Calculus Cheat Sheet All

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 ®¥
for every e > 0 there is a d > 0 such that can make f ( x ) as close to L as we want by
whenever 0 < x - a < d then f ( x ) - L < e . taking x large enough and positive.

“Working” Definition : We say lim f ( x ) = L There is a similar definition for lim f ( x ) = L
x ®a x ®-¥

if we can make f ( x ) as close to L as we want except we require x large and negative.
by taking x sufficiently close to a (on either side
of a) without letting x = a . Infinite Limit : We say lim f ( x ) = ¥ if we
x ®a

can make f ( x ) arbitrarily large (and positive)
Right hand limit : lim+ f ( x ) = L . This has by taking x sufficiently close to a (on either side
x ®a
the same definition as the limit except it of a) without letting x = a .
requires x > a .
There is a similar definition for lim f ( x ) = -¥
x ®a
Left hand limit : lim- f ( x ) = L . This has the
x ®a except we make f ( x ) arbitrarily large and
same definition as the limit except it requires negative.
x<a.
Relationship between the limit and one-sided limits
lim f ( x ) = L Þ lim+ f ( x ) = lim- f ( x ) = L lim+ f ( x ) = lim- f ( x ) = L Þ lim f ( x ) = L
x ®a x ®a x ®a x ®a x ®a x ®a

lim f ( x ) ¹ lim- f ( x ) Þ lim f ( x ) Does Not Exist
x ®a + x ®a x ®a

Properties
Assume lim f ( x ) and lim g ( x ) both exist and c is any number then,
x ®a x ®a

1. lim écf ( x ) ù = c lim f ( x ) é f ( x ) ù lim f ( x )
x ®a ë û provided lim g ( x ) ¹ 0
ú=
x ®a
x ®a 4. lim ê
x ®a g ( x )
û lim g ( x )
x ®a
ë x ®a
2. lim é f ( x ) ± g ( x ) ù = lim f ( x ) ± lim g ( x )
x ®a ë û x®a n
5. lim é f ( x ) ù = élim f ( x ) ù
n
x ®a
x ®a ë û ë x ®a û
3. lim é f ( x ) g ( x ) ù = lim f ( x ) lim g ( x ) 6. lim é n f ( x ) ù = n lim f ( x )
x ®a ë û x ®a x ®a x ®a ë û x®a

Basic Limit Evaluations at ± ¥
Note : sgn ( a ) = 1 if a > 0 and sgn ( a ) = -1 if a < 0 .
1. lim e x = ¥ & lim e x = 0 5. n even : lim x n = ¥
x®¥ x®- ¥ x ®± ¥

2. lim ln ( x ) = ¥ & lim ln ( x ) = - ¥ 6. n odd : lim x n = ¥ & lim x n = -¥
x ®¥ x ®0 - x ®¥ x ®- ¥

3. If r > 0 then lim
b
=0 7. n even : lim a x + L + b x + c = sgn ( a ) ¥
n
x ®± ¥
xr
x ®¥
8. n odd : lim a x n + L + b x + c = sgn ( a ) ¥
4. If r > 0 and x r is real for negative x x ®¥
b
then lim r = 0 9. n odd : lim a x n + L + c x + d = - sgn ( a ) ¥
x ®-¥
x ®-¥ x

Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins


Evaluation Techniques
Continuous Functions L’Hospital’s Rule
If f ( x ) is continuous at a then lim f ( x ) = f ( a ) f ( x) 0 f ( x) ± ¥
x ®a If lim = or lim = then,
x ®a g ( x ) 0 x ®a g ( x ) ±¥
Continuous Functions and Composition f ( x) f ¢( x)
f ( x ) is continuous at b and lim g ( x ) = b then lim = lim a is a number, ¥ or -¥
x ®a g ( x ) x ®a g ¢ ( x )
x ®a

x ®a
( x ®a
)
lim f ( g ( x ) ) = f lim g ( x ) = f ( b ) Polynomials at Infinity
p ( x ) and q ( x ) are polynomials. To compute
Factor and Cancel
p ( x)
lim
x 2 + 4 x - 12
= lim
( x - 2 )( x + 6 ) lim
x ®± ¥ q ( x )
factor largest power of x out of both
x®2 x - 2x
2 x®2 x ( x - 2)
p ( x ) and q ( x ) and then compute limit.
x+6 8
= lim
x®2 x
= =4
Rationalize Numerator/Denominator
2
lim
3x 2 - 4
= lim 2 5
( )
x 2 3 - 42
x
3 - 42
= lim 5 x = -
3

