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/ Calculus Cheat Sheet
Limits
Definitions Precise Definition : We say lim x➔a f ( x) = L if Limit at Infinity : We say lim f ( x) = L if we 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 < xl - 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 O and sgn (a)= -1 if a< O.
- limex^ =CXJ & lim ex^ =O (^) 5. neven: lim xn= (^) OO x ➔oo x ➔ - oo x ➔ ±<
- lim ln(x)=oo & lim ln(x)=- oo x➔oo x➔ o-
- If r > O then lim �= O x➔ooxr
- If r > O andxr^ is real for negativex
then lim �= O x➔-oo (^) x'
- n odd : lim x➔oo xn =oo & (^) x➔-oolim xn = -oo
7. n even : x➔±oolim axn^ + · · · + bx + e =sgn (a) oo
- nodd: limax x➔oo n+.. ·+bx+c=^ sgn(a)oo
9. n odd : x➔-oolim axn^ + · · · + ex + d =- sgn (a) 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 ]. x^2 +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 x^ ,➔m 9 X^ i^ - 81 -^ x ,➔m 9 X^2^ - 81 3+ -y,^ X
=hm------=hm------^.^ 9-x^.^ - x ➔^9 (x^2 -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 x➔a then 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
ofboth p ( x) and q ( x) then compute limit.
2 x^2 (3 ..i.) 3 ..i. lim
3 x^ - 4 = lim x
2 = lim __ x_i^ = _i x ➔ -^00 5x^ - 2x^2 x ➔ -^00 x^2 (f-2) x➔- 00 f- 2 2
Piecewise Function
lim g(x) where g(x) ={
x^2 + x➔ -2 (^) l-3x Compute two one sided limits, lim g ( x) = lim x^2 +5 = x➔-i- x➔-2- lim g ( x) = lim 1- 3x = x➔-2• x➔-2•
ifx <- jf X:?: -
One sided limits are different so lim g ( x) x➔- doesn't exist. Ifthe two one sided limits had been equa! then lim g ( x) would have existed x➔- and had the same value.
Some Continuous Functions Partial list ofcontinuous functions and the values ofx for which they are continuous.
- Polynomials for all x. (^) 7. cos ( x) and sin ( x) for all x.
- Rational function, except for x's that give ( ) ( )
division by zero. 8.^ tan^ x^ and sec^ x^ prov1ded
- .ef; (n odd) for ali x. (^) x * ... (^) ' _ 3 7r !!.. !!_^3 n (^) ...
- .ef; (n even) for ali x:?: O. 2'^ 2' 2 ' 2'
- ex^ for ali x. 9.^ cot(x)^ and csc (x)^ provided
- 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 Chain Rule Variants The chain rule applied to some specific functions.
- (^)! ([f(x)J) = n[f (x)T-
1 J'(x) (^) 5. (^)! (^) (cos[f(x)]) =-f'(x)sin[f(x)]
! ( ef(x)) =^ J'^ (X)^ ef(x)
�(ln[f(x)])=
f' ti dx j X
! (sin[/( x)^ ]) =^ f'^ (^ x )cos[f( x^ )]
!(tan[f(x)])= f'(x)sec
2 [J(x)]
d dx
(sec^ [/Cx)]) =^ f'(x) sec^ [f (x)) tan^ [f (x))
d f'(x) -(tan-^1 [/(x)])= (^2) dx (^) l+[f(x)]
Higher Order Derivatives The Second Derivative is denoted as The nth^ Derivative is denoted as
f" ( x) = jl^2 l ( x) = :J. and is defined as jln) ( x) = � and is defined as
f" ( x) = ( /' ( x) )' , i. e. the derivative of the (^) f(n) ( x) = ( jln-i) ( x) )' , i. e. the derivative of first derivative, f' ( x)^. (^) the (n-1)51 derivative, jln-1) ( x).
Implicit Differentiation Find y' if e^2 x-^9 Y + x^3 y^2 = sin ( y) + 1 lx. Remember y = y ( x) bere, so products/quotients of x and y
will use the product/quotient rule and derivatives of y will use the chain rule. The "trick" is to differentiate 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^ (2-9y')+3x^2 y^2 +2x^3 y y' = cos(y)y' +
2e2x-^9 y -9^ y'e^2 x-^9 y + 3x^2 y^2 + 2x^3 y y' = cos (y)y' (^) + 11
(2x^3 y-9e^2 x-^9 y -cos(y))y' =ll-2e2x-9 Y^ -3x^2 y^2
⇒ '^
11-2e2x-^9 y -3x^2 y^2 y= 2x^3 y-9e^2 x-^9 y -cos(y)
Increasing/Decreasing- Concave Up/Concave Down Criticai Points
x = e is a criticai point of f ( x) provided either
1. f' (e)= O or 2. f' (e) doesn't exist.
Increasing/Decreasing
1.^ If^ f'^ (^ x)^ > O for ali^ x^ in an interval^ I^ then
f ( x)^ is increasing on the interval I.
- If /'( x) < O for ali x in an intervalJ then
f (^ x)^ is decreasing on the interval^ I.
- If f' ( x) = O for ali x in an interval I then
f ( x)^ is constant on the interval I.
Concave Up/Concave Down
- If /" ( x) > O for ali x in an interval I then f ( x)^ is concave up on the interval I.
- If /" ( x) < O for ali x in an interval I then
f ( x)^ is concave down on the interval I.
Inflection Points x = e is a inflection point of f ( x) if the concavity changes at x = e.
Calculus Cheat Sheet Extrema Absolute Extrema
- (^) x = e is an absolute maximum of f ( x)
if / (e) � f ( x) for all x in the domain.
- 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,
- 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.
- Find all criticai points of f ( x) in [a, b].
