bab vi2 aplikasi integral

8/3/2019 Bab VI2 Aplikasi Integral

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7. APPLICATIONS OFTHE DEFINED INTEGRAL

1


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7.1 Area Between Two Curves

a. Let { })(0,|),( xfybxayxD =

a b

f(x)

D

Area of region D = ?

x

Steps :

1. Divided D into n pieces, the area of each piecesis approximated by area of rectangularwith height f(x) and length of base x

xxfA )(2. The area of D is approximated by sum area of rectangular .If , The area of D is

b

a

dxxf )(A =0x


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Example : Find the area of region that is boundedby parabola x axis, and x= 2.

3

8

3

12

0

3

2

0

2 =

== xdxxA

,2xy =

2xy =

2

Area of pieces

x

2x

xxA 2

Area of region


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b) Let { })()(,|),( xhyxgbxayxD =

x

xxgxhA ))()((

h(x)

g(x)

a b


Steps:

1. Divided D into n pieces, the area of eachpieces is approximated by area of rectangular

with height h(x)-f(x) and length of basex

2. The area of D is approximated by sum of area rectangular .

If , The area of D is

A = b

a

dxxgxh ))()((

D h(x)-g(x)

0x


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Example: Find the area of region that is bounded by y= x+4and parabola 22 = xy

242 =+ xx

22 = xy 06

2 = xx

0)2)(3( =+ xx

The straight line and parabola

intersects at

y=x+4-2 3

x = -2, x = 3x

)2()4(2 + xx

Area of pieces

xxxA + ))2()4(( 2


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++=+=3

2

3

2

22)6())2()4(( dxxxdxxxA

6

1256

2

1

3

13

2

23 =

++=

xxx

Area of region :


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Example : Find the area of region that is bounded by x axis,2

xy = And y = -x + 2Answer

Intersection points

22 += xx

2xy =

022 =+ xx

xxA 21

2

0)1)(2( =+ xx

x = -2, x = 1

y=-x+2

1

If pieces is vertical , then region must bedivided into 2 sub region

xx

xxA + )2(2

Area of pieces I

Area of pieces II


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===1

0

1

0

3

312

13

1|xdxxA

The area of region I

The area of region II

2

1

2

2

1

2

1

2 |22 xxdxxA +=+=

2

1)2()42(

21 =++=

The area of region

6

5

2

1

3

121

=+=+= AAA


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c). Let { })()(,|),( yhxygdycyxD =

y

d

c

dyygyh ))()((

h(y)

g(y)

c

dD


Langkah :

1. Divided D into n pieces, the area of eachpieces is approximated by area of rectangular

with height h(y)-f(y) and length of base y

yygyhA ))()((

y

A =

h(y)-g(y)

2. The area of D is approximated by sum area of rectangulars .

If , The area of D is0y


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23 yx =

231 yy =+

022 =+ yy

2

3 yx =Example: Find the area of region that is bounded byand 1= xy

Answer : The intersection point between

parabola and straight line are

0)1)(2( =+ yyy = -2 dan y = 1

1= xy

-2

1

y)1()3(

2 + yy

Area of pieces

yyyA += ))1()3(( 2


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The area of region is :

+=1

2

2))1()3(( dyyyL

+=1

2

2)2( dyyy

.2

92

2

1

3

11

2

23 =

+=

yyy


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7.2 NN ilai rata-rata

b

a

dxxfab

valueAverage )(1

Secara Geometri untuk 0)( xf , maka

Luas daerah dibawah kurvay =f(x), bxa

a b0

y=f(x)

Average value

off

gambar 6.9


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7.3 The Length of Curve

A parameter form of curve on

x = f(t)y = g(t) bta ,

Point A(f(a),g(a)) is called original point and B(f(b),g(b)) is calledterminal point of curve.

Definition : A curve is called smooth if

(1)

(i) 'f 'gand are continuous on [a,b]

(ii) 'f 'gand is not zero at the same time on (a,b)

2R


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Let a curve on parameter form (1), we will find length of that curve

Steps

1. Divided interval [a,b] into n subintervals

btttta no =


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2. Approximate length of curve

1iQ

iQis

iw

is length of arc ii QQ 1

iw length of line segment ii QQ 1

ii ws

The length of arc is approximated by length of linesegment

ix

iy

22

)()( ii yx +=2

1

2

1 )]()([)]()([ += iiii tgtgtftf

We apply mean value theorem to function f and g.There are , such that),(,

1 iiii

tttt

ttftftf iii = )(')()( 1

ttgtgtg iii = )(')()( 1


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Calculus I (MA 1114) - Faculty of ScienceTelkom Institut of Technology 28

where 1= iii ttt

so

22])('[])('[

iiiii

ttgttfw +=

iii ttgtf +=22

)]('[)]('[

The length of curve is approximated by length of polygonal arc

=

+n

i

iii ttgtfL1

22)]('[)]('[

If ||P|| 0, the length of curve is

dttgtfL

b

a

+=22

)]('[)]('[


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Remark:For curve y=f(x), bxa

dttgtfL

b

a +=22

)]('[)]('[ dtdt

dy

dt

dxb

a+= 22 ][][

dttgtfL

d

c +=22

)]('[)]('[

dx

dx

dydt

dx

dy

dt

dxb

a

b

a

+=

+=

22

21)1()(

For curve x=g(y), dyc

dtdt

dy

dt

dxd

c +=22

][][

dydy

dxdt

dy

dx

dt

dyd

c

d

c

+=

+=

22

211)(


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Example : Find length of curve

40;,23 == ttytx1.

2

3)(' ttx= tty 2)(', =

dtttL +=4

0

222)2()3(

The length of curve

dttt +=4

0

2449 +=

4

0

22)49( dttt

+=4

0

249 dttt

++=

4

0

22/12

18

)49()49(

t

tdtt

40

2/3232

181 |)49( += t )81080()84040( 27

1271

==


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2.

3/ 2 1

2 , 73 y x x= Answer :

2/13x

dx

dy=

( ) +=+=7

3/1

7

3/1

22/19131 dxxdxxL )91()91(

7

3/1

2/1

91 xdx ++=

31

2727

3/12/3

272 37)8512(|)91( ==+= x


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Problems

A. Sketch and find area of region that is bounded by

22and y x y x= = +1.

3, , 8and y x y x y= = =2.

3. y= x, y= 4x, y= -x+2

4. y= sin x, y= cos x, x= 0 , x= 2.

22 7 0 6 0 x y and y y x + = =5.


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E. Find length of curves

10,)2(3

1 2/32 += xxy

42,4

ln

2

2

= xxx

y

2/10),1ln(2 = xxy

90),3(3

1= yyyx

41;2/12,2332 =+= ttytx

== ttytx 0;5cos4,sin41.

2.

3.

4.

5.

6.

bab vi2 aplikasi integral

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