3- x 3- x 3+ x
x ®-¥ 5 x - 2 x 2 x ®-¥ x
( )
x -2
x ®- ¥
x -2 2
lim 2 = lim 2 Piecewise Function
x ®9 x - 81 x ®9 x - 81 3 + x
ì x 2 + 5 if x < -2
= lim
9- x
= lim
-1 lim g ( x ) where g ( x ) = í
( )
( x - 81) 3 + x x®9 ( x + 9 ) 3 + x ( ) î1 - 3x if x ³ -2
x ®-2
x ®9 2

Compute two one sided limits,
-1 1 lim- g ( x ) = lim- x 2 + 5 = 9
= =-
(18)( 6 ) 108 x ®-2 x ®-2

lim g ( x ) = lim+ 1 - 3 x = 7
Combine Rational Expressions x ®-2+ x ®-2

1æ 1 1ö 1 æ x - ( x + h) ö One sided limits are different so lim g ( x )
lim ç - ÷ = lim ç ÷ x ®-2
h ®0 h x + h
è x ø h ®0 h ç x ( x + h ) ÷
è ø doesn’t exist. If the two one sided limits had
been equal then lim g ( x ) would have existed
1 æ -h ö -1 1 x ®-2
= lim ç ÷ = lim =- 2
h ®0 h ç x ( x + h ) ÷ h®0 x ( x + h ) x and had the same value.
è ø

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 ) provided
3. n x (n odd) for all x. 3p p p 3p
x ¹ L , - , - , , ,L
4. n x (n even) for all x ³ 0 . 2 2 2 2
9. cot ( x ) and csc ( x ) provided
5. e x for all x.
6. ln x for x > 0 . x ¹ L , -2p , -p , 0, p , 2p ,L

Intermediate Value Theorem
Suppose that f ( x ) is continuous on [a, b] and let M be any number between f ( a ) and f ( b ) .
Then there exists a number c such that a < c < b and f ( c ) = M .



Derivatives
Definition and Notation
f ( x + h) - f ( x)
If y = f ( x ) then the derivative is defined to be f ¢ ( x ) = lim .
h ®0 h

If y = f ( x ) then all of the following are If y = f ( x ) all of the following are equivalent
equivalent notations for the derivative. notations for derivative evaluated at x = a .
df dy d df dy
f ¢ ( x ) = y¢ = = = ( f ( x ) ) = Df ( x ) f ¢ ( a ) = y ¢ x =a = = = Df ( a )
dx dx dx dx x =a dx x =a

Interpretation of the Derivative
If y = f ( x ) then, 2. f ¢ ( a ) is the instantaneous rate of
1. m = f ¢ ( a ) is the slope of the tangent change of f ( x ) at x = a .
line to y = f ( x ) at x = a and the 3. If f ( x ) is the position of an object at
equation of the tangent line at x = a is time x then f ¢ ( a ) is the velocity of
given by y = f ( a ) + f ¢ ( a )( x - a ) . the object at x = a .

