calculus note

8/18/2019 Calculus Note

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Theorem 1: Intermediate Value Theorem

Definition: A function y=f(x) that is continuous on a closed interval [a,!

ta"es on every value et#een f(a) and f()$

If u is a numer et#een

f (a) and f (), f (a) % u %f

()

Then there is a c ∈ (a, ) such

that

f (c) = u$

Al&eraic 'xamle:

[ ] [ ]

*, : *, +

( )

*

y x R

f x

x

x

=

=

=

=<


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imits

Definition: a limit is the value that a function or se-uence .aroaches.

as the inut or index aroaches some value

'stimatin& imits from /rahs

imits that exist:

A) 0ontinuous unction

2)Discontinuous unction


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imits that don3t exist:

A) 4um Discontinuity

2)Involvin& Infinity


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'stimatin& imits from Tales

imits that exist:

A)rom a 0ontinuous unction


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2)rom a Discontinuous unction

imits that do not exist:

A) 4um Discontinuity

2) Infinite Discontinuity


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0)5ori6ontal Asymtote

0alculatin& imits usin& Al&era

'valuatin& imits usin& 7ustitution:

A)8olynomial

( ) ( )

9 9

lim ; 9 () ;() 9 1 x

x x x→ + − + = + − + =2)


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( ) ( )

( ) ( )

9 9 9

9 = 9 lim lim lim

+ x x x

x x x x x

x x x x→ → →

+ − + − + = = = ÷ ÷ ÷ ÷− + − + 2)2y 0on>u&ate

?

? ?

?

9 *lim

? *

9 9 ?lim @ lim

? 9 ( ?)( 9)

1 1lim

=9

x

x x

x

x

x

x x x

x x x x

x

→

→ →

→

−= ÷ ÷−

− + −= ÷ ÷ ÷

÷ ÷− + − +

= ÷+ 0) 0ommon Denominator


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1$

9

9lim

9 x

x

x x→

− ÷− −

is(A) *

(2) 1

(0)

1

+

(D)∞

(')


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( ) ( )( ) ( )

( )( )

9

+ +;lim lim lim 9

+ x x x

x x x x x x

x x x x→ → →

− + + + + − ÷ ÷= = = ÷ ÷ ÷− + − +

0ontinuity

y Definition: The roadsE meet and the rid&esE is uilt #here the

roads meet$

The families of functions that are continuous: olynomial, exonential,sine, cosine, asolute value, radical, lo&arithmic, arcFtan, arcFcotan&ent,

arcsine, arccosine

0ontinuity #ith /rahs

0ontinuity at a 8oint:

I$ The function is defined at c $ That is, c is in the domain of

definition of f(x)$

II$

lim ( ) x c

f x→

'xists$

III$

lim ( ) ( ) x c

f x f c→

=

Gemovale Discontinuity:


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4um Discontinuity:

Infinite Discontinuity:


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0ontinuity #ith Al&era

0ontinuity at a oint:

( ) 1 f x x

= + 0ontinuous H x = * c

1) * *

lim ( ) 1 lim ( ) x x

f x f x→ − → +

= =

) (*) 1 f =

9) *(*) lim ( ) x f f x→=

Gemovale Discontinuity:


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( ) ( ) 1( ) 1

x x f x x if x

x

− += = + ≠

−Gemovale disc H x= c

1)

( )

lim ( ) lim 1 9 x x

f x x→ →

= + =

)

*()

* f dne= =

4um Discontinuity:

1 *( )

1 *

if x f x

if x

≥−


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35oital3s Gule is a method for comutin& a limit of the form

( )lim

( ) x c

f x

g x→

0 can e a numer,

∞or

−∞. The conditions for alyin& it are

1$ The functions f and & are differentiale in an oen interval

containin& c

$ / and &3 are non6ero in the oen interval , exect ossily at c

9$

J( )lim

J( ) x c

f x

g x→Is defined or is

∞or is

−∞.

+$ As

x c→

'K:

sin( +)( )

x f x

x

−=

−

sin( +) *lim

* x

x

x→

−=

−

sin( +) cos( +)

lim lim + 1 x x

x x x

x→ →

− −

= =−0ontinuous H x= c

cos( +) cos( +)lim + lim

1 1 x x

x x x x

→ − → +

− −= =

() +

cos( +)() lim +

1 x

f

x x f

→

=

−= =

ultile 0hoice Buestions C

1$ Lhich of the follo#in& isare true aout the function & if

( )

( )

=

x g x

x x

−=

+ −


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I$ & is continuous at x =

II$ the &rah of & has a vertical asymtote at x=F9

III$ the &rah of & has a hori6ontal asymtote at y=*

(a) I only () II only (c) III only (d) I and II only (e) II and III only0orrect ans#er (2)

/rah of &(x)

$ et

9

? +( )

?

x x x f x

x

− − −=

−$ Lhich of the follo#in&

statements is true

(a) f(x) has a removale discontinuity at x=F9

() f(x) has a >um discontinuity at x=9

(c) if f(9) = F9, then f(x) is continuous at x=9

(d) f(x) has non removale discontinuities at x=F9 and x=9

(e)9

lim ( ) x

f x→−

= ∞

0orrect ans#er (A)

( )( M)( 9)( )

( 9)( 9)

x x x f x

x x

+ − +=

− +Therefore there is a removale

discontinuity at x=F9

Theorem : Differentiaility and 0ontinuity


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Definition: If a function is differentiale at a oint x=a, then the function

is continuous at a oint x=a

DI'G'


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,

( ),

() *Q R

lim *

1, J , H

1,

x

x x y x

x x

yCont at x continous

y

x y LHD RHD Not Differentiable x

x

→

− ≥= − = − −


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Alternate orm:

Al&eraic 'xamle:

( ) ( )

( ) ( )lim

: ( )

