Calculus Cheat Sheet 5n6y1j
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Calculus Cheat Sheet
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
“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
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
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
taking x large enough and positive. x ®-¥
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 x® a same definition as the limit except it requires negative. x
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 x® a 4. lim ê provided lim g ( x ) ¹ 0 ú= x ®a g ( x ) x® a lim g ë û x® a ( x ) n
n 5. lim ëé f ( x )ûù = é lim 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 rel="nofollow"> 0 and sgn ( a ) = - 1 if a < 0 . 1. lim e x = ¥ & 2.
lim e x = 0
x®¥
x®- ¥
lim ln ( x ) = ¥
&
x ®¥
lim- ln ( x ) = - ¥
x ®0
b =0 x ®¥ x r 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
5. n even : lim x n = ¥ x ®± ¥
6. n odd : lim x n = ¥ & lim x n = -¥ x ®¥
x ®- ¥
7. n even : lim a x n + L + b x + c = sgn ( a ) ¥ x ®± ¥
8. n odd : lim a xn + L + b x + c = sgn ( a ) ¥ x ®¥
Evaluation Techniques Continuous Functions L’Hospital’s Rule f ( x) ± ¥ f ( x) 0 If f ( x ) is continuous at a then lim f ( x ) = f ( a ) x ®a = then, If lim = or lim x® a g ( x ) x® a g ( x ) ±¥ 0 Continuous Functions and Composition f ( x) f ¢ ( x) a is a number, ¥ or -¥ = lim lim f ( x ) is continuous at b and lim g ( x ) = b then x ®a g ( x ) x ®a g ¢ ( x ) x® a Polynomials at Infinity lim f ( g ( x ) ) = f lim g ( x ) = f ( b ) x® a x® a p ( x ) and q ( x ) are polynomials. To compute Factor and Cancel p ( x) lim factor largest power of x out of both ( x - 2)( x + 6) x2 + 4 x - 12 x®±¥ q ( x ) lim = lim 2 x ®2 x 2 ® x - 2x x ( x - 2) p ( x ) and q ( x ) and then compute limit. x+6 8 = lim = =4 x ®2 x 2 3 - 42 2 x 3 - 42 3x2 - 4 3 x lim = lim 2 5 = lim 5 x = Rationalize Numerator/Denominator 2 x®-¥ 5 x - 2 x x®- ¥ x x®-¥ 2 2 2 x x 3- x 3- x 3+ x = lim 2 lim 2 Piecewise Function x®9 x - 81 x®9 x - 81 3 + x ì x 2 + 5 if x < -2 -1 9- x lim g ( x ) where g ( x ) = í = lim = lim x®-2 î1 - 3 x if x ³ -2 x ®9 ( x 2 - 81) 3 + x x®9 ( x + 9 ) 3 + x Compute two one sided limits, -1 1 lim g ( x ) = lim- x 2 + 5 = 9 = =x®-2x®-2 (18 )( 6 ) 108 lim g ( x ) = lim+ 1 - 3 x = 7 x®-2+ x®-2 Combine Rational Expressions One sided limits are different so lim g ( x ) 1æ 1 1ö 1 æ x - ( x + h) ö x®-2 - ÷ = lim çç lim ç ÷ h ®0 h x + h x ø h®0 h è x ( x + h ) ÷ø è doesn’t exist. If the two one sided limits had been equal then lim g ( x ) would have existed -1 1 æ -h ö 1 x®-2 = lim çç =- 2 ÷ = lim and had the same value. h ®0 h x ( x + h ) ÷ h®0 x ( x + h ) x è ø
(
)
(
(
(
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)
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. n 3p p p 3p 3. 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 .
9. n odd : lim a xn + L + c x + d = - sgn ( a ) ¥
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
x ®-¥
© 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
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 line to y = f ( x ) at x = a and the
change of f ( x ) at x = a . 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 .
( c f )¢ = c f ¢ ( x )
2.
(f
3.
( f g )¢ =
± g )¢ = f ¢ ( x ) ± g ¢ ( x ) f ¢ g + f g ¢ – Product Rule
æ f ö¢ f ¢ g - f g ¢ 4. ç ÷ = – Quotient Rule g2 ègø
d 5. (c) = 0 dx d n 6. ( x ) = n x n-1 – Power Rule dx d f ( g ( x) ) = f ¢ ( g ( x) ) g¢ ( x) 7. dx This is the Chain Rule
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(
(
)
(
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)
(
(
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)
Higher Order Derivatives The Second Derivative is denoted as The nth Derivative is denoted as 2 d f dn f f ¢¢ ( x ) = f ( 2) ( x ) = 2 and is defined as f ( n) ( x ) = n and is defined as dx dx ¢ ¢ n n-1 ( ) f ¢¢ ( x ) = ( f ¢ ( x ) ) , i.e. the derivative of the f ( x ) = f ( ) ( x ) , i.e. the derivative of first derivative, f ¢ ( x ) . the (n-1)st derivative, f ( n -1) x .