Basic Properties and Formulas
If f ( x ) and g ( x ) are differentiable functions (the derivative exists), c and n are any real numbers,
d
1. ( c f )¢ = c f ¢ ( x ) 5. (c) = 0
dx
2. ( f ± g )¢ = f ¢ ( x ) ± g ¢ ( x ) 6.
d n
( x ) = n xn-1 – Power Rule
dx
3. ( f g )¢ = f ¢ g + f g ¢ – Product Rule d
7. ( )
f ( g ( x )) = f ¢ ( g ( x )) g¢ ( x )
æ f ö¢ f ¢ g - f g ¢ dx
4. ç ÷ = – Quotient Rule This is the Chain Rule
èg ø g2

Common Derivatives
d d d x
dx
( x) = 1
dx
( csc x ) = - csc x cot x
dx
( a ) = a x ln ( a )
d d d x
dx
( sin x ) = cos x
dx
( cot x ) = - csc2 x
dx
(e ) = ex
d d
dx
( cos x ) = - sin x
dx
( sin -1 x ) = 1 2 d
dx
1
( ln ( x ) ) = x , x > 0
1- x
d d
dx
( tan x ) = sec2 x d
( cos-1 x ) = - 1 2 dx
( ln x ) = 1 , x ¹ 0
x
dx 1- x
d d 1
dx
( sec x ) = sec x tan x d 1
( tan -1 x ) = 1 + x2 dx
( log a ( x ) ) = x ln a , x > 0
dx



Chain Rule Variants
The chain rule applied to some specific functions.
1.
d
dx ë ( û
n
ë ) û
n -1
é f ( x )ù = n é f ( x )ù f ¢ ( x ) 5.
d
dx ë û ( )
cos é f ( x ) ù = - f ¢ ( x ) sin é f ( x ) ù
ë û

2.
dx
e (
d f ( x)
)
= f ¢( x)e ( )
f x
6.
d
dx ë û ( )
tan é f ( x ) ù = f ¢ ( x ) sec 2 é f ( x ) ù
ë û
f ¢( x) d
3.
d
(
ln é f ( x ) ù =
ë û ) 7. ( sec [ f ( x)]) = f ¢( x) sec [ f ( x)] tan [ f ( x)]
dx f ( x) dx
d f ¢( x)
4.
d
(ë û )
sin é f ( x ) ù = f ¢ ( x ) cos é f ( x ) ù
ë û
8. tan -1 é f ( x ) ù =
ë û ( )
1 + é f ( x )ù
2
dx dx
ë û

Higher Order Derivatives
The Second Derivative is denoted as The nth Derivative is denoted as
d2 f dn f
f ¢¢ ( x ) = f ( 2) ( x ) = 2 and is defined as f ( n ) ( x ) = n and is defined as
dx dx
f ¢¢ ( x ) = ( f ¢ ( x ) )¢ , i.e. the derivative of the ¢
( )
f ( n ) ( x ) = f ( n -1) ( x ) , i.e. the derivative of
first derivative, f ¢ ( x ) . the (n-1)st derivative, f ( n-1) x . ( )
Implicit Differentiation
Find y ¢ if e 2 x -9 y + x3 y 2 = sin ( y ) + 11x . Remember y = y ( x ) here, 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¢ .

e 2 x -9 y ( 2 - 9 y¢ ) + 3 x 2 y 2 + 2 x3 y y¢ = cos ( y ) y¢ + 11
11 - 2e 2 x -9 y - 3x 2 y 2
2e 2 x -9 y
- 9 y¢e 2 x -9 y
+ 3x y + 2 x y y¢ = cos ( y ) y¢ + 11
2 2 3
Þ y¢ = 3
2 x y - 9e2 x -9 y - cos ( y )
( 2 x y - 9e x
3 2 -9 y
- cos ( y ) ) y¢ = 11 - 2e2 x -9 y - 3x 2 y 2

Increasing/Decreasing – Concave Up/Concave Down
Critical Points
x = c is a critical point of f ( x ) provided either Concave Up/Concave Down
1. If f ¢¢ ( x ) > 0 for all x in an interval I then
1. f ¢ ( c ) = 0 or 2. f ¢ ( c ) doesn’t exist.
f ( x ) is concave up on the interval I.
Increasing/Decreasing 2. If f ¢¢ ( x ) < 0 for all x in an interval I then
1. If f ¢ ( x ) > 0 for all x in an interval I then
f ( x ) is concave down on the interval I.
f ( x ) is increasing on the interval I.
2. If f ¢ ( x ) < 0 for all x in an interval I then Inflection Points
x = c is a inflection point of f ( x ) if the
f ( x ) is decreasing on the interval I.
concavity changes at x = c .
3. If f ¢ ( x ) = 0 for all x in an interval I then
f ( x ) is constant on the interval I.