J( ) lim lim

J(1) (1)

x a

x a x a

f x f a

x a

Ex f x x

x a x a x a f x a

x a x a

f

→

→ →

−−

=− +−

= = =− −

= =

/rahical Illustration:


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Derivatives to emori6e:

Tri& unctions:

sin( ) cos( ) csc( ) csc( ) cot( )

cos( ) sin( ) sec( ) sec( ) tan( )

tan( ) sec ( ) cot( ) csc ( )

d d x x x x x

dx dx

d d x x x x x

dx dx

d d x x x xdx dx

= = −

= − =

= = −

Inverse Tri& unctions:


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1 1

1 1

1 1

1 1sin ( ) csc ( )

1 1

1 1cos ( ) sec ( )1 1

1 1tan ( ) cot ( )

1 1

d d x x

dx dx x x x

d d x xdx dx x x x

d d x x

dx x dx x

− −

− −

− −

= = −− −

= − =− −

= = −+ +

'xonential and o&arithmic unctions

ln( )

1 1ln lo&

ln

x x x x

a

d d e e a a a

dx dx

d d x x

dx x dx x a

= =

= =

Derivative 7tructures:8roduct Gule:

J

( )

J( ) J( ) ( )

J( )

x

x x

x x

h x xe

h x x e x e

h x e xe

=

= +

= +Buotient Gule:


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1( )

9

( 9) [ 1! ( 1) [ 9!J( )

( 9)

()( 9) ( 1)(1) MJ( )

( 9) ( 9)

x f x

x

d d x x x xdx dx f x

x

x x f x

x x

+=

−

− + − + −=

−− − + −

= =− −

8o#er Gule:

1

1

1+

J

J 1

h

h

y x

y hx

y x y x

−

=

=

==0hain Gule:

(9 1)

J (9 1)(9)

J =(9 1) 1; =

y x

y x

y x x

= +

= += + = +


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Imlicit Gule:

( ) ( )

@ @1 @ *

x y xy

dy dy y x x y x ydx dx

dy dy xy x y xy

dx dx

dy xy y

dx x xy

+ =

+ + + = ÷ ÷

+ + +

− −=

+Derivative of an Inverse (not tri&):

1( ) ( ) ( )

J(9) Q

9 M

1 1 1J( ) J(M)= ?

1J(9) = J( )

J( )

f x x g x f x

g

x x

f x f x

so g g x f x

−= + ==

= + =

= = =+

= =

ultile 0hoice Buestions C9:

1$ ind dydx

1

1

x y

x

+=

−


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(a)( )

+

1

x

x

−

− ()

( )

+

1

x

x− (c)

( )

9

+

1

x

x

−

−(d)

( )

1

x

x−(e)

( )+

1 x−

0orrect Ans#er is 2

(1 ) (1 )( )J

(1 )

x x x x y

x

− − + −=

−

$ If

( ) dy

x y x ydx

+ = −, #hat is the value of

d y

dxat the oint (9,*)

(a) F1*9 () * (c) (d) 1*9 (e) Pndefined

0orrect Ans#er is A

( )( ) ( )(1 )

( )

( )( ) ( )(1 )

( )

1*9S *

9

dy x y

dx x y

dy dy x y x y

d y dx dx

dx x y

x y x y x y x y

d y x y x y

dx x y

d ywhen x y then

dx

−=

+

+ − − − +

= +− −

+ − − − ++ +=

+

−

= = =Derivative D


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, *( )

, *

x x f x x

x x

≥= = −


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I9( ) g x x=

1I9

9

J( )

9 9 g x x

x

−= =


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Vertical Tan&ent:

9( ) f x x=

I99

1 1J( )

9 9 f x x

x−= =


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Discontinuity:

, *( )

, *

x x f x

x x

+ ≥=

− =


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GB C1:

1?? A2 9

0onsider the curve defined y

9; 1+? x xy y− + + = −

(a) ind

dy

dx() Lrite an e-uation for the line tan&ent to the curve at the oint

(+,F1)$

(c) There is a numer k so that the oint (+$, k ) is on the curve$

Psin& the tan&ent line found in art (), aroximate the

value of k

(d) Lrite an e-uation that can e solved to find the actual value

of k so that the oint (+$, k ) is on the curve

(e) 7olve the e-uation found in art (d) for the value of k $

7olution:


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(a)

1= 9 *

1=

9

dy dy x y x y

dx dx

dy x y

dx x y

− + + + =

−=

+

()(+, 1)

=+ 9 1 9( +)

* 9

dy y x

dx −

+= = + = −

+

(c)1 9(+$ +) $+ $+ y y k + = − = − ≈ −

(d)

9;(+$) (+$) 1+?k k − + + = −

(e)$9M9k = −

ultile 0hoice Buestions C+:

1$( ) sin , J( ) If f x x x then f x= + =

(a)1 cos x+

() 1 cos x−

(c)cos x

(d)sin cos x x x−

(e)sin cos x x x+

0orrect ans#er is A

J( ) sin 1 cosdy dy

f x x x xdx dx

= + = +

$ The tan&ent to the curve

? * y xy− + =is vertical #hen

a$ y=*

$ y=9 S 9−


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c$ y=1

d$ y=F9S9

e$ none of these

0orrect ans#er is 0 ? *

*

J *

y xy

dy dy y y x

dx dx

dy y

dx y x

sub x y y y

is where y is vertical y y

− + =

− + =

=+

==

−

Theorem 9: ean Value Theorem

Definition: If y = f(x) is continuous at every oint of the closed interval

[a, ! and differentiale at every oint of its interior (a, ) then there is at

least one oint c at #hich

( ) ( )J( ) f b f a f cb a

−=−

/rahically sea"in&, if A and 2 are t#o oints on a differentiale curve

then some#here et#een oints A and 2 there is at least one tan&ent line


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to the curve that is arallel to

the chord A2$

Al&eraic 'xamle:

0onsider the function( ) 9 f x x=on the interval

[1, !

