Area by Integration

7/31/2019 Area by Integration

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TOPIC

APPLICATIONS

AREA BY INTEGRATION


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The area under a curve

Let us first consider

the irregular shape

shown opposite.

How can we find the

areaA of this shape?


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We can find an

approximation by

placing a grid of

squares over it.

By counting squares,

A > 33 andA < 60

i.e. 33


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By taking a finer mesh of

squares we could obtain a

better approximation forA.

We now study another way of

approximating toA, usingrectangles, in whichA can be

found by a limit process.


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The diagram shows part

of the curvey =f(x) from

x = a tox = b.

We will find an expression

for the areaA bounded bythe curve, thex-axis, and

the linesx = a andx = b.

A


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The interval [a,b] is

divided into n sections of

equal width, x.

n rectangles are then drawn

to approximate the areaAunder the curve.

x

A


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Dashed lines represent the

height of each rectangle.

Thus the area of the first rectangle = f(x1).x1

f(x1)The first rectangle

has height f(x1)and breadthx1.

The position of each line is

given by anx-coordinate,xn.

x1, x2,x3, x4, x5, x6

x1


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An approximation for the

area under the curve,

betweenx = a tox = b,can be found by

summing the areas of the

rectangles.

A =f(x1).x1 + f(x2).x2 + f(x3).x3 + f(x4).x4 + f(x5).x5 + f(x6).x6


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Using the Greek letter (sigma) to denote the sum of,

we have

6

1

).(i

iiixxfA

ni

iiixxfA

1

).(

For any number n rectangles, we then have


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ni

ii

xxfA1

).(

ax

bx

xxfA ).(


In order to emphasise that the sum extends over the

interval [a,b], we often write the sum as


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ax

bxx

xfA x).(lim0


By increasing the number n rectangles, we decrease their

breadth x.

As xgets increasingly smaller we say it tends to zero,

i.e. x 0.

So we define

Remember, we

met limits

before withDifferentiation


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ax

bxx

xfA x).(lim0


was simplified into the form that we are familiar with

today

The form

ax

bxx

xfA x).(lim0

b

a dxxfA )(

This reads

the areaA is equal to the integral off(x) from a to b.


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We have derived a method for finding the area under a curve

and a formal notation

b

adxxfA )(

We have seen the integration symbol before in connectionwith anti-differentiation, but we have not yet connected

finding the area under a curve with the process of integration.


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Let us remind

ourselves of where we

started.

Can we apply this method to calculate the area under a curve?


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In conclusion,

b

a dxxf )(

the areaA bounded by thex-axis, the linesx = a andx = b

and the curvey =f(x) is denoted by,


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Suggested Steps to Determine the Area of aPlane Figure by Integration:1. Determine the intersection points of the given

boundaries or equations.2.Graph the given functions.

3. Shade the area to be determined

4.Consider a thin rectangle anywhere within the region,horizontal or vertical element, to represent the entireregion.

5.Determine the dimensions of the rectangular element

and limits of integration. Apply the integral using theextreme points as the limit of integration to determinethe total area.

6.Set up the area of the element and evaluate the integral

throughout the region.


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0 1

232 xy

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

e.g. 10

223 dxx gives the area shaded on the graph

The limits of integration . ..

Definite integration results in a value.

Areas


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. . . give the boundaries of the area.

The limits of integration . . .

0 1

232 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 20

1

e.g. gives the area shaded on the graph


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0 1

23 2xy

Finding an area

the shaded area equals 3

The units are usually unknown in this type of question

10

223 dxxSince

3

1

0

xx 2

3


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SUMMARY

the curve ),(xfy the lines x = aand x = b

the x-axis and

PROVIDED thatthe curve lies on, or above, the x-axis between the values x = a

and x = b

The definite integral or gives the area

between

ba

dxxf )( ba

dxy


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xxy 22 xxy 22

Finding an area

0

1

2 2 dxxxAarea

A B

10

22 dxxxBarea

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


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xxy 22 A

Finding an area

0

1

22 dxxxA

22

3

230

1

xx

23 )1(

3

)1(0

13

1

1

1

3

4Area A


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xxy 22 B

Finding an area

10

22 dxxxB

23

1

03

xx

013

1

3

23

2Area B


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Area Under the Curve


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x24y Find the area of the region bounded by the line

and the coordinate axes.