Extrema
Absolute Extrema Relative (local) Extrema
1. x = c is an absolute maximum of f ( x ) 1. x = c is a relative (or local) maximum of
if f ( c ) ³ f ( x ) for all x in the domain. f ( x ) if f ( c ) ³ f ( x ) for all x near c.
2. x = c is a relative (or local) minimum of
2. x = c is an absolute minimum of f ( x )
f ( x ) if f ( c ) £ f ( x ) for all x near c.
if f ( c ) £ f ( x ) for all x in the domain.
1st Derivative Test
Fermat’s Theorem If x = c is a critical point of f ( x ) then x = c is
If f ( x ) has a relative (or local) extrema at
1. a rel. max. of f ( x ) if f ¢ ( x ) > 0 to the left
x = c , then x = c is a critical point of f ( x ) .
of x = c and f ¢ ( x ) < 0 to the right of x = c .
Extreme Value Theorem 2. a rel. min. of f ( x ) if f ¢ ( x ) < 0 to the left
If f ( x ) is continuous on the closed interval of x = c and f ¢ ( x ) > 0 to the right of x = c .
[ a, b] then there exist numbers c and d so that, 3. not a relative extrema of f ( x ) if f ¢ ( x ) is
1. a £ c, d £ b , 2. f ( c ) is the abs. max. in the same sign on both sides of x = c .
[ a, b] , 3. f ( d ) is the abs. min. in [ a, b] . 2nd Derivative Test
If x = c is a critical point of f ( x ) such that
Finding Absolute Extrema
To find the absolute extrema of the continuous f ¢ ( c ) = 0 then x = c
function f ( x ) on the interval [ a, b ] use the 1. is a relative maximum of f ( x ) if f ¢¢ ( c ) < 0 .
following process. 2. is a relative minimum of f ( x ) if f ¢¢ ( c ) > 0 .
1. Find all critical points of f ( x ) in [ a, b] . 3. may be a relative maximum, relative
2. Evaluate f ( x ) at all points found in Step 1. minimum, or neither if f ¢¢ ( c ) = 0 .
3. Evaluate f ( a ) and f ( b ) .
4. Identify the abs. max. (largest function Finding Relative Extrema and/or
value) and the abs. min.(smallest function Classify Critical Points
value) from the evaluations in Steps 2 & 3. 1. Find all critical points of f ( x ) .
2. Use the 1st derivative test or the 2nd
derivative test on each critical 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 is a number a < c < b such that f ¢ ( c ) = .
b-a

Newton’s Method
f ( xn )
If xn is the nth guess for the root/solution of f ( x ) = 0 then (n+1)st guess is xn +1 = xn -
f ¢ ( xn )
provided f ¢ ( xn ) exists.



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
you differentiate a function of t). Plug in known quantities and solve for the unknown quantity.
Ex. A 15 foot ladder is resting against a wall. Ex. Two people are 50 ft apart when one
The bottom is initially 10 ft away and is being starts walking north. The angle q changes at
pushed towards the wall at 1 ft/sec. How fast
4
0.01 rad/min. At what rate is the distance
is the top moving after 12 sec? between them changing when q = 0.5 rad?

We have q ¢ = 0.01 rad/min. and want to find
x¢ is negative because x is decreasing. Using
x¢ . We can use various trig fcns but easiest is,
Pythagorean Theorem and differentiating,
x x¢
x 2 + y 2 = 152 Þ 2 x x¢ + 2 y y¢ = 0 sec q = Þ sec q tan q q ¢ =
50 50
After 12 sec we have x = 10 - 12 ( 1 ) = 7 and
4 We know q = 0.05 so plug in q ¢ and solve.
x¢
so y = 152 - 7 2 = 176 . Plug in and solve sec ( 0.5 ) tan ( 0.5 )( 0.01) =
for y¢ . 50
7 x¢ = 0.3112 ft/sec
7 ( - 1 ) + 176 y¢ = 0 Þ y¢ =
4
ft/sec Remember to have calculator in radians!
4 176

Optimization
Sketch picture if needed, write down equation to be optimized and constraint. Solve constraint for
one of the two variables and plug into first equation. Find critical points of equation in range of
variables and verify that they are min/max as needed.
Ex. We’re enclosing a rectangular field with Ex. Determine point(s) on y = x 2 + 1 that are
500 ft of fence material and one side of the closest to (0,2).
field is a building. Determine dimensions that
will maximize the enclosed area.