(
)
Implicit Differentiation Find y¢ if e2 x -9 y + x3 y 2 = sin ( y ) + 11x . 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¢ . e2 x-9 y ( 2 - 9 y¢ ) + 3x 2 y 2 + 2 x 3 y y¢ = cos ( y ) y¢ + 11 2e 2 x -9 y - 9 y¢e 2x - 9 y + 3x 2 y 2 + 2 x 3 y y¢ = cos ( y ) y¢ + 11 3
d ( csc x ) = - csc x cot x dx d ( cot x ) = - csc2 x dx d ( sin -1 x ) = 1 2 dx 1- x d -1 ( cos x ) = - 1 2 dx 1- x d 1 tan -1 x ) = ( dx 1 + x2
)
( 2 x y - 9e x
Common Derivatives d ( x) = 1 dx d ( sin x ) = cos x dx d ( cos x ) = - sin x dx d ( tan x ) = sec 2 x dx d ( sec x ) = sec x tan x dx
)
( )
Basic Properties and Formulas If f ( x ) and g ( x ) are differentiable functions (the derivative exists), c and n are any real numbers, 1.
Chain Rule Variants The chain rule applied to some specific functions. n n-1 d d 1. é f ( x )ùû = n éë f ( x ) ùû f ¢ ( x ) 5. cos éë f ( x ) ùû = - f ¢ ( x ) sin éë f ( x ) ùû dx ë dx d f ( x) d 2. e = f ¢ ( x ) e f ( x) 6. tan éë f ( x )ùû = f ¢ ( x ) sec2 éë f ( x ) ùû dx dx d f ¢( x) d 7. ( sec [ f ( x)]) = f ¢( x ) sec [ f ( x)] tan [ f ( x )] 3. ln ëé f ( x ) ûù = dx dx f ( x) f ¢( x ) d d 8. tan -1 ëé f ( x ) ûù = sin éë f ( x ) ùû = f ¢ ( x ) cos éë f ( x ) ùû 2 4. dx 1 + ëé f ( x )ûù dx
d x ( a ) = a x ln ( a ) dx d x ( e ) = ex dx d ( ln ( x ) ) = 1x , x > 0 dx d ( ln x ) = 1x , x ¹ 0 dx d ( log a ( x )) = x ln1 a , x > 0 dx
2 -9 y
- cos ( y ) ) y ¢ = 11 - 2e2 x - 9 y - 3 x 2 y 2
Þ
y¢ =
11 - 2e 2 x -9 y - 3 x 2 y 2 2 x3 y - 9e2 x -9 y - cos ( y )
Increasing/Decreasing – Concave Up/Concave Down Critical Points Concave Up/Concave Down x = c is a critical point of f ( x ) provided either 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 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
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. 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
Absolute Extrema 1. x = c is an absolute maximum of f ( x ) if f ( c ) ³ f ( x ) for all x in the domain.
Calculus Cheat Sheet
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
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?
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 ) .
x¢ is negative because x is decreasing. Using Pythagorean Theorem and differentiating, x 2 + y 2 = 15 2 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
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 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.
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 n guess for the root/solution of f ( x ) = 0 then (n+1)st guess is xn+1 = xn th
f ( xn ) f ¢ ( xn )
provided f ¢ ( xn ) exists.
© 2005 Paul Dawkins
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 By 2nd deriv. test this is a rel. max. and so is the answer where after. Finally, find x. x = 500 - 2 (125 ) = 250 The dimensions are then 250 x 125.
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
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 to have calculator in radians!
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
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) = y2 - 3 y + 3 Differentiate and find critical point(s). f ¢ = 2y - 3 Þ y = 32 nd By the 2 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 .
© 2005 Paul Dawkins
Calculus Cheat Sheet
Calculus Cheat Sheet
Integrals Definitions Definite Integral: Suppose f ( x ) is continuous Anti-Derivative : An anti-derivative of f ( x )
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
on [ a , b ] . Divide [ a , b ] into n subintervals of
is a function, F ( x ) , such that F ¢ ( x ) = f ( x ) .
u Substitution : The substitution u = g ( x ) will convert
width D x and choose x from each interval.
Indefinite Integral : ò f ( x ) dx = F ( x ) + c
du = g ¢ ( x ) dx . For indefinite integrals drop the limits of integration.