- Evaluate f ( x) at ali points found in Step 1.
- Evaluate f (a) and f (b).
- Identify the abs. max. (largest function value) and the abs. min.(smallest function value) from the evaluations in Steps 2 & 3.
Relative (locai) Extrema
- x = e is a relative (or locai) maximum of f ( x) if f (e)� f ( x) for ali x near c.
- 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
- 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.
- a rei. min. of f ( x) if f' ( x) < O to the left of x = e and f' ( x) > Oto the right of x = e.
- 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
- is a relative maximum of f ( x) if f" (e)< O.
- is a relative minimum of f ( x) if /^11 (e) > O.
- may be a relative maximum, relative minimum, or neither if /^11 (e) = O.
Finding Relative Extrema and/or Classify Criticai Points
- Find ali criticai points of f ( x).
- 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 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 off ( 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^ X
n+l + C n "F- -
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 e u^ du = e u^ +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 csc^2 u du =-cotu +c
f tan u du = lnlsecul+e
f sec u du = ln isec u + tanul + e
f-a2+u2^1 -du = ..L a^ tan-^1 (!!..)+e a
f ,.!----, du = sin-i (;)+e -.;a^2 -u^2
Calculus Cheat Sheet Standard Integration Techniques Note that at many schools ali but the Substitution Rule tend to be taught in a Calculus II class.
u Substitution : The substitution u = g (^) ( x) will convert (^) f (^) b f (^) (g (^) ( x (^) )) g' ( x)dx= f^ g(b) (^) f ( u) du using a (^) g(a) du = g' (^) ( x) dx. For indefinite integrals drop the limits of integration.
Ex. fi
2 5x^2 cos(x^3 ) dx fi
2 5x^2 cos (x^3 ) dx = fi
8 tcos( u)du
u=x^3 ⇒ du=3x^2 dx ⇒ x^2 dx=½du (^) =tsin(u)j� =t( sin(8)-sin(l)) X= 1 ⇒ u = 13 = 1 :: X= 2 ⇒ u = 2 3 = 8
Integration by Parts: f udv = uv- f vdu and f: udv = uv (^) l!-f: vdu. Choose u and dv from
integrai and compute du by differentiating u and compute v using v = fdv. ..---------------------, Ex. f xe-x dx (^) Ex. (^) s:Inxdx
u = x dv = e-x ⇒ du = dx v = -e -x (^) u = In x dv = dx ⇒ du = l. dx v = x X f xe--" dx=-xe-x^ + f^ e-x dx= -xe-x -e-x +e J:1n xdx = x lnxl:- s: dx= (xln (x)-x t
= 5 In( 5 )- 3 ln ( 3 )-
Products and (some) Quotients of Trig Functions
For f sinn^ xcos^111 xdx we have the following: For f tann^ xsec^111 xdx we have the following:
1. n odd. Strip I sine out and convert rest to 1. n odd. Strip I tangent and 1 secant out and cosines using sin 2 x = 1-cos^2 x , then use convert the rest to secants using the substitution u = cosx. tan^2 x = sec^2 x-1, then use the substitution 2. m odd. Strip 1 cosine out and convert rest u = sec x.
to sines using cos^2 x = 1- sin 2 x, then use 2.^ m^ even.^ Strip 2 secants out and convert rest the substitution u =^ sm^.^ x. to tangents usmg sec.^ i^^ x^ =^1 + tan i^^ x,^ t enh
3. n and m both odd. Use either 1. or 2. use the substitution u = tan x. 4. 11 and m both even. Use double angle 3. n odd and m even. Use either I. or 2. and/or half angle formulas to reduce the 4. n even and m odd. Each integrai will be integrai into a form that can be integrated. dealt with differently.
Trig Formulas: sin(2x) = 2 sin( x )cos( x), cos^2 (x) = ½( l +cos(2x)), sin^2 (x) = ½(l-cos(2x))
Ex. f tan^3 xsec^5 xdx
f (^) tan^3 xsec^5 xdx = f tan^2 xsec^4 xtan xsecxdx
= (^) f(sec^2 x-l)sec^4 xtanxsecxdx
= J. 7 sec 7 x - .1 5 sec^5 x + e
(u = secx)
Ex. f sin
s (^) x dx cos^3 x f sin
(^5) x dx=f sin
(^4) xsinx dx=^ f^ (sin
(^2) x) (^2) sinx cos^3 x cos^3 x cos^3 x dx = f (l-cos
(^2) x)^2 sinx dx (u = COS x) cos^3 x = (^) **- f_** (J-u^2 )^2 du =^ - f l-2u^2 +u^4 du u3 u = ½ sec^2 x + 2 In icosxi-f cos^2 x + e
..
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 :=:;; x5b , the work done is W = f: F ( x) dx
Average Function Value: The average value of (^) f (^ X )^ on a5X5b 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 ofthe 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 ifthe limit exists and has a finite value and divergent ifthe 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+ fc
00 f ( x)dx provided BOTH integrals are convergent.
Discontinuous Integrand
1. Discont. at a: f^ ab f ( x)dx = lim^ f^ b^ f ( x)dx
t➔a• l 2.
Discont. at b :^ f^ b^ f ( x) dx =^ li� f^
I a^ f^ (^ x)dx 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
For 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 ] , ••• , [ xn-l, xn ] with x 0 = a and xn = 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 (^ x 0 ) + 2f^ (^ x,) ++2/ (x 2 ) + ..^ · + 2/ ( xn_,)^ +^ f^ (^ xn )]
Simpson's Rule: f: f ( x)dx�� [f (x 0 ) + 4f ( X1) + 2/ (x 2 )+ .. · + 2/ ( xn_ 2 ) + 4/ ( xn_,)^ + f^ (^ xn)]
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