Minimize f = d 2 = ( x - 0 ) + ( y - 2 ) and the
2 2

Maximize A = xy subject to constraint of constraint is y = x 2 + 1 . Solve constraint for
x + 2 y = 500 . Solve constraint for x and plug
x 2 and plug into the function.
into area.
x2 = y - 1 Þ f = x2 + ( y - 2)
2
A = y ( 500 - 2 y )
x = 500 - 2 y Þ
= y -1 + ( y - 2) = y 2 - 3 y + 3
2
= 500 y - 2 y 2
Differentiate and find critical point(s). Differentiate and find critical point(s).
A¢ = 500 - 4 y Þ y = 125 f ¢ = 2y -3 Þ y=3 2
nd
By 2 deriv. test this is a rel. max. and so is By the 2nd derivative test this is a rel. min. and
the answer we’re after. Finally, find x. so all we need to do is find x value(s).
x = 500 - 2 (125 ) = 250 x 2 = 3 - 1 = 1 Þ x = ± 12
2 2

The dimensions are then 250 x 125. The 2 points are then ( 1
2 2 )
, 3 and - ( 1
2 )
,3 .
2



Integrals
Definitions
Definite Integral: 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 D x and choose x from each interval. Indefinite Integral : ò f ( x ) dx = F ( x ) + c
*
i
¥
where F ( x ) is an anti-derivative of f ( x ) .
ò a f ( x ) dx = n å f ( x ) D x .
b *
Then lim i
®¥
i =1

Fundamental Theorem of Calculus
Part I : If f ( x ) is continuous on [ a, b ] then Variants of Part I :
d u( x)
f ( t ) dt = u ¢ ( x ) f éu ( x ) ù
dx ò a
x
g ( x ) = ò f ( t ) dt is also continuous on [ a, b ] ë û
a
d x d b
f ( t ) dt = -v¢ ( x ) f év ( x ) ù
dx ò v( x )
and g ¢ ( x ) = òa f ( t ) dt = f ( x ) . ë û
dx
Part II : f ( x ) is continuous on [ a, b ] , F ( x ) is d u( x)
f ( t ) dt = u ¢ ( x ) f [ u ( x ) ] - v¢ ( x ) f [ v ( x ) ]
dx ò v( x )
an anti-derivative of f ( x ) (i.e. F ( x ) = ò f ( x ) dx )
b
then ò f ( x ) dx = F ( b ) - F ( a ) .
a

Properties
ò f ( x ) ± g ( x ) dx = ò f ( x ) dx ± ò g ( x ) dx ò cf ( x ) dx = c ò f ( x ) dx , c is a constant
b b b b b
òa f ( x ) ± g ( x ) dx = ò f ( x ) dx ± ò g ( x ) dx òa cf ( x ) dx = c ò f ( x ) dx , c is a constant
a a a
a b b
òa f ( x ) dx = 0 òa f ( x ) dx = ò f ( t ) dt
a
b a
ò a f ( x ) dx = -òb f ( x ) dx
b b
ò f ( x ) dx £ ò
a a
f ( x ) dx
b a
If f ( x ) ³ g ( x ) on a £ x £ b then ò f ( x ) dx ³ ò g ( x ) dx
a b
b
If f ( x ) ³ 0 on a £ x £ b then ò f ( x ) dx ³ 0
a
b
If m £ f ( x ) £ M on a £ x £ b then m ( b - a ) £ ò f ( x ) dx £ M ( b - a )
a

