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 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 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 e = f ¢ ( x ) e f ( x) tan éë f ( x ) ùû = f ¢ ( x ) sec 2 éë f ( x ) ùû 2. 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. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.



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 provided f ¢ ( xn ) exists.


f ( xn ) f ¢ ( xn )



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 = 0.5 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. 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) = 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

®¥

* i

=1

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 ò a f ( x ) ± g ( x ) dx = ò a f ( x ) dx ± ò a g ( x ) dx

Properties

ò cf ( x ) dx = c ò f ( x ) dx , c is a constant b b ò a cf ( x ) dx = c ò a f ( x ) dx , c is a constant

a

b

ò a f ( x ) dx = 0 b

ò c dx = c ( b - a ) ò f ( x ) dx £ ò f ( x ) a

a

ò a f ( x ) dx = -òb f ( x ) dx b

c

b

a

a

c

b

b

a

a

dx

ò f ( x ) dx = ò f ( x ) dx + ò f ( x ) dx for any value of c. 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

b

b

a

a

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 © 2005 Paul Dawkins


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



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

inner radius : a - g ( x )

inner radius: a + f ( x )

radius : a - y 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. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.



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
