x
-x-x-x--xx
xedx=-xe+edx=-xcee-+ 555
b òò 5
lnxdx=xlnx-dx=-xln xx
then fxdx=-FbFa. ( () )
() () () òò
3
ò 333
a
=5ln5--3ln32
() ()
Properties
fx±gxdx=±fxdxgxdx cfxdx=cfxdx, c is a constant
( ) ( ) ( ) ( ) ( ) ( ) Products and (some) Quotients of Trig Functions
òòò òò
bbbbb nm nm
For sinxcos xdx we have the following : For tanxsec xdx we have the following :
fx±gxdx=±fxdxgxdx cfxdx=cfxdx, c is a constant ò ò
() () () () () ()
òòòòò
aaaaa 1. n odd. Strip 1 sine out and convert rest to 1. n odd. Strip 1 tangent and 1 secant out and
a bb 22 convert the rest to secants using
fxdx=0 fxdx= ftdt cosines using sinxx=-1cos , then use
() () ()
ò òò
a aa 22
ba bb the substitution ux=cos. tanxx=-sec1, then use the substitution
fxdx=- fxdx 2. m odd. Strip 1 cosine out and convert rest ux=sec .
() () fxdx£ fxdx
òò () ()
ab òò
aa 22 2. m even. Strip 2 secants out and convert rest
ba to sines using cosxx=-1sin , then use
If fx³gx ona££xbthen fxdx³ gxdx 22
( ) ( ) () () the substitution ux= sin . to tangents using secxx=+1tan , then
òò
ab 3. n and m both odd. Use either 1. or 2. use the substitution ux= tan .
If fx³0 on a££xb then b fxdx³0 4. n and m both even. Use double angle 3. n odd and m even. Use either 1. or 2.
( ) òa ()
b and/or half angle formulas to reduce the 4. n even and m odd. Each integral will be
If m££fxMon a££xb then mb-a£fxdx£-Mba integral into a form that can be integrated. dealt with differently.
( ) ( ) () ( )
ò
a 2 1 2 1
Trig Formulas : sin2x=2sinxxcos , cosxx=+1cos2 , sinxx=-1cos2
( ) ( ) ( ) ( ) 2 ( ( )) ( ) 2 ( ( ))
Common Integrals
35 sin5 x
òkdx=+kxc òcosudu=+sinuc òtanudu=+lnsecuc Ex. òtanxsec xdx Ex. ò 3 dx
cos x
nn+1
1 3524 5422
sin x
òxdx=x+cn,1¹- òsinudu=-+cosuc òsecudu=lnsecu++tanuc tanxsecxdx=tanxsecxtanxsecxdx sinxsinxxsin (sin)x
n+1 òò dx==dxdx
333
òòò
-1 1 2 1 1 -1 u cosxcosxxcos
xdx=dx=+ln xc secudu=+tanuc duc=+tan 24 22
( ) =-secx1secxtanxsecxdx sin x
x 22 aa (1-cos)x
òò ò ò ( )
au+ ò ==dxuxcos
3 ( )
ò
11 secutanudu=+secuc 1 -1 u 24 cos x
dx=lnax++bc duc=+sin
ò a ò ( ) =u-=1uduuxsec 22
axb+ ò 22 a ( ) ( ) (1)-u24
au- ò 12-+uu
=-du=- du
33
òò
cscucotudu=-+cscuc 75 uu
lnudu=uln u-+uc 11
ò ( ) ò =secx-+sec xc
75 22
11
2 =secx+2lncosx-+cos xc
uu cscudu=-+cotuc 22
òeeduc=+ ò
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins
Calculus Cheat Sheet Calculus Cheat Sheet
Trig Substitutions : If the integral contains the following root use the given substitution and Applications of Integrals
formula to convert into an integral involving trig functions. b
Net Area : fxdx represents the net area between fx and the
() ()
ò
222 a 222 a 222 a a
a-bxxÞ=sinq bx-axÞ=secq a+bxxÞ=tanq
b b b x-axis with area above x-axis positive and area below x-axis negative.