Common Integrals
ò k dx = k x + c ò cos u du = sin u + c ò tan u du = ln sec u + c
ò x dx = n+1 x + c, n ¹ -1 ò sin u du = - cos u + c ò sec u du = ln sec u + tan u + c
n 1 n +1

ò x dx = ò x dx = ln x + c ò sec u du = tan u + c ò a + u du = a tan ( a ) + c
-1 1 2 1 u
1 -1
2 2

ò a x + b dx = a ln ax + b + c
1 1
ò sec u tan u du = sec u + c ò a - u du = sin ( a ) + c
1
2 2
u -1

ò ln u du = u ln ( u ) - u + c ò csc u cot udu = - csc u + c
ò e du = e + c ò csc u du = - cot u + c
u u 2



Standard Integration Techniques
Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class.

( )
ò a f ( g ( x ) ) g ¢ ( x ) dx = ò g (a ) f ( u ) du
b g b
u Substitution : The substitution u = g ( x ) will convert using

du = g ¢ ( x ) dx . For indefinite integrals drop the limits of integration.

cos ( x3 ) dx cos ( x3 ) dx = ò
2 2 8
Ex. ò 1 5x
2
ò 1 5x
2 5
1 3
cos ( u ) du
u = x 3 Þ du = 3x 2 dx Þ x 2 dx = 1 du ( sin (8) - sin (1) )
8
3 = 5 sin ( u ) 1 =
3
5
3
x = 1 Þ u = 1 = 1 :: x = 2 Þ u = 2 = 8
3 3

b b b
Integration by Parts : ò u dv = uv - ò v du and ò a u dv = uv a
- ò v du . Choose u and dv from
a

integral and compute du by differentiating u and compute v using v = ò dv .

ò xe
-x 5
Ex. dx Ex. ò3 ln x dx
u=x dv = e- x Þ du = dx v = -e - x u = ln x dv = dx Þ du = 1 dx v = x
x
ò xe dx = - xe + ò e dx = - xe - e
-x -x -x -x -x
+c
ln x dx = x ln x 3 - ò dx = ( x ln ( x ) - x )
5 5 5 5
ò3 3 3

= 5ln ( 5) - 3ln ( 3) - 2

Products and (some) Quotients of Trig Functions
For ò sin n x cos m x dx we have the following : For ò tan n x sec m x dx we have the following :
1. n odd. Strip 1 sine out and convert rest to 1. n odd. Strip 1 tangent and 1 secant out and
cosines using sin x = 1 - cos x , then use
2 2 convert the rest to secants using
the substitution u = cos x . 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 = sin x . to tangents using sec2 x = 1 + tan 2 x , then
3. n and m both odd. Use either 1. or 2. use the substitution u = tan x .
4. n and m both even. Use double angle 3. n odd and m even. Use either 1. or 2.
and/or half angle formulas to reduce the 4. n even and m odd. Each integral will be
integral into a form that can be integrated. dealt with differently.
Trig Formulas : sin ( 2 x ) = 2sin ( x ) cos ( x ) , cos 2 ( x ) = 2 (1 + cos ( 2 x ) ) , sin ( x ) = 2 (1 - cos ( 2 x ) )
1 2 1

Ex. ò tan 3 x sec5 x dx Ex. sin5 x
ò cos x dx 3

ò tan x sec5 xdx = ò tan 2 x sec 4 x tan x sec xdx
3 5 4 2 2
(sin x ) sin x
ò cos x dx = ò cos x dx = ò cos x dx
sin x sin x sin x
3 3 3

= ò ( sec2 x - 1) sec 4 x tan x sec xdx (1- cos x ) sin x 2 2
=ò
cos x
dx 3 ( u = cos x )
= ò ( u 2 - 1) u 4 du ( u = sec x ) 2 2
= - ò (1-u ) du = - ò 1-2u +u du
2 4
u 3 u 3
= 1 sec7 x - 1 sec5 x + c
7 5
= 1 sec2 x + 2 ln cos x - 1 cos 2 x + c
2 2