( ) ( ) M 9 M1;

1 +

f b f a

b a

− −= = =

− −

Theorem +: Golle3s Theorem:

Definition: If f(x) is continuous on [a, ! and differentiale on (a, )$ If

f (a) =f () then there is at least one oint c in (a, ) #here f (c)=*

( ) ( ) ( ) ( ) * f b f a when f b f ab a

− = =−


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Al&eraic examle:

9( ) + ( ) (*) * If f x x x and f f = − − = =, find the value of c that

satisfies the conclusion of Golle3s Theorem$

J( ) *

J( ) 9 +

J( ) *

9 + * 1$1M 1$1M

f c

f x x

f x

x x and

=

= −=

− = = −

0=F1$1M c c is on the interval of [F, *!

AG0IG0

/rahical:

AG0:


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( ) ( ) f b f a RC

b a

−=

−

IG0:

( ) ( )lim x

f x h f x IRC h→∞

+ −=

Al&eraic AG0:

8olynomial:

9( ) 1 int [,9!

(9) () ; ?1?

9 9

f x x over erval

f f RC

= +− −

= = =− −


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( ) sin int [ , != +

( ) ( ) = =+ =

+ =

f x x over erval

f f RC

π π

π π

π π π

=

− −= =−

8iece#ise:

, *( ) int [ 1,1!

, *

(1) ( 1)1

1 1

x x x f x over erval

x x

f f RC

− − <= −

+ ≥− −

= =− −

7ecant:

( ) sin int ,= +

= = 1( )

=

1 = =

( ) =

f x x over erval

m f

y x

π π

π

π

π

π

=

−= =

−

− = −

Al&eraic IG0


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8olynomial:

* *

( ) ( )lim lim

J

h h

y x

x h x h x h xh h

y x

→ →

=

+ − += ==


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Tan&ent e-uation


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/rahs:

Tales:


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ind AGO0 et#een x= and x=+

(+) () +; 9*?

+

f f − −= =

−Aroximate IGO0 at x=+

() (9) 9; 9=J(+) 1

9

f f f

− −≈ = =

−

ultile 0hoice Buestions C:

1$ Lhat is the Instantaneous Gate of 0han&e of y=ln(+)

(a) * () U (c)1 (d) e (e) non existent

0orrect ans#er is 2

*ln(+ ) ln(+) 1lim

+hhh→ + − =

$

The tale &ives the values of function f that is differentiale on the

interval [*, 1!$ Lhat is the est aroximation of f3 ($)

(a) $ ()1$*; (c)$?9 (d)1$MM (e) $;;0orrect ans#er is 0

($9) ($1)J($)

$9M $1M1

f f f

−≈

−


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GB C:

**M A29

The #ind chill is the temerature, in de&rees ahrenheit (°), a human

feels ased on the air temerature, in de&rees ahrenheit, and the #indvelocity v, in miles er hour (mh)$ If the air temerature is 9°, then

the #ind chill is &iven y

$1=( ) $= $1! v v= −and is valid for

=*v≤ ≤

(a) ind L3 (*)$ Psin& correct units , exlain the meanin& of

L3(*) in terms of the #ind chill

() ind the avera&e rate of chan&e of ! over the interval =*v≤ ≤

$

ind the value of v at #hich the instantaneous rate of chan&e ofL is e-ual to the avera&e rate of chan&e of L over interval

=*v≤ ≤

(c) Over the time interval* +t ≤ ≤

hours, the air temerature is a

constant 9°$ At time t=*, the #ind velocity is v = * mh$ If

the #ind velocity increases at a constant rate of mh er hour,

#hat is the rate of chan&e of the #ind chill #ith the resect to

time at t=9 hoursQ Indicate units of measure$

7olution

(a)

$;+J(*) $1@$1= @ * $;! −= − = − #hen v=* mh, the

#ind chill is decreasin& at $;°mh

() The AG0 of L over the interval

=*v≤ ≤is

(=*) ()$9

=*

! ! −= −

−

(=*) ()J( ) 9$*11

=*

! ! ! v when v

−= =

−


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(c)

$1=( ) $= $1(* )! v t = − + 9;$?

t

d!

dt == −

°hr

The irst Derivative:

Increasin&Decreasin&:

f(x) is increasin& #hen f 3(x) *

f(x) is decreasin& #hen f 3(x) % *

axin:

If f (c) = * or D


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The 7econd Derivative:

0oncave P0oncave Do#n:

f 3(x) is increasin& or f (x) * then f(x) is concave u

f (x) is decreasin& or f (x)%* then f(x) is concave do#n

8oint of Inflection:

If f (c) = * and chan&es from W to X OG X to W at x=c , then f(c) has a

oint of inflection$

OG

f (x) has an extremum then f(x) has a oint of inflection$


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9

( ) 1 ( 9, 9)

J( ) 9 *

JJ( ) =

JJ(*) *

H * I JJ( )

f x x x on

f x x

f x x

f

#$I x b c f x sign changes from to

= − + −

= − =

==

= − +

ultile 0hoice Buestions C:

1$ The &rah of

+ 9 9 1= + +; y x x x= − + +is concave do#n for

(a)x % * ()x *(c) x % F or x F9 (d) x%9 or x (e)9 % x %

0orrect ans#er is e

+ 9 9

9 1= + +; J 1 +; +;

JJ 1(9 ; +) * JJ *

9

JJ

9

y x x x y x x x

y x x y when x and x

x is negative from to on y so y is CD

= − + + = − +

= − + = = = =

$


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The &rah of the function f is sho#n aove$ or #hich of the follo#in&

values of x if f (x) ositive and increasin&

(a)a () (c)c (d)d (e)e

e c f is increasin& thus f (x) is ositive and f is concave u thus f (x)

is increasin&

Theorem : 'xtreme Value TheoremDefinition: if a realFvalued function f is continuous in the closed and

ounded interval [a,!, then f must attain a maximum and a minimum,

each at least once$

That is, there exist numers c and d in [a,! such that


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( ) ( ) ( ) [ , ! f c f x f d for all x a b≥ ≥ ∈

Al&eraic 'xamle:

ind the in and ax values of

+ 9( ) 9 1 f x x x= − −on [F,!

