( ) Limits

Transcription

( ) Limits

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
whenever 0 < x - a < d then f ( x ) - L < e .
can make f ( x ) as close to L as we want by
taking x large enough and positive.
“Working” Definition : We say lim f ( x ) = L
There is a similar definition for lim f ( x ) = L
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 .
except we require x large and negative.
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 hand limit : lim- f ( x ) = L . This has the
x ®a
x ®-¥
Infinite Limit : We say lim f ( x ) = ¥ if we
x ®a
can make f ( x ) arbitrarily large (and positive)
by taking x sufficiently close to a (on either side
of a) without letting x = a .
There is a similar definition for lim f ( x ) = -¥
x ®a
except we make f ( x ) arbitrarily large and
negative.
same definition as the limit except it requires
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 éëcf ( x ) ùû = c lim f ( x )
x ®a
x ®a
2. lim éë f ( x ) ± g ( x ) ùû = lim f ( x ) ± lim g ( x )
x ®a
x®a
x ®a
3. lim éë f ( x ) g ( x ) ùû = lim f ( x ) lim g ( x )
x ®a
x ®a
x ®a
f ( x)
é f ( x ) ù lim
4. lim ê
= x ®a
provided lim g ( x ) ¹ 0
ú
x ®a
x ®a g ( x )
g ( x)
ë
û lim
x ®a
n
n
5. lim éë f ( x ) ùû = élim f ( x ) ù
x ®a
ë x ®a
û
6. lim é n f ( x ) ù = n lim f ( x )
û
x ®a ë
x®a
Basic Limit Evaluations at ± ¥
Note : sgn ( a ) = 1 if a > 0 and sgn ( a ) = -1 if a < 0 .
1. lim e x = ¥ &
x®¥
2. lim ln ( x ) = ¥
x ®¥
lim e x = 0
x®- ¥
&
lim ln ( x ) = - ¥
x ®0 +
b
=0
xr
4. If r > 0 and x r is real for negative x
b
then lim r = 0
x ®-¥ x
3. If r > 0 then lim
x ®¥
5. n even : lim x n = ¥
x ®± ¥
6. n odd : lim x n = ¥ & lim x n = -¥
x ®¥
x ®- ¥
7. n even : lim a x + L + b x + c = sgn ( a ) ¥
n
x ®± ¥
8. n odd : lim a x n + L + b x + c = sgn ( a ) ¥
x ®¥
9. n odd : lim a x n + L + c x + d = - sgn ( a ) ¥
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
x ®-¥
© 2005 Paul Dawkins
Evaluation Techniques
Continuous Functions
L’Hospital’s Rule
f ( x) 0
f ( x) ± ¥
If f ( x ) is continuous at a then lim f ( x ) = f ( a )
x ®a
If lim
= or lim
=
then,
x ®a g ( x )
x ®a g ( x )
0
±¥
Continuous Functions and Composition
f ( x)
f ¢( x)
lim
=
lim
a is a number, ¥ or -¥
f ( x ) is continuous at b and lim g ( x ) = b then
x ®a g ( x )
x ®a g ¢ ( x )
(
)
x ®a
lim f ( g ( x ) ) = f lim g ( x ) = f ( b )
x ®a
x ®a
Polynomials at Infinity
p ( x ) and q ( x ) are polynomials. To compute
Factor and Cancel
( x - 2 )( x + 6 )
x 2 + 4 x - 12
lim
= lim
2
x®2
x®2
x - 2x
x ( x - 2)
p ( x)
factor largest power of x in q ( x ) out
x ®± ¥ q ( x )
lim
x+6 8
= lim
= =4
x®2
x
2
Rationalize Numerator/Denominator
3- x
3- x 3+ x
lim 2
= lim 2
x ®9 x - 81
x ®9 x - 81 3 +
x
9- x
-1
= lim
= lim
2
x ®9
( x - 81) 3 + x x®9 ( x + 9 ) 3 + x
(
)
(
of both p ( x ) and q ( x ) then compute limit.
