CHAPTER 2_Integration Part 4 dnd

7/29/2019 CHAPTER 2_Integration Part 4 dnd

1/30

How do we findArea under a curve ?

If a function fis continuous and f(x) 0 on

[a,b], then the area of the region underthegraph offfrom a to b is;

Integrate f continuously

from x = a to x = b

b

adxxfAAREA,

a b


2/30

Solution

Find the area of the region bounded by thecurve and x-axis for xin [0, 4].12 xy

2

4

0

3

4

0

2

unit3

76

043

64

3

1Area

xx

dxx


3/30

Solution

Calculate the area between and thex-axis from x= -1 to x= 2.

2

7 xy

Step 1:

Step 2:

We sketch the region under the curve & boundedby the x axis.

2

33

2

1

3

2

1

2

unit18320

334

3

17

3

814

3

117

3

227

37

7

x

x

dxxA


4/30

SolutionDetermine the area of the shaded region.

342 xxy

a.

2

23

23

3

1

23

3

1

2

unit3

4

3

40

323

19189

13123

13332

3

3

323

34

xxx

dxxxA


5/30

SolutionDetermine the area of the shaded region.

b.565

23

xxxy

c.

xy

1

d.xy


6/30

Area of region below x axis ?

If the area of the region is below the x

axis then;

Area is alwayspositive.

b

a

dxxfAArea )(,

Example:

Find the area of the

region bounded by

the curve y= -x

andx-axis forxin

[0, 4].2

4

0

2

4

0

4

0

unit8

02

16

2

x

dxx

dxx-Area


7/30

Solution

Find the area of the region bounded by thecurve and x-axis for xin [-1, 0].

2

0

1-

4

0

1-

3

unit4

1

4

1

4

x

dxxArea

0

3xy


8/30

How do we findArea between 2 curves ?

Iff & gare continuous and f(x) g(x)for

every x in [a, b] then the area of the regionbounded by the graphs off& gfrom a to b

is;

b

a

dxxgxfAAREA,


9/30

How do we findArea between 2 curves ?

Iff & gare continuous and f(y) g(y)for

every y in [c, d] then the area of the regionbounded by the graphs off& gfrom cto d

is;

d

c

dyygyfAAREA,


10/30

Solution

1. Calculate the area bounded by thegraphs 27and2 xyxy

Step 1:

Step 2:

Determine the intersectionpoints.

y = x2- 4

y = 2 - x

3and2

032

06

24

2

2

xx

xx

xx

xx

Identify the upper boundary& lower boundary functions.

2

3

2 42 dxxxA


11/30

2

3232

2

3

32

2

3

2

2

3

2

unit6

125

2

27

3

22

3

3

2

336

3

2

2

226

326

6

42

xxx

dxxx

dxxxA

Solution

1. Calculate the area bounded by thegraphs 27and2 xyxy

Step 3:

y = x2- 4

y = 2 - x

Integrate the area formula.


12/30

Solution

2. Calculate the area bounded by thegraphs of .30for8and 22 xxyxy

Step 1:

Step 2:

There are 2 different areawith different intersectionpoints.

Identify the upperboundary & lowerboundary functions.

28 xy

2xy

AB

Step 3: The area formula;

dxxxdxxx

BA

3

2

222

0

22 88

Area


13/30

2

333

3

2

32

0

3

3

2

22

0

2

3

2

222

0

22

unit346

3

326

3

32

283

2238

3

320

3

2228

8

3

2

3

28

8228

88Area

xxx

x

dxxdxx

dxxxdxxxBA

Solution

2. Calculate the area bounded by the

graphs of .30for8and 22 xxyxy

28 xy

2xy

AB

Step 3: The area formula;


14/30

Solution

3. Evaluate each of the following areabounded by the given curves.

Step 1:

Step 2:

intersection

points:

The areaformula:

322 xxy 42 xy

1;1

011

01

4232

2

2

xx

xx

x

xxx

2

1

1

3

1

1

2

1

1

2

unit3

4

3

11

3

11

3

1

3242

xx

dxx

dxxxxA

a.