EXAMPLE

Ox

y

(0,4)

(2,0)

(x,y)y = 4 - 2x

L = y

w = dx

Ox

y

(0,4)

(2,0)

(x,y)

y = 4 - 2x

L = x

w = dy

Using vertical stripping Using horizontal stripping

The rectangular strip has a partial


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Ox

y

(0,4)

(2,0)

(x,y)y = 4 - 2x

L = y

w = dx

2

0

)24(x

x

dxxA

2

0

2

0

24 xdxdxA

0

2

2

24

2

xxA02)2(4

2 A

dA = y dx, y = 4

2x

squnitsA 4

Using Vertical Stripping:The rectangular strip has a partialarea:

dA = area of the element

Integrate throughout the region:

Th t l t i hUsing Horizontal Stripping:


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The rectangular strip has apartial area:

dA = area of the element

)4(2

1yx ,xdydA

dyyA

4

0)4(

2

1

4

0

4

02

12 ydydyA

04

2212

2

yyA

0

4

4)4(2

2

A

Ox

y

(0,4)

(2,0)

(x,y)

y = 4 - 2x

L = x

w = dy

squnitsA 4

Using Horizontal Stripping:

Integrate throughout the region:


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Area Between two Curves


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Area Between two CurvesFinding the limits of integration for area between two curves

Step 1: Sketch the region and draw a vertical line segmentthrough the region at the arbitrary point on the x-axis,connecting the top and bottom boundaries

Step 2. The y-coordinate of thetop endpoint of the linesegment sketched instep 1 will be f(x), thebottom one g(x), andthe length of the linesegment will bef(x) g(x). This is theintegrand in 1.


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Step 3.

To determine the limits of integration,

imagine moving the line segment left andthen right. The leftmost position at which

the line segment intersects the region is x=a

and the rightmost is x=b.


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d

cdyyvywA )]()([


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xy 42 Determine the area of the region bounded by the curve

, the lines x=0 , and y=2.

EXAMPLE

Ox

y

y = 2

y2

= 4x

(x,y)

x = 0

w = dy

L = x

(1,2)

Using horizontal stripping,

2

0

xdyA4

yx

2

but

2

o

2

dy4

yA

dyy4

1

A

2

0

2

2

0

3

3

y

4

1A

0212

1A 3

3

2

12

8A

Point of intersection of y = 2 and


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Using vertical stripping,

Ox

y

y = 2

y2 = 4x

(x,y)

x = 0

w = dx

(1,2)L = 2- y

4

yx

2

4

2x

2

1x

2),1(

Point of intersection of y = 2 andthe curve:

If y = 2:

dx)y2(dA

1

0

dx)y2(A x4y

x2y

x2y

1

0dxx22A

1

0dxx22A

squnitsA 3

2

2Fi d th f th i b d d b th


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2xy .xy

Find the area of the region bounded by the curveand.

O

V(0,0)

y = x

y = -x2

(-1,-1)

x

y

(x,yL)

(x,yC)

L = yC - yL

w = dx

By vertical stripping

Intersection points of the line

and the curve: (0,0) and (-1,1)

dxyydA LC

dxyyA LC 2

C xy xyL

dxxxA

1

0

2

dxxxA 01 2

6

1A


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Figure 6.1.10 (p. 417)

Find the area of the region enclosed by and2

yx .2 xy

EXAMPLE


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Find the area bounded by the given curves:1. y = x2, y = x

2. y = x + 7, y = 9

x23. x = 1 y2, x = y214. 4ay = x2, y = x + 3a5. y = x3, y = 4x26. y = (x 1)3, y = x2 x 17. x2 2x + 2y + 5 = 0, x2 2x + y + 1 = 08. x + 2y = 2, y x = 1, 2x + y = 79. y3 = x2, 2x + y + 1 = 0, x y = 410. y2 = x, y = x3, x + y = 2

EXAMPLE

EXAMPLE


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1. Find the values of k such that the area of theregion bounded by the parabolas y = x2 k2 and y =k2 x2 is 576 square units.

2. Find the area of the region bounded by theparabola y = x2, the tangent line to this parabola at(1, 1) and the x-axis.

3. Find the number b such that the line y = b dividesthe region bounded by the curves y = x2 and y = 4into the regions with equal area.

EXAMPLE

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Area by Integration