+ 9

9

( ) 9 1

J( ) + ?

?*

+

f x x x

f x x x

x and x

= − −

= −

= = 2ecause x=?+ is not in the interval [Y,!,

the only critical oint occurs at x = * #hich is (*,Y1)$ The function

values at the endoints of the interval are f()=Y? and f(Y)=9?Z hence,

the maximum function value 9? at x = Y, and the minimum function

value is Y? at x = $

'xtrema

Asolute vs$ Gelative 'xtrema:

et f e a function defined on some interval and c e a numer in thatinterval$

An asolute maximum of the function f occurs at (c,f(c)) if( ) ( ) f x f c≤

for every x value in the interval$


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An asolute minimum of the function f occurs at (c,f(c)) if ( ) ( ) f x f c≥

for every x value in the interval$

The oint (c,f(c)) is a relative maximum of a function f if there exists an

oen interval (a,) in the domain of f containin& c such that( ) ( ) f x f c≤

for all x in (a,)

The oint (c,f(c)) is a relative maximum of a function f if there exists an

oen interval (a,) in the domain of f containin& c such that ( ) ( ) f x f c≥ for all x in (a,)

9 Lays to ind 'xtrema:

0omarison of unction values:


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46

9

( ) 1 [ 9,9!

J( ) 9

f x x x on

f x x

= − + −

= −

( ) $*;;M ( ) $*;;M

9 9

( 9) * (9)

f f

f f

= − − =

− = − =

As ax Value = at x =9 Gelative ax =F$*;;M at x=

9

As in value = F* at x=F9 Gelative in=$*;;M at

9−

As ax c hi&hest y value Gelative ax c f (x) &oes W to X

As ix c lo#est y value Gelative in c f (x) &oes X to W


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irst Derivative Test:

If f (c) = * or D


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%F (x)

1$ 3(c) = *

A =

− = − < = −

GB C9:


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1?;? 209

0onsider the function f is defined y( ) cos [*, ! x f x e x with the domain π =

(a) ind the asolute maximum and minimum values of f(x)() ind the intervals on #hich f is increasin&

(c) ind the xFcoordinate of each oint of inflection on &rah of f

7olution:

(a)

( ) cos [*, !

J( ) sin cos [cos sin !

J( ) * sin cos , ,+ +

x

x x x

f x e x with the domain

f x e x e x e x x

f x when x x x

π

π π

=

= − + = −= = =

ax:

e π in:

I +

e π −

()

Increasin& on

[*, !,[ , !

+ +

π π π

(c)

JJ( ) [ sin cos ! [cos sin ! sinJJ( ) * *, ,

x x x

f x e x x e x x e x f x when x

#$I at x

π π

π

= − − + − = −= =

=


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otion on a ine:

ovin& eftGi&ht:

Velocity = v (t) = s3 (t)8article ovin& Gi&ht: v (t) *

8article ovin& eft: v (t) % *

8article at Gest: v (t) = *

0han&e Direction:

The article chan&es direction #hen v (t) or s3 (t) chan&es direction from

ositive to ne&ative or ne&ative to ositive

( ) 9 1 s t t t = − −

T 7(t) V(t)=tF9

7eed=( )v t V3(t)=A(t

)

A(t)=

8article3s otion

* F1 F9 9 ovin& eftAccel$ To Gi&ht

7lo#in& Do#n

1 F+ F1 1 ovin& eft

Accel Gi&ht

7lo#in& Do#n

1$ F9$ * * 0han&es

Direction

F9 1 1 ovin& Gi&htAccel$ Gi&ht

7eedin& u

9 F1 9 9 ovin& Gi&ht

Accel$ Gi&ht


51/95

51

7eedin& u

7eed u7lo# Do#n:

A(t) % * A(t) * A(t)=*

V(t) % * 7eedin& P 7lo#in& Do#n 0onstant

Velocity eft

V(t) * 7lo#in& Do#n 7eedin& P 0onstant

Velocity Gi&ht

V(t) = * 7toed

Accel eft

7toed

Accel Gi&ht

7toed


52/95

52

Dislacement vs$ Total Distance:

9

( ) [*, !

( ) 9

( ) * $;1=+? S

(*) ($;1=+?) * () 1*

: ( ) ( )

1* ;

tan : ( ) ( )

1* 1

x t t t

v t t

v t when t changes sign

v v v

Dis%lacment s f t t f t

units dis%laced

&otal Dis ce f t t f t

units traveled

= − += −= =

= − = =

∆ = + ∆ −− + =

+ ∆ +

+ =


53/95

53

GB C+

** A2 9

A article moves alon& the xFaxis so that its velocity v at time t, for

* t ≤ ≤is &iven y

( ) ln( 9 9)v t t t = − +$ The article is at the osition

x=; at time t=*

(a) ind the acceleration of the article at time t=+

() ind all times t in the oen interval* t < <

at #hich the

article chan&es direction$ Durin& #hich time intervals , for * t ≤ ≤

does the article travel to the left

(c) ind the osition of the article at time t=

(d) ind the avera&e seed of the article over the interval* t ≤ ≤

7olution:

(a) a(+)=v (+)=

M

()

( ) ln( 9 9)

( ) *

9 9 1

( )( 1) * 1,

( ) * * 1

( ) * 1

( ) *

v t t t

v t

t t

t t t

v t for t

v t for t

v t for t

= − +=

− + =− − = =

> < <

< < <> < <

The article chan&es direction H t=1,

The article travels to the left #hen 1 % t %


54/95

54

(c)