* 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
Ex.
2
ò 1 5x
u= x
3
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 éë v ( x )ùû and g ¢ ( x ) = f ( t ) dt = f ( x ) . dx ò v( x) dx ò a 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 ) then ò f ( x ) dx = F ( b ) - F ( a ) . b
2
cos ( x3 ) dx
2
ò 1 5x
Properties ò cf ( x) dx = c ò f ( x) dx , c is a constant
ò a cf ( x ) dx = c ò a f ( x ) dx , c is a constant b
ò a f ( x ) dx = 0
b
ò a f ( x ) dx = ò a f ( t ) dt
a
b
ò a f ( x) dx = - òb f ( x) dx a
b
ò f ( x ) dx £ ò b
b
a
a
If f ( x ) ³ g ( x ) on a £ x £ b then
ò f ( x) dx ³ ò g ( x ) dx
If f ( x ) ³ 0 on a £ x £ b then
f ( x ) dx ³ 0
b a
b
a
a
b
f ( x ) dx
If m £ f ( x ) £ M on a £ x £ b then m ( b - a ) £ ò f ( x ) dx £ M ( b - a ) b
a
2
+1
-1
1
1
u
1
u
8 5 1 3
cos ( u ) dx
= 53 sin ( u ) 1 = 53 ( sin (8 ) - sin (1) ) 8
1 3
x = 1 Þ u = 13 = 1 :: x = 2 Þ u = 2 3 = 8 Integration by Parts : ò u dv = uv - ò v du and
b
ò a u dv = uv
b a
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
-x
u=x
dx
dv = e
Ex. -x
Þ
du = dx v = -e
-x
-x -x -x -x -x ò xe dx = - xe + ò e dx = - xe - e + c
5
ò3 ln x dx
u = ln x
dv = dx Þ du = 1x dx v = x
ò3 ln x dx = x ln x 3 - ò3 dx = ( x ln ( x ) - x ) 3 5
5
5
5
Products and (some) Quotients of Trig Functions For ò tan n x sec m x dx we have the following : For ò sin n x cos m x dx we have the following : 1. n odd. Strip 1 sine out and convert rest to 1. cosines using sin 2 x = 1 - cos 2 x , then use the substitution u = cos x . 2. m odd. Strip 1 cosine out and convert rest 2. to sines using cos2 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 ) =
Common Integrals
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
cos ( x3 ) dx = ò
using
= 5ln ( 5 ) - 3ln ( 3) - 2
ò f ( x) ± g ( x) dx = ò f ( x) dx ± ò g ( x) dx b b b ò a f ( x ) ± g ( x ) dx = ò a f ( x ) dx ± ò a g ( x ) dx
ò
2
g b
Þ du = 3 x dx Þ x dx = du 2
a
b
b
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 = secu + 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 1 u ò a - u du = sin ( a ) + c 1
2
1
-1
2
-1
2
2
Ex.
ò tan
3
x sec5 x dx = ò ( sec2 x - 1) sec 4 x tan x sec xdx
( u = sec x )
= 17 sec 7 x - 15 sec5 x + c
2
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
© 2005 Paul Dawkins
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 (1-u ) 1 2 u + u du = -ò du = - ò u u sin 5 x
Ex.
3 5 2 4 ò tan x sec xdx = ò tan x sec 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 sec 2 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 ) )
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
3
5
3
2
4
3
2
3
2
2
3
2 2
3
2
4
3
= 12 sec 2 x + 2 ln cos x - 12 cos2 x + c © 2005 Paul Dawkins
Calculus Cheat Sheet
Calculus Cheat Sheet
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 x2 - a 2 Þ x = ab sec q
cos 2 q = 1 - sin 2 q Ex.
òx
16 2
4- 9 x2
tan 2 q = sec 2 q - 1
ó õ
dx
x = sin q Þ dx = cos q d q 2 3
a 2 + b 2 x 2 Þ x = ab tan q
2 3
sec 2 q = 1 + tan 2 q
16 4 sin 2 q ( 2cosq ) 9
( 23 cos q ) dq = ò sin122 q dq
Recall
2
= ò 12 csc dq = -12 cot q + c
2
Use Right Triangle Trig to go back to x’s. From 3x x 2 = x . Because we have an indefinite substitution we have sin q = 2 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 x =í î- x if x < 0
From this we see that cot q =
òx
4 - 9x 2 = 2 cos q .