Trig Substitutions : If the integral contains the following root use the given substitution and
formula to convert into an integral involving trig functions.
a 2 - b 2 x 2 Þ x = a sin q
b
a
b 2 x 2 - a 2 Þ x = b sec q a 2 + b 2 x 2 Þ x = a tan q
b
cos 2 q = 1 - sin 2 q tan 2 q = sec 2 q - 1 sec2 q = 1 + tan 2 q

òx ó ( 2 cos q ) dq = ò sin2 q dq
16
Ex. dx 16 12
2
4 -9 x 2 õ 4 sin 2 q ( 2cosq )
9
3

x = 2 sin q Þ dx = 2 cos q dq
3 3 = ò 12 csc 2 dq = -12 cot q + c
4 - 9x 2 = 4 - 4sin q = 4 cos q = 2 cos q
2 2
Use Right Triangle Trig to go back to x’s. From
Recall x 2 = x . Because we have an indefinite substitution we have sin q = 32x so,
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,
ì x if x ³ 0 4 -9 x 2
x =í From this we see that cot q = 3x
. So,
î- x if x < 0
4 -9 x 2
òx dx = - 4 +c
16
In this case we have 4 - 9x = 2 cos q .
2 2
4 -9 x 2 x

P( x )
Partial Fractions : If integrating ò Q( x) dx where the degree of P ( x ) is smaller than the degree of
Q ( x ) . Factor denominator as completely as possible and find the partial fraction decomposition of
the rational expression. Integrate the partial fraction decomposition (P.F.D.). For each factor in the
denominator we get term(s) in the decomposition according to the following table.

Factor in Q ( x ) Term in P.F.D Factor in Q ( x ) Term in P.F.D
A A1 A2 Ak
ax + b ( ax + b )
k
+ +L +
ax + b ax + b ( ax + b ) 2
( ax + b )
k

Ax + B A1 x + B1 Ak x + Bk
+L +
( ax + bx + c )
2 k
ax 2 + bx + c ax + bx + c ( ax 2 + bx + c )
2 k
ax + bx + c
2

7 x2 +13 x
+ Bx +C = A( x2 + 4) + ( Bx + C ) ( x -1)
Ex. ò ( x -1)( x 2
+4)
dx 7 x2 +13 x
2
( x -1)( x + 4 )
= A
x -1 x2 + 4 ( x -1)( x 2 + 4 )

7 x2 +13 x Set numerators equal and collect like terms.
ò ( x -1)( x2 + 4 )
dx = ò x4 1 + 3xx2+16 dx
- +4
7 x 2 + 13x = ( A + B ) x 2 + ( C - B ) x + 4 A - C
= ò x41 +
-
3x
x2 + 4
+ 16
x2 + 4
dx Set coefficients equal to get a system and solve
= 4 ln x - 1 + 2 ln ( x 2 + 4 ) + 8 tan -1 ( x )
3 to get constants.
2
A+ B = 7 C - B = 13 4A - C = 0
Here is partial fraction form and recombined.
A=4 B=3 C = 16

An alternate method that sometimes works to find constants. Start with setting numerators equal in
previous example : 7 x 2 + 13x = A ( x 2 + 4 ) + ( Bx + C ) ( x - 1) . Chose nice values of x and plug in.
For example if x = 1 we get 20 = 5A which gives A = 4 . This won’t always work easily.



Applications of Integrals
b
Net Area : ò a 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 general formulas for the two main cases for each are,
b d
y = f ( x) Þ A = ò éupper function ù
ë û - élower
ë function ù dx
û & x = f ( y) Þ A = ò é right function ù
ë û - éleft
ë function ù dy
û
a c
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.

d
A = ò f ( y ) - g ( y ) dy
b c b
A = ò f ( x ) - g ( x ) dx c A = ò f ( x ) - g ( x ) dx + ò g ( x ) - f ( x ) dx
a a c

Volumes of Revolution : The two main formulas are V = ò A ( x ) dx and V = ò A ( y ) dy . Here is
some general information about each method of computing and some examples.
Rings Cylinders
(
A = p ( outer radius ) - ( inner radius)
2 2
) A = 2p ( radius ) ( width / height )
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 f ( x ) , Vert. Axis use f ( y ) , Horz. Axis use f ( y ) , Vert. Axis use f ( x ) ,
g ( x ) , A ( x ) and dx. g ( y ) , A ( y ) and dy. g ( y ) , A ( y ) and dy. g ( x ) , A ( x ) and dx.