1

*

*

( ) (*) ln( 9 9)

() ; ln( 9 9) ;$9=;

s t s u u du

s u u du

= + − +

= + − + =

∫ ∫

(d)

*

1( ) $9M*

v t dt =∫

Alications of Differential 0alculus

ineari6ation:

'-uation of tan&ent at x=a

1 1( )

( ) J( )( )

( ) J( )( )

( ) ( ) ( ) J( )( )

y y m x x

y f a f a x a

y f a f a x a

f x L x f a f a x a

− = −− = −= = −

≈ = = −

ind the ineari6ation of f(x) at x=*


55/95

55

1I

( ) 1

1 1J( ) (1 )

1

(*) 1 * 1 1

1 1J(*)

1 *

1

1 ( *)

1( ) 1

f x x

f x x x

f

f

y x

y L x x

−

= +

= + =+

= + = =

= =+

− = −

= = +

Differential:

J( )

J( )

J( )

dy f x

dx

dy f x dx

Estimated Change dy f a dx

=

==

bsolute Error ctual Estimate= −

'stimate the chan&e in the function value #hen x chan&es from to $*


56/95

56

9

9 $* $*

9 9

(9 9)

(9() 9)($*) $+

y x x dx

dy x

dxdy x dx

dy E'&I"&ED CHN(E

= − − = =

= −

= −

= − =

Actual 0han&e:

9

9

() 9()

($*) $* 9($*) $+=

$+= $+=

$+= $+ $*1

y

y

Error

= − == − =− == − =

Gelated Gates 8rolems

:

:

:

S

)ind Rate

!hen Not Rate

(iven Rate

) ( relatein an e*uation


57/95

57

1$ A hot air alloon is risin& and is trac"ed y a ran&e finder **ft$

from the lift off oint$ Lhen the ran&e finder elevation an&le is+

π

the an&le is increasin& at a rate of $1+ radians er minute$ 5o# fastis the alloon risin& at that momentQ

:

:+

: $1+ I min

dh )ind

dt

!hen

d (iven rad dt

π θ

θ

=

=

tan ** tan**

**sec

** sec ($1+) 1+*+

hh

d dh

dt dt

dh

dt

θ θ

θ θ

π

= =

= ÷

= = ÷

The alloon is risin& at a rate of 1+*ftmin #hen+

π θ =

$ Later runs into the tan", #hich is a cone #ith the radius of ft and

the hei&ht of 1*ft, at a rate of

9? I min ft $ 5o# fast is the #ater

level risin& #hen the #ater is ft deeQ


58/95

58

9

:

: =

: ? I min

dh )ind

dt

!hen h ft

dv(iven ft dt

=

=

9

1

1*

1

9

1 1

9

1

1

r r h

h

v r h

v h h

v h

π

π

π

= =

=

= ÷

=

1

+1

? (=)+

1

dv dhh

dt dt dh

dt

dh

dt

π

π

π

=

=

=

The #ater level is risin& at a rate of

1

π ftmin #hen the #ater level

is ft$

Otimi6ation 8rolems:

1$ A cylindrical tin can is to hold * cuic centimeters of >uice$

5o# should the can e constructed (dimensions: r and h) in


59/95

59

order to minimi6e the amount of material needed in its

construction$

Z + r h ' r rhπ π π = = +

*

*

+ r h r h

hr

π π

π

= =

=

in: 7urface Area

*

1**

1**J + *

1$??=

*9$??

(1$??=)

J(1$??=) *

JJ(1$??=) *

min H 1$??=

' r rh

' r r r

' r r

' r

r r

h

'EC$ND DERI+&E&E'&

'

'

r

π π

π π π

π

π

π

= + = + ÷

= +

= − =

=

= = ÷

=

>∴ =

$ Le #ant to construct a ox #hose ase len&th is 9 times the

ase #idth$ The material used to uild the to and ottom cost


60/95

60

1* I ft and the material used to uild the sides costs

= I ft $ If

the ox must have a volume of

9* ft determine the dimensions

that #ill minimi6e the cost to uild the ox$

1*( ) =( ) =* +;

** 99

C lw wh lh w wh

lwh w h hw

= + + = +

= = =


61/95

61

( )

9 9

9 9

:* ;**( ) =* +; =*

9

1* ;**J( ) 1* ;** JJ( ) 1* 1=**

;**1* ;** * 1$;;1

1*

1$;;1

9 9(1$;;1) :$=+=9:*

+$M*:*9 1$;;1

(1$;;1) [=9M$=*

C w w w ww w

wC w w w C w ww

w w

w

l w

h

C

− −

= + = + ÷

−= − = = +

− = = =

=

= = == =

=

/rah f(x) from f3(x) 8rolems:

1$


62/95

62

(a) & increasin& : &3* : (F,) , (,9$)

() & decreasin& : &3%* : (FM,F),(9$,+)(c)ax: &3=* and chan&es si&n W to F : x= 9$

(d)in: &3=* and chan&es si&n X to W : x=F

(e)0oncave u : &3 increasin& , &33*: [FM,F! [F,*! [,9!

(f) 0oncave do#n : &3 decreasin& , &33%* : [*,! [9,+!

