Pertemuan 13

Definite Integration and Areas


u dv = v du

Integral Parsial

Contoh.

D I

f′(x) gdx

f″(x) (g(x))dx dx dst dst

f(x) g(x) dx


Caranya baris (1) kolom D dikalikan dengan baris ke (2) kolom I dimulai dengan tanda + , kemudian baris ke (2) D dengan baris ke (3) I dengan tanda ( - ) dan seterusnya samapai ada 0. Sebaliknya bila kolom D tidak 0 maka caranya sebagai berikut :


Contoh.

x2 ex

2x ex

2 ex

0 ex

x2 ex dx = x2 ex – 2x ex +2 ex

CaraCarasederhanasederhana


Contoh.

cos(x) ex dx = ½ ex sin 2x + ¼ ex cos 2x + ¼

ex cos 2x dx = 4/3 (½ ex sin 2x + ¼ ex cos 2x ) + C

B = ½ ex sin 2x + ¼ ex cos 2xIntegral menjadi A = B + ¼ A atau A = 4/3 B

A = ex cos 2x dx


D I

x


0 1

23 2 xy

It can be used to find an area bounded, in part, by a curve

e.g.

1

0

2 23 dxx gives the area shaded on the graph

The limits of integration . . .

Definite integration results in a value.

Areas


. . . give the boundaries of the area.

The limits of integration . . .

0 1

23 2 xy

It can be used to find an area bounded, in part, by a curve

Definite integration results in a value.

Areas

x = 0 is the lower limit( the left hand

boundary )x = 1 is the upper limit(the right hand

boundary )

dxx 23 2

0

1

e.g.

gives the area shaded on the graph


0 1

23 2 xy

Cari luas yg diarsir

Luas yang diarsir = 3

The units are usually unknown in this type of question

1

0

2 23 dxxKaren

a

31

0

xx 23


SUMMARY

• curve

),(xfy

• garis x = a dan x = b

• x-axis dan

PROVIDED that the curve lies on, or above, the x-axis

between the values x = a and x = b

Definite integral or

merupakan luas diantara

b

a

dxxf )( b

a

dxy


xxy 22 xxy 22

Finding an area

0

1

2 2 dxxxA area

A B

1

0

2 2 dxxxB area

For parts of the curve below the x-axis, the definite integral is negative, so


xxy 22

A

Finding an area

0

1

2 2 dxxxA

2

2

3

23 0

1

xx

2

3

)1(3

)1(0

1

3

1

1

1

3

4Area A


xxy 22

B

Finding an area

1

0

2 2 dxxxB

2

3 1

03

xx

013

1

3

2

3

2Area B

Definite Integration and Areas SUMMAR

Y Luas selalu positip

Definite integral positive untuk luas yang ada di atas sumbu – x dan negatip untuk luas yang ada di bawah sumbu - x

To find an area, we need to know whether the curve crosses the x-axis between the boundaries.• For areas above the axis, the definite

integral gives the area.• For areas below the axis, we need to

change the sign of the definite integral to find the area.

Definite Integration and AreasExercis

e Find the areas described in each question.

1. The area between the curve the x-axis and the lines x = 1 and x = 3.

2xy

2. The area between the curve , the x-axis and the x = 2 and x = 3.

)3)(1( xxy

Coba sebelum Coba sebelum melihat melihat

jawabnyajawabnya


B

)3)(1( xxy

A

2xy

1.

2.

Solutions: 3

1

3

1

32

3

xdxxA

3

232

3

3

2

23

xx

x

3

2

2 34 dxxxB

3

2 B

32

33

83

)1(

3

)3(

Definite Integration and AreasExtensi

onThe area bounded by a curve, the y-axis and the lines y = c and y = d is found by switching the xs and ys in the formula.

So,become

sb

a

dxy

d

c

dyx

d

c

dyx

e.g. To find the area between the curve , the y-axis and the lines y = 1 and y = 2, we need

xy

3

7

2

1

2 dyy


22 xxy

xy

Harder Arease.g.1 Find the coordinates of the points of

intersection of the curve and line shown. Find the area enclosed by the curve and line.

22 xxx

Solution: The points of intersection are given by

02 xx 0)1( xx10 xx or


22 xxy

xy 00 yx

xy Substitute in

11 yx The area required is the area under the curve between 0 and 1 . . . . . . minus the area under the line (a

triangle )

3

2

32

1

0

32

1

0

2

x

xdxxx

Area of the triangle 2

1)1)(1(

2

1

Area under the curve

Required area 6

1

2

1

3

2

Method 1

0 1


22 xxy

xy Instead of finding the 2 areas and then subtracting, we can subtract the functions before doing the integration.

1

0

321

0

2

32

xx

dxxxArea

We get

Method 2

xxx 222xx

0

3

1

2

16

1

0 1


6y

22 xy

Exercise Find the points of intersection of the

following curves and lines. Show the graphs in a sketch, shade the region bounded by the graphs and find its area.

22 xy 6y(a) ; (b) ; 2xy24 xy

Solution:

(a) 622 x

42 x2x

( y = 6 for both points )


6y

22 xy

Shaded area = area of rectangle – area under curve

3

164

3

84

3

8

Area under curve

2

2

32

2

2 23

2

x

xdxx

Shaded area3

1624

3218


2xy

24 xy

,02 yx

Area of the triangle

2x 1xor

Substitute in : 2xy31 yx

Area under the curve

1

2

31

2

2

344

xxdxx 9

3321

(b) ; 2xy24 xy

022 xx0)1)(2( xx

Shaded area = area under curve – area of triangle

29

29

242 xx


3xy

The symmetry of the curve means that the integral from 1 to +1 is 0.

If a curve crosses the x-axis between the limits of integration, part of the area will be above the axis and part below.

3xy e.g. between 1 and +1

To find the area, we could integrate from 0 to 1 and, because of the symmetry, double the answer.For a curve which wasn’t symmetrical, we

could find the 2 areas separately and then add.

Definite Integration and AreasYou don’t need to know how the formula for

area using integration was arrived at, but you do need to know the general ideas.

The area under the curve is split into strips.

The area of each strip is then approximated by 2 rectangles, one above and one below the curve as shown.

The exact area of the strip under the curve lies between the area of the 2 rectangles.


Using 10 rectangles below and 10 above to estimate an area below a curve, we have . . .Greater accuracy would be given with 20 rectangles below and above . . .For an exact answer we let the number of rectangles approach infinity. The exact area is “squashed”

between 2 values which approach each other. These values become the definite integral.


Pertemuan 13

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