In this case we have
16 2
4 -9 x2
dx = -
4 -9 x2 3x
4 4- 9 x 2 x
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,
2
4 - 9x = 4 - 4 sin q = 4 cos q = 2 cosq 2
Applications of Integrals b
y = f ( x) Þ A = ò
b a
é ù ë upper function û
+c
P( x )
ò Q( x) dx where the degree of P ( x ) is smaller than the degree 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.
Ex.
ò
ò
Term in P.F.D Factor in Q ( x )
ax + b
A ax + b
ax 2 + bx + c
Ax + B ax + bx + c
7 x2 +13 x ( x -1)( x 2 + 4 )
7 x2 +13 x ( x -1)( x 2 + 4 )
( ax
2
4 x -1
+ bx + c )
Term in P.F.D A1 A2 Ak + +L+ k ax + b ( ax + b )2 ( ax + b )
k
k
7 x 2 +13 x
dx
dx = ò
=ò
2
( ax + b )
Ak x + Bk A1 x + B1 +L + k ax 2 + bx + c ( ax 2 + bx + c )
( x -1)( x 2 + 4 )
4 x -1
+
+
3x x2 +4
3 x +16 x2 + 4
+
dx
16 x2 +4
dx
= 4 ln x - 1 + 32 ln ( x 2 + 4 ) + 8 tan -1 ( x2 ) Here is partial fraction form and recombined.
=
A
x -1
+C + Bx = x2 + 4
é right function ù ë û
- éëleft function ùû dy
. So,
Q ( x ) . Factor denominator as completely as possible and find the partial fraction decomposition of
Factor in Q ( x )
d 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.
A = ò f ( x ) - g ( x ) dx b
a
Partial Fractions : If integrating
- éëlower function ùû dx & x = f ( y ) Þ A = ò
A = ò f ( y ) - g ( y ) dy d
A = ò f ( x ) - g ( x ) dx + ò g ( x ) - f ( x ) dx
c
c
b
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 ) 2 - ( inner radius ) 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
inner radius : a - g ( x )
inner radius: a + f ( x )
radius : a - y width : f ( y ) - g ( y )
A( x 2 +4)+ ( Bx +C ) ( x -1) ( x -1)( x 2 + 4 )
Set numerators equal and collect like . 7 x 2 + 13 x = ( 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 + 13 x = 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. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
© 2005 Paul Dawkins
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. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
© 2005 Paul Dawkins
Calculus Cheat Sheet
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 =
in a £ x £ b , the work done is W = ò F ( x ) dx b
a
ò f ( x ) dx
b 1 b-a a
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 + ds =
dy dx dx dy
2
2
( dxdt )
( )
2
dx if y = f ( x ) , a £ x £ b
ds =
dy if x = f ( y ) , a £ y £ b
ds = r 2 + ( ddrq ) d q if r = f (q ) , a £ q £ b
2
+
dy dt
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 ¥
f ( x ) dx = lim ò f ( x ) dx t
1.
ò
3.
ò ¥ f ( x ) dx = ò ¥ f ( x ) dx + ò
t ®¥
a
2.
a
¥
c
¥
-
-
c
ò ¥ f ( x ) dx = lim ò f ( x ) dx b
b
-
t ®-¥
t
f ( x ) dx provided BOTH integrals are convergent.
Discontinuous Integrand 1. Discont. at a: ò f ( x ) dx = lim+ ò f ( x ) dx b
2. Discont. at b : ò f ( x ) dx = lim- ò f ( x ) dx
b
t ®a
a
b
t
t
t ®b
a
a
ò f ( x ) dx = ò f ( x ) dx + ò f ( x ) dx provided both are convergent.
3. Discontinuity at a < c < b :
b
c
b
a
a
c
Comparison Test for Improper Integrals : If f ( x ) ³ g ( x ) ³ 0 on [ a, ¥ ) then, 1. If ò
¥ a
¥
f ( x ) dx conv. then ò g ( x ) dx conv.
Useful fact : If a > 0 then
¥
òa
¥ a
f ( x ) dx divg.
dx converges if p > 1 and diverges for p £ 1 .