(&)8OI: x= * , ,9

$


63/95

63

(a) f(x) increasin& : f3 * : (a,c) (e, ∞

) (aove xFaxis)

() f(x) decreasin& : f3%* (−∞

,a) (c,e) (elo# xFaxis)

(c)f(x) max : f3=* And chan&es from W to F : x=c(d)f(x) min : f3=* And chan&es from X to W : x=a,e

(e)f(x) concave u : f3 increasin& :(−∞

,),(d,∞

)

(f) f(x) concave do#n: f3 decreasin& : (,d)

(&)8OI : maxmin of f3: x= ,d

GB C


64/95

64

1?; A2

The alloon sho#n is in the shae of a cylinder #ither the hemisherical

ends of the same radius as that cylinder$ The alloon is ein& inflated at

the rate of =1π cuic centimeters er minute$ AT the instant the radius

of the cylinder is 9 centimeters$ The volume of the alloon is1++π

cuic

centimeters and the radius of the cylinder is increasin& at the rate of

centimeters er minute$ (The volume of a cylinder isr hπ

and the

volume of a shere is

9+

9r π

)

(a) At this instant, #hat is the hei&ht of the cylinder

() At this instant, ho# fast is the hei&ht of the cylinder increasin&Q

7olution:

(a)

9

9

+

9

+1++ (9) 9

91

+ r h r

h

h

π π

π π π

= +

= +

=

At this instant, the hei&ht is 1 centimeters$

()

+

=1 (9) (9)(1)() + (9) ()

d+ dh dr dr r rh r

dt dt dt dt

dh

dt

dh

dt

π π π

π π π π

= + +

= + +

=

At this instant, the hei&ht is increasin& at the rate of

centimeters er minute$


65/95

65

Theorem : undamental Theorem of 0al$:

8art 1:

Definition: If f is continuous on [a,! then the function

( ) ( ) x

a f x f t dt = ∫

has a derivative at every art in [a,! and

( ) ( )df d

f t dt f xdx dx

= =

Al&eraic 'xamle:

ind the derivative of the function of &(x)

+

9

+

( )

J( )

x

g x t dt

g x x

==∫


66/95

66

8art :

Definition: If f is continuous on [a,! and if is any antiFderivative of f

on [a,! then

( ) ( ) ( )b

a f t dt ) b ) a= −∫

Al&eraic 'xamle:

[ ]**

I

*

cos( ) sin( ) sin( ) sin(*) sin( )

cos( ) sin 1

x xt dt t x x

t dt π π

= = − =

= = ÷

∫

∫


67/95

67

[ ] **1

*

1arctan( ) arctan( )

1

1arctan(1)

1 +

x xdt t x

t

dt x

π

= =+

= =+

∫

∫

GB C

** A2 +

et f e a function that is continuous on the interval [*,+)$ The function f

is t#ice differentiale exect x= $ The function f and its derivatives

have the roerties indicated in the tale aove, #here D


68/95

68

maximum or a relative minimum at each of these values$ 4ustify

your ans#er

(d) or the function & defined in art (c) m find all values of x, for

* %x % +, at #hich the &rah of & has a oint of inflection$

4ustify your ans#er$

7olution:

(a) f has a relative maximum at x= ecause f chan&es from

ositive to ne&ative at x=

()

(c)J( ) ( ) * 1,9 g x f x at x= = =

&3 chan&es from ne&ative to ositive at x=1 so & has arelative minimum at x=1$ &3 chan&es from ositive to

ne&ative at x=9 so & has a relative maximum at x=9

(d) The &rah of & has aoint of inflection at x= ecause

&33=f3 chan&es si&n at x=


69/95

69

The Difference et#een \AntiFDerivatives:

9

+

1

+

dy x

dx

x y x c

= +

= + +

Indefinite Inte&rals:

*

9

( ) ( )

J( ) ( )

cos( ) sin( )

9

x

g x f t dt

g x f x

x dx x C

x x x x x C

=

=

= +

+ − = + − +

∫

∫

∫ Definite Inte&rals:


70/95

70

11

9 9

9

J( ) ( )

( ) ( ) ( ) ( )

+= + = ;+

b b

aa

) x f x

f x dx ) x ) b ) a

x x dx x x x−−

=

= = −

− + = − + = − + − − − − = ÷ ÷ ÷

∫

∫ Accumulation unction:

*( ) ( )

() +

t

s t f x dx

s

=

=

∫

ultile 0hoice Buestions CM:

1$

;

* 1

dx

x+∫

(a)1()9

(c)

(d)+

(e)

0orrect ans#er is D

;;

* * 1 = +

1

dx x

x

= + = − =

+

∫

$

1

* 1 x x dx is− +∫

(a)F1

()F1


71/95


72/95

72

1 1

1 1

1 1

1 1sin ( ) csc ( )

1 1

1 1cos ( ) sec ( )1 1

1 1tan ( ) cot ( )

1 1

x C dx x C dx x x x

x C dx x dx x x x

x C dx x dx x x

− −

− −

− −

+ = + = −− −

+ = − =− −

+ = = −+ +

∫ ∫

∫ ∫

∫ ∫ 'xonential and o&arithmic unctions:

ln( )

1 1ln lo&

ln

x x x x

a

e C e dx a C a a dx

x C dx x C dx x x a

+ = + =

+ = + =

∫ ∫ ∫ ∫

8o#er Gule:

Lhat is

9 x dx∫ Q

Le can use the 8o#er Gule, #here n=9

1

+9

( 1)

+

nn x

x dx C n

x x dx C

+

= ++

= +

∫ ∫


73/95

73

Lhat is

x dx∫ Q

Le can use the 8o#er Gule, #here n=1

1

1$$

( 1)

1$

nn x x dx C

n

x x dx C

+

= ++

= +

∫

∫ PF7ustitution:

Inte&rate( ) ( )

M x x x dx+ +∫

( ) ( ) ( )

;;

M M

:

( :)

: : :

; ;

u x x

du x dx

x xu x x x dx u du C C

= +

= +

++ + = = + = +∫ ∫

ultile 0hoice Buestions C;:

1$

sec tan x xdx =

∫ (a)

sec x C +

()tan x C +


74/95

74

(c)

tan

xC +

(d)

sec

x

C +

(e)

sec tan

x xC +

0orrect ans#er is A

sec tan sec

sec sec tan

x xdx x C

dy x x x

dx

= +

=

∫

$

;