(
(
)
Piecewise Function
)
-1
1
=(18)( 6 ) 108
Combine Rational Expressions
1æ 1
1ö
1 æ x - ( x + h) ö
lim ç
- ÷ = lim çç
÷
h ®0 h x + h
x ø h®0 h è x ( x + h ) ÷ø
è
1 æ -h ö
1
-1
= lim çç
=
lim
=
÷
h ®0 h x ( x + h ) ÷
h®0 x ( x + h )
x2
è
ø
=
)
x 2 3 - 42
3 - 42
3x 2 - 4
3
x
lim
= lim 2 5
= lim 5 x = ®¥
x ®-¥ 5 x - 2 x 2
x ®-¥ x
x
2
x -2
x -2
ì x 2 + 5 if x < -2
lim g ( x ) where g ( x ) = í
x ®-2
î1 - 3x if x ³ -2
Compute two one sided limits,
lim- g ( x ) = lim- x 2 + 5 = 9
x ®-2
x ®-2
x ®-2+
x ®-2
lim g ( x ) = lim+ 1 - 3 x = 7
One sided limits are different so lim g ( x )
x ®-2
doesn’t exist. If the two one sided limits had
been equal then lim g ( x ) would have existed
x ®-2
and had the same value.
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
8. tan ( x ) and sec ( x ) provided
division by zero.
3p p p 3p
3. n x (n odd) for all x.
x ¹ L , - , - , , ,L
2
2 2 2
4. n x (n even) for all x ³ 0 .
9. cot ( x ) and csc ( x ) provided
5. e x for all x.
x ¹ L , -2p , -p , 0, p , 2p ,L
6. ln x for x > 0 .
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
equivalent notations for the derivative.
df dy d
f ¢ ( x ) = y¢ =
=
= ( f ( x ) ) = Df ( x )
dx dx dx
If y = f ( x ) then,
If y = f ( x ) all of the following are equivalent
notations for derivative evaluated at x = a .
df
dy
f ¢ ( a ) = y ¢ x =a =
=
= Df ( a )
dx x =a dx x =a
Interpretation of the Derivative
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
time x then f ¢ ( a ) is the velocity of
equation of the tangent line at x = a is
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,
1.
( c f )¢ = c f ¢ ( x )
2.
( f ± g )¢ = f ¢ ( x ) ± g ¢ ( x )
3.
( f g )¢ =
æ f
4. ç
èg
d
(c) = 0
dx
d n
6.
x ) = n x n-1 – Power Rule
(
dx
d
7.
f ( g ( x )) = f ¢ ( g ( x )) g¢ ( x )
dx
This is the Chain Rule
5.
f ¢ g + f g ¢ – Product Rule
ö¢ f ¢ g - f g ¢
– Quotient Rule
÷ =
g2
ø
(
)
Common Derivatives
d
( x) = 1
dx
d
( sin x ) = cos x
dx
d
( cos x ) = - sin x
dx
d
( tan x ) = sec2 x
dx
d
( sec x ) = sec x tan x
dx
d
( csc x ) = - csc x cot x
dx
d
( cot x ) = - csc2 x
dx
d
1
sin -1 x ) =
(
dx
1 - x2
d
1
cos -1 x ) = (
dx
1 - x2
d
1
tan -1 x ) =
(
dx
1 + x2
d x
a ) = a x ln ( a )
(
dx
d x
e ) = ex
(
dx
d
1
ln ( x ) ) = , x > 0
(
dx
x
d
1
( ln x ) = x , x ¹ 0
dx
d
1
log a ( x ) ) =
, x>0
(
dx
x ln a
Chain Rule Variants
The chain rule applied to some specific functions.
n
n -1
d
d
1.
éë f ( x ) ùû = n éë f ( x ) ùû f ¢ ( x )
5.
cos éë f ( x ) ùû = - f ¢ ( x ) sin éë f ( x ) ùû
dx
dx
d f ( x)
d
f x
e
tan éë f ( x ) ùû = f ¢ ( x ) sec 2 éë f ( x ) ùû
2.
= f ¢( x)e ( )
6.
dx
dx
d
¢
f ( x)
d
7.
( sec [ f ( x)]) = f ¢( x) sec [ f ( x)] tan [ f ( x)]
3.
ln éë f ( x ) ùû =
dx
dx
f ( x)
f ¢( x)
d
d
tan -1 éë f ( x ) ùû =
8.