1 p
x
¥
a
2. If ò g ( x ) dx divg. then ò
a
Approximating Definite Integrals For given integral
b
òa
f ( x ) dx and a n (must be even for Simpson’s Rule) define Dx = b -n a 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 ëé f ( x ) + 2 f ( x ) + +2 f ( x ) + L + 2 f ( x ) + f ( x ) ûù b
a
0
1
2
n -1
n
Dx ò f ( x ) dx » 3 ëé f ( x ) + 4 f ( x ) + 2 f ( x ) +L + 2 f ( x ) + 4 f ( x ) + f ( x )ûù b
a
0
1
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
2
n -2
n -1
n
© 2005 Paul Dawkins
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
“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
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
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
taking x large enough and positive. x ®-¥
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 x® a same definition as the limit except it requires negative. x
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 x® a 4. lim ê provided lim g ( x ) ¹ 0 ú= x ®a g ( x ) x® a lim g ë û x® a ( x ) n
n 5. lim ëé f ( x )ûù = é lim 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 rel="nofollow"> 0 and sgn ( a ) = - 1 if a < 0 . 1. lim e x = ¥ & 2.
lim e x = 0
x®¥
x®- ¥
lim ln ( x ) = ¥
&
x ®¥
lim- ln ( x ) = - ¥
x ®0
b =0 x ®¥ x r 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
5. n even : lim x n = ¥ x ®± ¥
6. n odd : lim x n = ¥ & lim x n = -¥ x ®¥
x ®- ¥
7. n even : lim a x n + L + b x + c = sgn ( a ) ¥ x ®± ¥
8. n odd : lim a xn + L + b x + c = sgn ( a ) ¥ x ®¥
Evaluation Techniques Continuous Functions L’Hospital’s Rule f ( x) ± ¥ f ( x) 0 If f ( x ) is continuous at a then lim f ( x ) = f ( a ) x ®a = then, If lim = or lim x® a g ( x ) x® a g ( x ) ±¥ 0 Continuous Functions and Composition f ( x) f ¢ ( x) a is a number, ¥ or -¥ = lim lim f ( x ) is continuous at b and lim g ( x ) = b then x ®a g ( x ) x ®a g ¢ ( x ) x® a Polynomials at Infinity lim f ( g ( x ) ) = f lim g ( x ) = f ( b ) x® a x® a p ( x ) and q ( x ) are polynomials. To compute Factor and Cancel p ( x) lim factor largest power of x out of both ( x - 2)( x + 6) x2 + 4 x - 12 x®±¥ q ( x ) lim = lim 2 x ®2 x 2 ® x - 2x x ( x - 2) p ( x ) and q ( x ) and then compute limit. x+6 8 = lim = =4 x ®2 x 2 3 - 42 2 x 3 - 42 3x2 - 4 3 x lim = lim 2 5 = lim 5 x = Rationalize Numerator/Denominator 2 x®-¥ 5 x - 2 x x®- ¥ x x®-¥ 2 2 2 x x 3- x 3- x 3+ x = lim 2 lim 2 Piecewise Function x®9 x - 81 x®9 x - 81 3 + x ì x 2 + 5 if x < -2 -1 9- x lim g ( x ) where g ( x ) = í = lim = lim x®-2 î1 - 3 x if x ³ -2 x ®9 ( x 2 - 81) 3 + x x®9 ( x + 9 ) 3 + x Compute two one sided limits, -1 1 lim g ( x ) = lim- x 2 + 5 = 9 = =x®-2x®-2 (18 )( 6 ) 108 lim g ( x ) = lim+ 1 - 3 x = 7 x®-2+ x®-2 Combine Rational Expressions One sided limits are different so lim g ( x ) 1æ 1 1ö 1 æ x - ( x + h) ö x®-2 - ÷ = lim çç lim ç ÷ h ®0 h x + h x ø h®0 h è x ( x + h ) ÷ø è doesn’t exist. If the two one sided limits had been equal then lim g ( x ) would have existed -1 1 æ -h ö 1 x®-2 = lim çç =- 2 ÷ = lim and had the same value. h ®0 h x ( x + h ) ÷ h®0 x ( x + h ) x 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 8. tan ( x ) and sec ( x ) provided division by zero. n 3p p p 3p 3. 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 .
9. n odd : lim a xn + L + c x + d = - sgn ( a ) ¥
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
x ®-¥
© 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
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 line to y = f ( x ) at x = a and the
change of f ( x ) at x = a . 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 .
( c f )¢ = c f ¢ ( x )
2.
(f
3.
( f g )¢ =
± g )¢ = f ¢ ( x ) ± g ¢ ( x ) f ¢ g + f g ¢ – Product Rule
æ f ö¢ f ¢ g - f g ¢ 4. ç ÷ = – Quotient Rule g2 ègø
d 5. (c) = 0 dx d n 6. ( x ) = n x n-1 – Power Rule dx d f ( g ( x) ) = f ¢ ( g ( x) ) g¢ ( x) 7. dx This is the Chain Rule
(
(
)
(
(
)
(
)
)
(
(
)
)
Higher Order Derivatives The Second Derivative is denoted as The nth Derivative is denoted as 2 d f dn f f ¢¢ ( x ) = f ( 2) ( x ) = 2 and is defined as f ( n) ( x ) = n and is defined as dx dx ¢ ¢ n n-1 ( ) f ¢¢ ( x ) = ( f ¢ ( x ) ) , i.e. the derivative of the f ( x ) = f ( ) ( x ) , i.e. the derivative of first derivative, f ¢ ( x ) . the (n-1)st derivative, f ( n -1) x .