22 22 22
cosqq=-1sin tanqq=-sec1 secqq=+1tan
16 16 2 12
Ex. dx ó cosqddqq=
ò2 ( )
2 3 2
x49-x 4 2 ò Area Between Curves : The general formulas for the two main cases for each are,
õ sinq
sinqq2cos
9 ( ) b d
22
x=sinqÞ=dxdcosqq 2 éù éù
éù éù
y=fxÞA=-upper functionlower function dx & x=fyÞA=-right functionleft function dy
33 =12cscdcqq=-+12cot () ëûëû() ëûëû
ò ò
ò a c
222
=4-4sinq==4cosqq2cos
49-x Use Right Triangle Trig to go back to x’s. From If the curves intersect then the area of each portion must be found individually. Here are some
2 substitution we have sinq = 3x so, sketches of a couple possible situations and formulas for a couple of possible cases.
Recall xx=. Because we have an indefinite 2
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,
xxif 0³ 2
ì From this we see that cotq = 49- x . So,
x=
í 3x
-0 Ex. Axis : ya=£0 Ex. Axis : ya=>0 Ex. Axis : ya=£0
2 k
ax++bxc ax++bxc
2 ( ) 2
ax++bxc ax++bxc ax++bxc
( )
2 2 2
7xx+13 7xx+13 A BxC+ Ax(++4)(Bx+-Cx)()1
Ex. dx =+=
2 222
ò x-1
(xx-+14)() (x-1)(x+4)x+4(xx-+14)()
2
7xx+13 4 3x+16 Set numerators equal and collect like terms.
2dx=+dx
2
x-1
òò
(xx-+14)() x+4 22
7x+134x=A+Bx+C-Bx+-AC
()()
4 3x 16
=++dx
x-1 22
ò Set coefficients equal to get a system and solve
xx++44
21-
3 xto get constants.
=4lnxx-1+ln++48tan ()
( )
22 radius :ay-
A+B=7C-B=1340AC-= outer radius :a-fx outer radius: a+gx radius : ay+
Here is partial fraction form and recombined. () ()
width : fy-gy
A=4BC==316 ()()
inner radius : a-gx inner radius: a+fx width : fy-gy
() () ()()
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
22
previous example : 7x+13x=Ax+41+Bx+-Cx . Chose nice values of x and plug in.
()()
() ya=£0 case with a=0. For vertical axis of rotation (xa=>0 and xa=£0) interchange x and
For example if x=1 we get 205=A which gives A=4. This won’t always work easily. y to get appropriate formulas.
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins
Calculus Cheat Sheet
Work : If a force ofFxmoves an object Average Function Value : The average value
( )
b 1b
of fx on a££xb is f= fxdx
( ) avg ò ()
ina££xb, the work done is W= Fxdx ba- a
òa()
Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are,
b b b
L= ds SA= 2pyds (rotate about x-axis) SA= 2pxds (rotate about y-axis)
òa òa òa
where ds is dependent upon the form of the function being worked with as follows.
dy 2 dx 2 dy 2
ds=1+dx if ,y=fxa££xb ds=+dt if x=ft,,y=gta££tb
( ) () () ( ) () ()
dx dt dt
dx 2 2dr 2
ds=r+dq if ,r=fqqab££
ds=1+dy if ,x=fya££yb ( ) ()
( ) () dq
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
¥ t bb
1. fxdx=lim fxdx 2. fxdx=lim fxdx
() () () ()
òò òò
aa -¥ t
t®¥ t®-¥
¥¥c
3. fxdx=+fxdxfxdx provided BOTH integrals are convergent.
() () ()
òòò
--¥¥c
Discontinuous Integrand
bb bt
1. Discont. at a: fxdx=lim fxdx 2. Discont. at b : fxdx=lim fxdx
() () () ()
òò òò
+ -
at aa
ta® tb®
bcb
3. Discontinuity at a< 0 then ¥ 1 dx converges if p >1 and diverges for p £1.
òa xp
Approximating Definite Integrals
b ba-
For given integral fxdx and a n (must be even for Simpson’s Rule) define D=x and
òa() n
divide ab, into n subintervals xx, , xx, , … , xx, with xa= and xb= then,
[ ] [ ] [ ] [ ]
01 12 n-1 n 0 n
b *
***
éù
Midpoint Rule : fxdx»Dxfx+fx++Lfx , x is midpoint xx,
() ( ) ( ) ( ) [ i ]
12ni i-1
ò
a ëû
b Dx
Trapezoid Rule : fxdx»éùfx+2fx++22fx+L++fxfx
() () () () ( ) ()
0121nn-
òa 2ëû
b Dx
Simpson’s Rule : fxdx»éùfx+4fx+2fx+L+24fx++fxfx
() () () () ( ) ( ) ()
012n--21nn
ëû
ò
a3
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins
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