2
4.
sin éë f ( x ) ùû = f ¢ ( x ) cos éë f ( x ) ùû
dx
1 + éë f ( x ) ùû
dx
)
(
(
)
(
(
)
(
)
)
(
(
)
)
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
2 x -9 y
3 2
¢
Find y if e
+ x y = 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
2e
2 x -9 y
- 9 y¢e
( 2 x y - 9e x
3
2 x -9 y
2 -9 y
+ 3x y + 2 x y y¢ = cos ( y ) y¢ + 11
2
2
3
- cos ( y ) ) y¢ = 11 - 2e2 x -9 y - 3x 2 y 2
Þ
11 - 2e 2 x -9 y - 3x 2 y 2
y¢ = 3
2 x y - 9e2 x -9 y - cos ( y )
Increasing/Decreasing – Concave Up/Concave Down
Critical Points
x = c is a critical point of f ( x ) provided either
1. f ¢ ( c ) = 0 or 2. f ¢ ( c ) doesn’t exist.
Increasing/Decreasing
1. If f ¢ ( x ) > 0 for all x in an interval I then
f ( x ) is increasing on the interval I.
2. If f ¢ ( x ) < 0 for all x in an interval I then
f ( x ) is decreasing on the interval I.
3. If f ¢ ( x ) = 0 for all x in an interval I then
Concave Up/Concave Down
1. If f ¢¢ ( x ) > 0 for all x in an interval I then
f ( x ) is concave up on the interval I.
2. If f ¢¢ ( x ) < 0 for all x in an interval I then
f ( x ) is concave down on the interval I.
Inflection Points
x = c is a inflection point of f ( x ) if the
concavity changes at x = c .
f ( x ) is constant on the interval I.
Absolute Extrema
1. x = c is an absolute maximum of f ( x )
if f ( c ) ³ f ( x ) for all x in the domain.
Extrema
Relative (local) Extrema
1. x = c is a relative (or local) maximum of
f ( x ) if f ( c ) ³ f ( x ) for all x near c.
2. x = c is an absolute minimum of f ( x )
if f ( c ) £ f ( x ) for all x in the domain.
Fermat’s Theorem
If f ( x ) has a relative (or local) extrema at
x = c , then x = c is a critical point of f ( x ) .
Extreme Value Theorem
If f ( x ) is continuous on the closed interval
[ a, b] then there exist numbers c and d so that,
1. a £ c, d £ b , 2. f ( c ) 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.
1. Find all critical points of f ( x ) in [ a, b] .
2. Evaluate f ( x ) at all points found in Step 1.
3. Evaluate f ( a ) and f ( b ) .
4. Identify the abs. max. (largest function
value) and the abs. min.(smallest function
value) from the evaluations in Steps 2 & 3.
2. x = c is a relative (or local) minimum of
f ( x ) if f ( c ) £ f ( x ) for all x near c.
1st Derivative Test
If x = c is a critical point of f ( x ) then x = c is
1. a rel. max. of f ( x ) if f ¢ ( x ) > 0 to the left
of x = c and f ¢ ( x ) < 0 to the right of x = c .
2. a rel. min. of f ( x ) if f ¢ ( x ) < 0 to the left
of x = c and f ¢ ( x ) > 0 to the right of x = c .
3. not a relative extrema of f ( x ) if f ¢ ( x ) is
the same sign on both sides of x = c .
2nd Derivative Test
If x = c is a critical point of f ( x ) such that
f ¢ ( c ) = 0 then x = c
1. is a relative maximum of f ( x ) if f ¢¢ ( c ) < 0 .
2. is a relative minimum of f ( x ) if f ¢¢ ( c ) > 0 .
3. may be a relative maximum, relative
minimum, or neither if f ¢¢ ( c ) = 0 .
Finding Relative Extrema and/or
Classify Critical Points
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 )
then there is a number a < c < b such that f ¢ ( c ) =
f (b) - f ( a )
.
b-a
Newton’s Method
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 )
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
0.01 rad/min. At what rate is the distance
pushed towards the wall at 14 ft/sec. How fast
between them changing when q = 0.5 rad?
is the top moving after 12 sec?
x¢ is negative because x is decreasing. Using
Pythagorean Theorem and differentiating,
x 2 + y 2 = 152 Þ 2 x x¢ + 2 y y¢ = 0
After 12 sec we have x = 10 - 12 ( 14 ) = 7 and
so y = 152 - 7 2 = 176 . Plug in and solve
for y¢ .
7
7 ( - 14 ) + 176 y¢ = 0 Þ y¢ =
ft/sec
4 176
We have q ¢ = 0.01 rad/min. and want to find
x¢ . We can use various trig fcns but easiest is,
x
x¢
sec q =
Þ sec q tan q q ¢ =
50
50
We know q = 0.05 so plug in q ¢ and solve.
x¢
sec ( 0.5 ) tan ( 0.5 )( 0.01) =
50
x¢ = 0.3112 ft/sec
Remember to have calculator in radians!