(
)
Implicit Differentiation Find y¢ if e2 x -9 y + x3 y 2 = sin ( y ) + 11x . 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¢ . e2 x-9 y ( 2 - 9 y¢ ) + 3x 2 y 2 + 2 x 3 y y¢ = cos ( y ) y¢ + 11 2e 2 x -9 y - 9 y¢e 2x - 9 y + 3x 2 y 2 + 2 x 3 y y¢ = cos ( y ) y¢ + 11 3
d ( csc x ) = - csc x cot x dx d ( cot x ) = - csc2 x dx d ( sin -1 x ) = 1 2 dx 1- x d -1 ( cos x ) = - 1 2 dx 1- x d 1 tan -1 x ) = ( dx 1 + x2
)
( 2 x y - 9e x
Common Derivatives d ( x) = 1 dx d ( sin x ) = cos x dx d ( cos x ) = - sin x dx d ( tan x ) = sec 2 x dx d ( sec x ) = sec x tan x dx
)
( )
Basic Properties and Formulas If f ( x ) and g ( x ) are differentiable functions (the derivative exists), c and n are any real numbers, 1.
Chain Rule Variants The chain rule applied to some specific functions. n n-1 d d 1. é f ( x )ùû = n éë f ( x ) ùû f ¢ ( x ) 5. cos éë f ( x ) ùû = - f ¢ ( x ) sin éë f ( x ) ùû dx ë dx d f ( x) d 2. e = f ¢ ( x ) e f ( x) 6. tan éë f ( x )ùû = f ¢ ( x ) sec2 éë f ( x ) ùû dx dx d f ¢( x) d 7. ( sec [ f ( x)]) = f ¢( x ) sec [ f ( x)] tan [ f ( x )] 3. ln ëé f ( x ) ûù = dx dx f ( x) f ¢( x ) d d 8. tan -1 ëé f ( x ) ûù = sin éë f ( x ) ùû = f ¢ ( x ) cos éë f ( x ) ùû 2 4. dx 1 + ëé f ( x )ûù dx
d x ( a ) = a x ln ( a ) dx d x ( e ) = ex dx d ( ln ( x ) ) = 1x , x > 0 dx d ( ln x ) = 1x , x ¹ 0 dx d ( log a ( x )) = x ln1 a , x > 0 dx
2 -9 y
- cos ( y ) ) y ¢ = 11 - 2e2 x - 9 y - 3 x 2 y 2
Þ
y¢ =
11 - 2e 2 x -9 y - 3 x 2 y 2 2 x3 y - 9e2 x -9 y - cos ( y )
Increasing/Decreasing – Concave Up/Concave Down Critical Points Concave Up/Concave Down x = c is a critical point of f ( x ) provided either 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 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
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. 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
Absolute Extrema 1. x = c is an absolute maximum of f ( x ) if f ( c ) ³ f ( x ) for all x in the domain.
Calculus Cheat Sheet
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
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?
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 ) .
x¢ is negative because x is decreasing. Using Pythagorean Theorem and differentiating, x 2 + y 2 = 15 2 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
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 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.
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 n guess for the root/solution of f ( x ) = 0 then (n+1)st guess is xn+1 = xn th
f ( xn ) f ¢ ( xn )
provided f ¢ ( xn ) exists.
© 2005 Paul Dawkins
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 By 2nd deriv. test this is a rel. max. and so is the answer where after. Finally, find x. x = 500 - 2 (125 ) = 250 The dimensions are then 250 x 125.
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
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 to have calculator in radians!
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
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) = y2 - 3 y + 3 Differentiate and find critical point(s). f ¢ = 2y - 3 Þ y = 32 nd By the 2 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 .
© 2005 Paul Dawkins
Calculus Cheat Sheet
Calculus Cheat Sheet
Integrals Definitions Definite Integral: Suppose f ( x ) is continuous Anti-Derivative : An anti-derivative of f ( x )
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
on [ a , b ] . Divide [ a , b ] into n subintervals of
is a function, F ( x ) , such that F ¢ ( x ) = f ( x ) .
u Substitution : The substitution u = g ( x ) will convert
width D x and choose x from each interval.
Indefinite Integral : ò f ( x ) dx = F ( x ) + c
du = g ¢ ( x ) dx . For indefinite integrals drop the limits of integration.