Ex. Axis : y = a > 0 Ex. Axis : y = a £ 0 Ex. Axis : y = a > 0 Ex. Axis : y = a £ 0

outer radius : a - f ( x ) outer radius: a + g ( x ) radius : a - y radius : a + y
inner radius : a - g ( x ) inner radius: a + f ( x ) width : f ( y ) - g ( 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 £ 0 case with a = 0 . For vertical axis of rotation ( x = a > 0 and x = a £ 0 ) interchange x and
y to get appropriate formulas.



Work : If a force of F ( x ) moves an object Average Function Value : The average value
b
b of f ( x ) on a £ x £ b is f avg = 1
ò f ( x ) dx
in a £ x £ b , the work done is W = ò F ( x ) dx b-a a
a

Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are,
b b b
L = ò ds SA = ò 2p y ds (rotate about x-axis) SA = ò 2p x ds (rotate about y-axis)
a a a
where ds is dependent upon the form of the function being worked with as follows.

( ) ( dx ) ( )
2 2
dx if y = f ( x ) , a £ x £ b dt if x = f ( t ) , y = g ( t ) , a £ t £ b
dy 2 dy
ds = 1 + dx
ds = dt
+ dt

1+ ( ) ds = r 2 + ( dq ) dq if r = f (q ) , a £ q £ b
2 dr 2
ds = dx
dy
dy if x = f ( y ) , a £ y £ b
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 b b
1. ò f ( x ) dx = lim ò f ( x ) dx 2. ò¥ f ( x ) dx = lim ò f ( x ) dx
a t ®¥ a - t ®-¥ t
¥ c ¥
3. ò ¥ f ( x ) dx = ò ¥ f ( x ) dx + ò
- - c
f ( x ) dx provided BOTH integrals are convergent.
Discontinuous Integrand
b b b t
1. Discont. at a: ò f ( x ) dx = lim ò f ( x ) dx
+
2. Discont. at b : ò f ( x ) dx = lim ò f ( x ) dx
-
a t ®a t a t ®b a
b c b
3. Discontinuity at a < c < b : ò f ( x ) dx = ò f ( x ) dx + ò f ( x ) dx provided both are convergent.
a a c

Comparison Test for Improper Integrals : If f ( x ) ³ g ( x ) ³ 0 on [ a, ¥ ) then,
¥ ¥ ¥ ¥
1. If ò f ( x ) dx conv. then ò g ( x ) dx conv. 2. If ò g ( x ) dx divg. then ò f ( x ) dx divg.
a a a a
¥
Useful fact : If a > 0 then òa dx converges if p > 1 and diverges for p £ 1 .
1
xp

Approximating Definite Integrals
b
For given integral ò a f ( x ) dx and a n (must be even for Simpson’s Rule) define Dx = b-a
n and

divide [ a, b] into n subintervals [ x0 , x1 ] , [ x1 , x2 ] , … , [ xn -1 , xn ] with x0 = a and xn = b then,

ò f ( x ) dx » Dx é f ( x ) + f ( x ) + L + f ( x )ù , xi is midpoint [ xi -1 , xi ]
b
* * * *
Midpoint Rule : ë 1 û 2 n
a

b Dx
Trapezoid Rule : ò f ( x ) dx » 2 é f ( x ) + 2 f ( x ) + +2 f ( x ) + L + 2 f ( x ) + f ( x )ù
a ë 0 1 û 2 n -1 n

b Dx
Simpson’s Rule : ò f ( x ) dx » 3 é f ( x ) + 4 f ( x ) + 2 f ( x ) + L + 2 f ( x ) + 4 f ( x ) + f ( x )ù
a ë 0 1 2 û n-2 n -1 n


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