* 1

dx

x=

+∫ (a)1

()9

(c)(d)+

(e)

0orrect ans#er is D

?; ?1I 1I

* 1 1 +

1

1 (;) ? (*) 1

dxu u

x

u x du dx u u

− = = = +

= + = = =

∫ ∫

8roerties of Definite Inte&rals:


75/95

75

1$ $ Le can interchan&e the limits on any

definite inte&ral, all that #e need to do is tac" a minus si&n onto the

inte&ral #hen #e do$

$ $ If the uer and lo#er limits are the same then

there is no #or" to do, the inte&ral is 6ero$

9$ , #here c is any numer$ 7o, as #ith

limits, derivatives, and indefinite inte&rals #e can factor out a

constant$

+$ $ Le can rea"

u definite inte&rals across a sum or difference$

$ #here c is any numer$

This roerty is more imortant than #e mi&ht reali6e at first$ One of

the main uses of this roerty is to tell us ho# #e can inte&rate a

function over the ad>acent intervals, [a,c! and [c,b!$


76/95

76

1$

1

* 1 x x dx is− +∫

(a) F1

() F1(c) ]

(d) 1

(e)


77/95

77

1 1 1sin( 9) sin cos cos( 9)

1

9

x dx udu u C x

u x du dx du dx

+ = = − + = − + +

= + = =

∫ ∫

Arox$ Areas: GA, GGA, GA

rom '-uation

The velocity of an o>ect in motion is modeled y the function

( ) +v t t t = −$ ind the aroximate total distance traveled in the first +

seconds usin& Giemann 7ums #ith + suintervals$

GA:

[ ] [ ]

+ *1

+

(*) (1) () (9) 1 * 9 + 9 1*

x

LR" x v v v v

−∆ = =

∆ + + + = + + + =


78/95

78

GGA:

[ ] [ ]

+ *1

+

(1) () (9) (+) 1 9 + 9 * 1*

x

RR" x v v v v

−∆ = =

∆ + + + = + + + =

GA:

[ ] [ ]

+ *1

+

($) (1$) ($) (9$) 1 1$M 9$M 9$M 1$M 11

-

"R" - v v v v

−= =

+ + + = + + + =


79/95

79

rom Tale:

The Tale elo# sho#s the velocity of a model train en&ine movin&

alon& a trac" for 1* seconds$ 'stimate the distance traveled y the

en&ine usin& 1* suintervals of the len&th 1 usin& GA and GGA

Time (sec) Velocity (insec) Time

(sec)

Velocity

(insec)

* * 11

1 1 M

;

9 1* ?

+ 1* *

1

[ ] [ ]

[ ] [ ]

1

(*) (1) ()$$$ (?) 1 * 1 1* : 1 11 = = ;M

(1) () (9)$$$ (1*) 1 1 1* : 1 11 = = * ;M

t

LR" t v v v v in

RR" t v v v v in

∆ =

= ∆ + + + = + + + + + + + + + =

= ∆ + + + = + + + + + + + + + =


80/95

80

[ ] [ ]

(*) () (+)$$$ (1*) * : 11 * ;*

t

"R" t v v v v in

∆ =

= ∆ + + + = + + + + + =

rom /rah:

[ ]( ) 1 *, f x x on= +

To estimate, divide u intosuintervals

et3s use + intervals

GGA:


81/95

81

GGA uses the ri&ht

endoint of each interval to

define the hei&ht of the

rectan&le

The #idth of each rectan&le

is ]

GA:

GA uses the left endoint

of each su interval to define

the hei&ht of the rectan&le$


is ]


82/95

82

GA:

GA uses the midoint of

each su interval to define

the hei&ht of the rectan&le


is ]

ore rectan&les = etter estimate

Trae6oidal Arox$:

rom '-uation:

4ust li"e in the other rules #e rea" u the interval [a,!

into n suintervals of #idth


83/95

83

Psin& n=+ aroximate the value of the follo#in& inte&ral:

*

xe dx∫

( )

*

* $ 1 1$

*

* 1

+

1I *$=++?*

x

x

e dx

x

e dx e e e e e

−∆ = =

≈ + + + + =

∫

∫

rom Tale

T (sec) * 1 9* 9 * *

V(t)

(ftsec)

* 9* * 1+ 1* * 1*

A car travels on a strai&ht trac"$ Durin& the time interval * * ^ ^t

seconds, the car3s velocity v, measured in feet er second is a continuous

function$

Aroximate

=*

9*( )v t dt ∫

usin& a trae6oidal aroximation #ith 9

suintervals determined y the tale


84/95

84

( ) ( ) ( ) ( ) ( )

=*

9*( )

1 1 11+ 1* 1* 1 1* 1* 1;

v t dt

ft = + + + =

∫

otion on a ine usin& Inte&ration

8osition:

J( ) ( )

( ) ( )

x t v t so

v t dt x t C

=

= +∫

Dislacement vs$ Total Distance:

Dislacement:

( )b

av t dt ∫

Total Distance:


85/95

85

( )b

av t dt ∫

ind the total distance traveled y a ody and the odyJs dislacement

for a ody #hose velocity is v (t) = sin 9t on the time interval

*

t π ≤ ≤

Dislacement:

I I

**=sin 9 cos 9 cos 9( ) cos 9(*)

t dt t

π π π

= − = − − − =∫

Total Distance:

I9 I I9 I

* I9* I9

I

*

=sin 9 * *,9

(*, ) ( ) * ( , ) ( ) *9 9

=sin 9 = sin 9 cos 9 cos 9 =

= sin 9 =

t t

v t v t

t dt t dt t t

$R

t dt

π π π π

π π

π

π

π π π

= =

> <

− = − + =

=

∫ ∫

∫

Differential '-uations:

Definition: A differential e-uation is the relationshi satisfied y the

function and its derivative$

7olvin& Differential '-uations:1$ ind the articular solution to y=f(x) to the differential e-uation

1dy y

dx x

−=

#ith the initial condition f()=*


86/95

86

( )

1

1 1

1

1 1

1

1ln 1

1

1

( ) 1 , *

C c C x x

x

dy dx

y x

x y C C

x

y e e e k e

k e

f x e x

−

− −+

− ÷

=−

−− = + = +

−− = = = ±

= −

= − >

∫ ∫

$ ind the articular solution to y=&(x) to the differential e-uation

dy xydx

−=

#ith the initial condition f(F1)=

1

+

1 1 1Z

+ +

+

1

dy xdx

y

xC

y

C C

y x

= −

− = − +

− = − + = −

=+

∫ ∫

Dra#in& a 7loe ield


87/95

87

Analysis of a 7loe ield

indin& the solution at (1, 1)

ultile 0hoice Buestions C1*:


88/95

88

1$ The acceleration a(t) of a ody movin& in a strai&ht line is &iven

terms of time t y a=;Ft$ If the velocity of the ody is at t=1

and if s(t) is the distance of the ody from the ori&in at the time

t, #hat is s(+)Fs()

(a) *

() +

(c) ;

(d) 9

(e) +

0orrect ans#er is D

+ + 9

( ) ( ) ; = ; 9

(1) ;(1) 9(1) *

; 9 * + * 9

v t a t dt tdt t t C

v C c

t t dt t t t

= = − = − += − + = ∴ =

− + = − + =

∫ ∫

∫ $

Lhat differential e-uation is onthe sloefieldQ

(a) xy

() yx

(c) xy

(d) yW

(e) xWy

0orrect ans#er is A2c hori6ontal asymtotes at y=x and hori6ontal

asymtotes

GB CM:


89/95

89

0onsider the differential e-uation

( 1)dy

x ydx

= −

(a) On the axes rovided, s"etch a sloe field for &iven differential

e-uations at the t#elve oints indicated

() Lhile the sloe field in art (a) is dra#n at only 1 oints, it is

defined at every oint in the xyFlane$ Descrie all oints in the

xyFlane for #hich the sloes are ositive

(c) ind the articular solution y=f(x) to the &iven differential

e-uation #ith the initial condition f(*)=9

7olution:

(a)

() 7loes are ositive at the oints (x,y) #here* 1 x and y≠ >

(c)

9

9

9

9

1

9

1

9

*

1

9

( 1)

1ln 1

9

1

1 ,

1

xC

xC

x

dy x dx

y

y x C

y e e

y .e . e

.e .

y e

=−

− = +

− =

− = = ±

= =

= +

∫ ∫


90/95

90

Alications of Inte&rals:

Avera&e value of a function:

( )

( )

I 9

11

( ) = cos 1,

1 1 = = cos sin( ) 1$=*??9

M 9 ( 1)

avg

f t t t t on

f t t t t t

π

π π π − −

= − + −

= − + = − + = − ÷ − −

∫

Areas:

To find the area under the curve

you do

( ) ( )

( ) ( )

( ) ( )b

a

b

a

f x g x dx $R

/%%er )unc Lower )unc dx

= −

= −

∫

∫

To find the area in et#een a curve

you do

( ) ( )

( ) ( )

( ) ( )d

c

d

c

f y g y dx $R

Right )unc Left )unc dx

= −

= −

∫

∫


91/95

91

'xamle:

( ) ( )

*

1 9$*? x x xe dx−= + − =

∫ Volumes:

Dis":

ind the Volume &enerated y revolvin& the re&ion ounded y

9 , 9, 1*, 9 y x the line y and the line x above the line y= = = =

( ) 1*

19 9 *+$9*?+ x dxπ = − =∫


92/95

92

Lasher:

ind the Volume &enerated y revolvin& the re&ion ounded y

y x x= + + and the line + y x= − + aout the xFaxis

$=1

9$=1

$=1

9$=1

+

( )

(( +) ( ) ) M;$?9

R x

r x x

+ R r dx

+ x x x dx

π

π

−

−

= − +

= + +

= −

= + − + + =

∫

∫

7hells:

Gectan&le arallel to axis of rotation ( )( )+ shell rad shell height π =


93/95

93

ind an exression for the volume &enerated y revolvin& the re&ion

ounded y

9 y x x y x= + =and revolved aout the xFaxis

( )( )( )

9

1$9;1?==9

*

( )

( ) $9=

r x h x x x

+ x x x x dxπ

= = + −= + − =∫

0ross 7ections:

ind the cross sections of

+

y x

y

= +=

usin& 7-uares, '-uilateral Trian&les,

and 7emi 0ircles

7-uares:

( )( )

+ ( )

=$*99

s x x

s x

+ x dx−

= − + = −

= = −

= − =∫

'BT:


94/95

94

( )

( )

+ ( )

9 9

+ +

9 $=1M+

s x x

s x

+ x dx−

= − + = −

= = −

= − =∫

7emiF0ircle:

( )

( )

+ ( )

; ;

$9=?9;

s x x

d x

+ x dx

π π

π −

= − + = −

= = −

= − =∫

GB C;

7olution

(a)

( ) ( )( )1 1

* *( ) ( ) 1 9 1 1$199 rea f x g x dx x x x x dx= − = − − − =∫ ∫


95/95

95

()

( ) ( )( )

( )( )

1

*

1

*

( ) ( )

( 9( 1) ) (1 ) 1=$M1?

+olume g x f x dx

x x x x dx

π

π

= − − − =

− − − − − =

∫

∫

(c)

( )

( ) ( )( )

1

*

1

*

( ) ( )

1 9 1 1

+olume h x g x dx

kx x x x dx

= −

− − − =

∫

∫

calculus note

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