* 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
Ex.
2
ò 1 5x
u= x
3
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 éë v ( x )ùû and g ¢ ( x ) = f ( t ) dt = f ( x ) . dx ò v( x) dx ò a 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 ) then ò f ( x ) dx = F ( b ) - F ( a ) . b
2
cos ( x3 ) dx
2
ò 1 5x
Properties ò cf ( x) dx = c ò f ( x) dx , c is a constant
ò a cf ( x ) dx = c ò a f ( x ) dx , c is a constant b
ò a f ( x ) dx = 0
b
ò a f ( x ) dx = ò a f ( t ) dt
a
b
ò a f ( x) dx = - òb f ( x) dx a
b
ò f ( x ) dx £ ò b
b
a
a
If f ( x ) ³ g ( x ) on a £ x £ b then
ò f ( x) dx ³ ò g ( x ) dx
If f ( x ) ³ 0 on a £ x £ b then
f ( x ) dx ³ 0
b a
b
a
a
b
f ( x ) dx
If m £ f ( x ) £ M on a £ x £ b then m ( b - a ) £ ò f ( x ) dx £ M ( b - a ) b
a
2
+1
-1
1
1
u
1
u
8 5 1 3
cos ( u ) dx
= 53 sin ( u ) 1 = 53 ( sin (8 ) - sin (1) ) 8
1 3
x = 1 Þ u = 13 = 1 :: x = 2 Þ u = 2 3 = 8 Integration by Parts : ò u dv = uv - ò v du and
b
ò a u dv = uv
b a
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
-x
u=x
dx
dv = e
Ex. -x
Þ
du = dx v = -e
-x
-x -x -x -x -x ò xe dx = - xe + ò e dx = - xe - e + c
5
ò3 ln x dx
u = ln x
dv = dx Þ du = 1x dx v = x
ò3 ln x dx = x ln x 3 - ò3 dx = ( x ln ( x ) - x ) 3 5
5
5
5
Products and (some) Quotients of Trig Functions For ò tan n x sec m x dx we have the following : For ò sin n x cos m x dx we have the following : 1. n odd. Strip 1 sine out and convert rest to 1. cosines using sin 2 x = 1 - cos 2 x , then use the substitution u = cos x . 2. m odd. Strip 1 cosine out and convert rest 2. to sines using cos2 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 ) =
Common Integrals
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
cos ( x3 ) dx = ò
using
= 5ln ( 5 ) - 3ln ( 3) - 2
ò f ( x) ± g ( x) dx = ò f ( x) dx ± ò g ( x) dx b b b ò a f ( x ) ± g ( x ) dx = ò a f ( x ) dx ± ò a g ( x ) dx
ò
2
g b
Þ du = 3 x dx Þ x dx = du 2
a
b
b
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 = secu + 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 1 u ò a - u du = sin ( a ) + c 1
2
1
-1
2
-1
2
2
Ex.
ò tan
3
x sec5 x dx = ò ( sec2 x - 1) sec 4 x tan x sec xdx
( u = sec x )
= 17 sec 7 x - 15 sec5 x + c
2
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
© 2005 Paul Dawkins
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 (1-u ) 1 2 u + u du = -ò du = - ò u u sin 5 x
Ex.
3 5 2 4 ò tan x sec xdx = ò tan x sec 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 sec 2 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 ) )
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
3
5
3
2
4
3
2
3
2
2
3
2 2
3
2
4
3
= 12 sec 2 x + 2 ln cos x - 12 cos2 x + c © 2005 Paul Dawkins
Calculus Cheat Sheet
Calculus Cheat Sheet
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 x2 - a 2 Þ x = ab sec q
cos 2 q = 1 - sin 2 q Ex.
òx
16 2
4- 9 x2
tan 2 q = sec 2 q - 1
ó õ
dx
x = sin q Þ dx = cos q d q 2 3
a 2 + b 2 x 2 Þ x = ab tan q
2 3
sec 2 q = 1 + tan 2 q
16 4 sin 2 q ( 2cosq ) 9
( 23 cos q ) dq = ò sin122 q dq
Recall
2
= ò 12 csc dq = -12 cot q + c
2
Use Right Triangle Trig to go back to x’s. From 3x x 2 = x . Because we have an indefinite substitution we have sin q = 2 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 x =í î- x if x < 0
From this we see that cot q =
òx
4 - 9x 2 = 2 cos q .