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
Maximize A = xy subject to constraint of
x + 2 y = 500 . Solve constraint for x and plug
into area.
A = y ( 500 - 2 y )
x = 500 - 2 y Þ
= 500 y - 2 y 2
Differentiate and find critical point(s).
A¢ = 500 - 4 y Þ y = 125
nd
By 2 deriv. test this is a rel. max. and so is
the answer we’re after. Finally, find x.
x = 500 - 2 (125 ) = 250
The dimensions are then 250 x 125.
2
constraint is y = x 2 + 1 . Solve constraint for
x 2 and plug into the function.
2
x2 = y - 1 Þ f = x2 + ( y - 2)
= y -1 + ( y - 2) = y 2 - 3 y + 3
Differentiate and find critical point(s).
f ¢ = 2y -3
Þ y = 32
By the 2nd derivative test this is a rel. min. and
so all we need to do is find x value(s).
x 2 = 32 - 1 = 12 Þ x = ± 12
2
The 2 points are then
(
1
2
)
(
, 32 and -
1
2
)
, 32 .
Integrals
Definitions
Anti-Derivative : An anti-derivative of f ( x )
Definite Integral: Suppose f ( x ) is continuous
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
Then
¥
where F ( x ) is an anti-derivative of f ( x ) .
f (x )D x .
ò a f ( x ) dx = nlim å
i
b
®¥
=1
*
i
Fundamental Theorem of Calculus
Variants of Part I :
Part I : If f ( x ) is continuous on [ a, b ] then
d u( x)
x
f ( t ) dt = u ¢ ( x ) f éëu ( x ) ùû
g ( x ) = ò f ( t ) dt is also continuous on [ a, b ]
dx ò a
a
d b
d x
f ( t ) dt = -v¢ ( x ) f éëv ( x ) ùû
and g ¢ ( x ) =
f
t
dt
=
f
x
.
()
( )
ò
v( x )
ò
dx
a
dx
d u( x)
Part II : f ( x ) is continuous on [ a, b ] , F ( x ) is
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
ò f ( x ) ± g ( x ) dx = ò f ( x ) dx ± ò g ( x ) dx
b
b
b
f
x
±
g
x
dx
=
f
x
dx
±
(
)
(
)
(
)
òa
òa
ò a g ( x ) dx
a
Properties
ò cf ( x ) dx = c ò f ( x ) dx , c is a constant
b
b
cf
x
dx
=
c
(
)
òa
ò a f ( x ) dx , c is a constant
b
f ( x ) dx = 0
òa
b
òa
a
ò a f ( x ) dx = -òb f ( x ) dx
a
b
b
a
a
ò f ( x ) dx £ ò
If f ( x ) ³ g ( x ) on a £ x £ b then
If f ( x ) ³ 0 on a £ x £ b then
b
f ( x ) dx = ò f ( t ) dt
b
a
a
b
f ( x ) dx
ò f ( x ) dx ³ ò g ( x ) dx
b
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
ò k dx = k x + c
n
n
1
ò x dx = n+1 x + c, n ¹ -1
ò x dx = ò x dx = ln x + c
ò a x + b dx = a ln ax + b + c
ò ln u du = u ln ( u ) - u + c
ò e du = e + c
+1
-1
1
1
u
1
u
Common Integrals
ò cos u du = sin u + c
ò sin u du = - cos u + c
ò sec u du = tan u + c
ò sec u tan u du = sec u + c
ò csc u cot udu = - csc u + c
ò csc u du = - cot u + c
2
ò tan u du = ln sec u + c
ò sec u du = ln sec u + tan u + c
u
ò a + u du = a tan ( a ) + c
u
1
ò a - u du = sin ( a ) + c
1
2
1
2
-1
-1
2
2
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
u Substitution : The substitution u = g ( x ) will convert
g b
using
du = g ¢ ( x ) dx . For indefinite integrals drop the limits of integration.
Ex.
2
ò 1 5x
2
cos ( x3 ) dx
2
ò 1 5x
2
cos ( x3 ) dx = ò
u = x 3 Þ du = 3x 2 dx Þ x 2 dx = 13 du
5
cos
1 3
8
= 53 sin ( u ) 1 =
x = 1 Þ u = 1 = 1 :: x = 2 Þ u = 2 = 8
3
8
3
Integration by Parts : ò u dv = uv - ò v du and
b
ò a u dv = uv
b
a
5
3
( u ) du
( sin (8) - sin (1) )
b
- ò v du . Choose u and dv from
a
integral and compute du by differentiating u and compute v using v = ò dv .