15/30

Solution


Step 1:

Step 2:

intersection

points:

The areaformula:

b.xy

xy


16/30

Solution


Step 1:

Step 2:

intersection

points:

The areaformula:

c.164 xy

1022 xy


17/30

Solution


Step 1:

Step 2:

intersection

points:

The areaformula:

f.

12 xy

27 xy


18/30

How to determine a volume of solid ?

The volume of a solid of known

integrable cross sectional areaA(x) froma to b is;

b

a

dxxAVVOLUME,


19/30

What is a solid of revolution ?

When the region is revolved about x axis / y

axis, it is said to generate a solid of

revolution.

Rotation with respect to the x axis:

b

a

dxxfV 2

3600


20/30


When the region is revolved about x axis / y

axis, it is said to generate a solid of

revolution.

Rotation with respect to the y axis:

d

c dyyfV

2

360


21/30

Solution

a) Calculate the volume obtained by rotatingthe region enclosed by the curve onthe interval [0, 1], through 360 about x axis & y axis.

Aboutx axis :

xy

xy

3

1

0

21

0

2

1

0

2

unit2

1

21

2

x

dxxV


22/30

Solution

a) Calculate the volume obtained by rotatingthe region enclosed by the curve onthe interval [0, 1], through 360 about x axis & y axis.

Abouty axis :

xy

3

51

0

5

1

0

22

unit5

1

5

1

5

y

dxyV

Step 1:

Solve for x; When x = 1; y = 1:

2yx

yx

Step 2:


23/30

Solution

Evaluate each of the following:

Aboutx axis :

axis.-xtheabout2][0,intervaltheon1curvetheunder

regiontherevolvingbyformedsolidtheofVolumea.

2 xy

3

35

2

0

35

2

0

24

2

0

22

unit15

206

23

16

5

32

023

22

5

2

3

2

5

12

1

xxx

dxxx

dxxV


24/30

Solution

Evaluate each of the following:

Aboutx axis :


regiontherevolvingbyformedsolidtheofVolumeb.

xy


regiontherevolvingbyformedsolidtheofVolumec.

2 xy


regiontherevolvingbyformedsolidtheofVolumed.

2xy


25/30


The Washer Method:

Rotation about the x

axis / y

axis :

Outer

radius f(x)

Inner

radius

g(x)3600

Outer

radius

f(y)

Inner

radius

g(y)

d

cdyygyfV 22

b

adxxgxfV 22


26/30

Solution

b) Calculate the volume obtained by rotatingthe region enclosed by the curve andthe line y = x through 360 about x axis &

y axis.

Aboutx axis :

2xy

y = x2y = x

1;0

01

02

2

xx

xx

xx

xx

Step 1:

Step 2:

Determine theintersection points.

Identify the innerradius & outer radius.

xy

xy

:Outer

:Inner2


27/30

Solution

b) Calculate the volume obtained by rotatingthe region enclosed by the curve andthe line y = x through 360 about x axis &

y axis.

Aboutx axis :

2xy

y = x2y = x

Step 2: The volume:

3

1

0

53

1

0

42

1

0

222

unit15

2

5

1

3

1

53

xx

dxxx

dxxxV


28/30

Solution

Calculate the volume of the solid formed byrevolving the given region about prescribed axis.

Aboutx axis :

axis.-xtheabout4

andcurvesthebyboundedregionThea.x

yxy

Step 1:

Step 2:

Sketch the graph &region.

Intersection point:

xy

4

xy

16;0

016

016

1642

2

xx

xx

xx

xx

xx


29/30

3

32

16

0

32

16

0

2

16

0

22

unit3

128

48

4096128

48

16

2

16

3162

16

4

xx

dxx

x

dxx

xV

Solution


Aboutx axis :

axis.-xtheabout4

andcurvesthebyboundedregionThea.x

yxy

xy

4

xy

Step 3:


30/30

Solution


Aboutx axis :

axis.-xtheabout8and2curvesthebyboundedregionTheb. 2 xyxy

Step 1:

Step 2:

Sketch the graph &region.

Intersection point:

Step 3: The volume:

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CHAPTER 2_Integration Part 4 dnd

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