In this case we have
16 2
4 -9 x2
dx = -
4 -9 x2 3x
4 4- 9 x 2 x
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,
2
4 - 9x = 4 - 4 sin q = 4 cos q = 2 cosq 2
Applications of Integrals b
y = f ( x) Þ A = ò
b a
é ù ë upper function û
+c
P( x )
ò Q( x) dx where the degree of P ( x ) is smaller than the degree 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.
Ex.
ò
ò
Term in P.F.D Factor in Q ( x )
ax + b
A ax + b
ax 2 + bx + c
Ax + B ax + bx + c
7 x2 +13 x ( x -1)( x 2 + 4 )
7 x2 +13 x ( x -1)( x 2 + 4 )
( ax
2
4 x -1
+ bx + c )
Term in P.F.D A1 A2 Ak + +L+ k ax + b ( ax + b )2 ( ax + b )
k
k
7 x 2 +13 x
dx
dx = ò
=ò
2
( ax + b )
Ak x + Bk A1 x + B1 +L + k ax 2 + bx + c ( ax 2 + bx + c )
( x -1)( x 2 + 4 )
4 x -1
+
+
3x x2 +4
3 x +16 x2 + 4
+
dx
16 x2 +4
dx
= 4 ln x - 1 + 32 ln ( x 2 + 4 ) + 8 tan -1 ( x2 ) Here is partial fraction form and recombined.
=
A
x -1
+C + Bx = x2 + 4
é right function ù ë û
- éëleft function ùû dy
. So,
Q ( x ) . Factor denominator as completely as possible and find the partial fraction decomposition of
Factor in Q ( x )
d 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.
A = ò f ( x ) - g ( x ) dx b
a
Partial Fractions : If integrating
- éëlower function ùû dx & x = f ( y ) Þ A = ò
A = ò f ( y ) - g ( y ) dy d
A = ò f ( x ) - g ( x ) dx + ò g ( x ) - f ( x ) dx
c
c
b
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 ) 2 - ( inner radius ) 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
inner radius : a - g ( x )
inner radius: a + f ( x )
radius : a - y width : f ( y ) - g ( y )
A( x 2 +4)+ ( Bx +C ) ( x -1) ( x -1)( x 2 + 4 )
Set numerators equal and collect like . 7 x 2 + 13 x = ( 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 + 13 x = 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. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
© 2005 Paul Dawkins
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. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
© 2005 Paul Dawkins
Calculus Cheat Sheet
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 =
in a £ x £ b , the work done is W = ò F ( x ) dx b
a
ò f ( x ) dx
b 1 b-a a
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 + ds =
dy dx dx dy
2
2
( dxdt )
( )
2
dx if y = f ( x ) , a £ x £ b
ds =
dy if x = f ( y ) , a £ y £ b
ds = r 2 + ( ddrq ) d q if r = f (q ) , a £ q £ b
2
+
dy dt
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 ¥
f ( x ) dx = lim ò f ( x ) dx t
1.
ò
3.
ò ¥ f ( x ) dx = ò ¥ f ( x ) dx + ò
t ®¥
a
2.
a
¥
c
¥
-
-
c
ò ¥ f ( x ) dx = lim ò f ( x ) dx b
b
-
t ®-¥
t
f ( x ) dx provided BOTH integrals are convergent.
Discontinuous Integrand 1. Discont. at a: ò f ( x ) dx = lim+ ò f ( x ) dx b
2. Discont. at b : ò f ( x ) dx = lim- ò f ( x ) dx
b
t ®a
a
b
t
t
t ®b
a
a
ò f ( x ) dx = ò f ( x ) dx + ò f ( x ) dx provided both are convergent.
3. Discontinuity at a < c < b :
b
c
b
a
a
c
Comparison Test for Improper Integrals : If f ( x ) ³ g ( x ) ³ 0 on [ a, ¥ ) then, 1. If ò
¥ a
¥
f ( x ) dx conv. then ò g ( x ) dx conv.
Useful fact : If a > 0 then
¥
òa
¥ a
f ( x ) dx divg.
dx converges if p > 1 and diverges for p £ 1 .
1 p
x
¥
a
2. If ò g ( x ) dx divg. then ò
a
Approximating Definite Integrals For given integral
b
òa
f ( x ) dx and a n (must be even for Simpson’s Rule) define Dx = b -n a 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 ëé f ( x ) + 2 f ( x ) + +2 f ( x ) + L + 2 f ( x ) + f ( x ) ûù b
a
0
1
2
n -1
n
Dx ò f ( x ) dx » 3 ëé f ( x ) + 4 f ( x ) + 2 f ( x ) +L + 2 f ( x ) + 4 f ( x ) + f ( x )ûù b
a
0
1
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.
2
n -2
n -1
n
© 2005 Paul Dawkins
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