Ex.
ò xe
u=x
ò xe
-x
-x
dx
Ex.
dv = e- x Þ
du = dx v = -e - x
dx = - xe + ò e dx = - xe - e
-x
-x
-x
-x
+c
5
ò3 ln x dx
u = ln x
5
ò3
dv = dx Þ du = 1x dx v = x
ln x dx = x ln x 3 - ò dx = ( x ln ( x ) - x )
5
5
5
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.
2
2
cosines using sin x = 1 - cos x , then use
the substitution u = cos x .
2. m odd. Strip 1 cosine out and convert rest
2.
to sines using cos 2 x = 1 - sin 2 x , then use
the substitution u = sin x .
3. n and m both odd. Use either 1. or 2.
4. n and m both even. Use double angle
3.
and/or half angle formulas to reduce the
4.
integral into a form that can be integrated.
Trig Formulas : sin ( 2 x ) = 2sin ( x ) cos ( x ) , cos 2 ( x ) =
Ex. ò tan 3 x sec5 x dx
ò tan
3
sin5 x
ò cos x dx
(sin x ) sin x
sin x
sin x sin x
ò cos x dx = ò cos x dx = ò cos x dx
(1- cos x ) sin x
=ò
dx
( u = cos x )
cos x
= - ò (1-u ) du = - ò 1-2u +u du
u
u
Ex.
x sec5 xdx = ò tan 2 x sec 4 x tan x sec xdx
= ò ( sec2 x - 1) sec 4 x tan x sec xdx
= ò ( u 2 - 1) u 4 du
n odd. Strip 1 tangent and 1 secant out and
convert the rest to secants using
tan 2 x = sec 2 x - 1 , then use the substitution
u = sec x .
m even. Strip 2 secants out and convert rest
to tangents using sec2 x = 1 + tan 2 x , then
use the substitution u = tan x .
n odd and m even. Use either 1. or 2.
n even and m odd. Each integral will be
dealt with differently.
2
1
1
2 (1 + cos ( 2 x ) ) , sin ( x ) = 2 (1 - cos ( 2 x ) )
( u = sec x )
= 17 sec7 x - 15 sec5 x + c
3
5
3
2
4
3
2
3
2
2
3
2 2
2
3
3
4
= 12 sec2 x + 2 ln cos x - 12 cos 2 x + c
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 = ab sin q
b 2 x 2 - a 2 Þ x = ba sec q
cos 2 q = 1 - sin 2 q
Ex.
òx
16
2
4 -9 x 2
tan 2 q = sec 2 q - 1
ó
õ
dx
x = 23 sin q Þ dx = 23 cos q dq
sec2 q = 1 + tan 2 q
16
4 sin 2 q ( 2cosq )
9
Recall x 2 = x . 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 q ’s and remove absolute value
bars based on that and,
ì x if x ³ 0
x =í
î- x if x < 0
Use Right Triangle Trig to go back to x’s. From
substitution we have sin q = 32x so,
From this we see that cot q =
òx
4 - 9x = 2 cos q .
2
Partial Fractions : If integrating
( 23 cos q ) dq = ò sin122 q dq
= ò 12 csc 2 dq = -12 cot q + c
2
2
4 - 9x 2 = 4 - 4sin q = 4 cos q = 2 cos q
In this case we have
a 2 + b 2 x 2 Þ x = ab tan q
16
2
4 -9 x 2
dx = - 4
4 -9 x 2
3x
4 -9 x 2
x
. So,
+c
P( x )
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 )
Ex.
ò
ò
Term in P.F.D Factor in Q ( x )
ax + b
A
ax + b
ax 2 + bx + c
Ax + B
2
ax + bx + c
( ax + b )
( ax
2
( x -1)( x
7 x2 +13 x
( x -1)( x2 + 4 )
2
Ak x + Bk
A1 x + B1
+L +
k
2
ax + bx + c
( ax 2 + bx + c )
k
7 x2 +13 x
dx
+4)
2
( x -1)( x + 4 )
dx = ò x4-1 + 3xx2++164 dx
= ò x4-1 +
Ak
A1
A2
+
+L +
2
k
ax + b ( ax + b )
( ax + b )
k
+ bx + c )
7 x2 +13 x
Term in P.F.D
3x
x2 + 4
+
16
x2 + 4
dx
= 4 ln x - 1 + 23 ln ( x 2 + 4 ) + 8 tan -1 ( x2 )
Here is partial fraction form and recombined.
=
A
x -1
+C
+ Bx
=
x2 + 4
A( x2 + 4) + ( Bx + C ) ( x -1)
( x -1)( x 2 + 4 )
Set numerators equal and collect like terms.
7 x 2 + 13x = ( A + B ) x 2 + ( C - B ) x + 4 A - C
Set coefficients equal to get a system and solve
to get constants.
A+ B = 7
C - B = 13
4A - C = 0
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
Net Area :
b
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,
y = f ( x) Þ A = ò
b
a
éupper function ù
ë
û
- éëlower
function ùû dx
& x = f ( y) Þ A = ò
d
c
é right function ù
ë
û
- éëleft
function ùû 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.
b
A = ò f ( x ) - g ( x ) dx
a
d
A = ò f ( y ) - g ( y ) dy
c
b
a
c
A = ò f ( x ) - g ( x ) dx + ò g ( x ) - f ( x ) dx
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
2
2
A = 2p ( radius ) ( width / height )
A = p ( outer radius ) - ( inner radius)
(
)
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
of f ( x ) on a £ x £ b is f avg =
b
in a £ x £ b , the work done is W = ò F ( x ) dx
a
b
1
b-a a
ò f ( x ) dx
Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are,
b
b
L = ò ds
b
SA = ò 2p y ds (rotate about x-axis)
a
SA = ò 2p x ds (rotate about y-axis)
a
a
where ds is dependent upon the form of the function being worked with as follows.
( )
1+ ( )
ds = 1 +
dy
dx
ds =
dx
dy
( dxdt )
( )
2
dx if y = f ( x ) , a £ x £ b
ds =
2
dy if x = f ( y ) , a £ y £ b
ds = r 2 + ( ddrq ) dq if r = f (q ) , a £ q £ b
2
+
dy
dt
2
dt if x = f ( t ) , y = g ( t ) , a £ t £ b
2
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
1.
3.
ò
¥
t
f ( x ) dx = lim ò f ( x ) dx
t ®¥
a
2.
a
¥
c
¥
-
-
c
ò ¥ f ( x ) dx = ò ¥ f ( x ) dx + ò
b
f ( x ) dx = lim
ò¥
-
t ®-¥
b
ò f ( x ) dx
t
f ( x ) dx provided BOTH integrals are convergent.
Discontinuous Integrand
b
b
b
1. Discont. at a: ò f ( x ) dx = lim+ ò f ( x ) dx
a
t ®a
3. Discontinuity at a < c < b :
t
2. Discont. at b : ò f ( x ) dx = lim- ò f ( x ) dx
t
a
b
c
b
a
a
c
t ®b
a
ò f ( x ) dx = ò f ( x ) dx + ò f ( x ) dx provided both are convergent.
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.
a
a
Useful fact : If a > 0 then
¥
òa
1
xp
¥
¥
a
a
2. If ò g ( x ) dx divg. then ò f ( x ) dx divg.
dx converges if p > 1 and diverges for p £ 1 .
Approximating Definite Integrals
For given integral
b
ò a f ( x ) dx and a n (must be even for Simpson’s Rule) define Dx = b-na
and
divide [ a, b] into n subintervals [ x0 , x1 ] , [ x1 , x2 ] , … , [ xn -1 , xn ] with x0 = a and xn = b then,
Midpoint Rule :
Trapezoid Rule :
Simpson’s Rule :
ò f ( x ) dx » Dx éë f ( x ) + f ( x ) + L + f ( x )ùû , xi
b
*
1
a
*
2
*
n
*
is midpoint [ xi -1 , xi ]
Dx
ò f ( x ) dx » 2 éë f ( x ) + 2 f ( x ) + +2 f ( x ) + L + 2 f ( x ) + f ( x )ùû
b
a
0
1
2
n -1
n
Dx
ò f ( x ) dx » 3 éë f ( x ) + 4 f ( x ) + 2 f ( x ) + L + 2 f ( x ) + 4 f ( x ) + f ( x )ùû
b
a
0
1
2
n-2
n -1
n

( ) Limits

Transcription

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