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Area/Sigma Nota.on

Objec.ve: To define area for plane regions with curvilinear boundaries. To use Sigma Nota.on to find areas.

The Area Problem

•  Formulas for the areas of polygons are well known. However, the problem of finding formulas for regions with curved boundaries caused difficul>es for early mathema>cians.

The Area Problem

•  Formulas for the areas of polygons are well known. However, the problem of finding formulas for regions with curved boundaries caused difficul>es for early mathema>cians.

•  The first real progress was made by Archimedes who obtained areas of regions with curved bounds by the method of exhaus>on.

The Area Problem

•  This method, when applied to a circle, consists of inscribing a succession of regular polygons in the circle and allowing the number of sides to increase indefinitely. As the number of sides increases, the polygons tend to “exhaust” the region inside the circle, and the area of the polygons become beBer and beBer approxima>ons of the exact area.

The Rectangle Method

•  We will now use Archimedes’ method of exhaus>on with rectangles to find the area under a curve in the following way:

•  We will now use Archimedes’ method of exhaus>on with rectangles in the following way:

•  Divide the interval [a, b] into n equal subintervals, and over each subinterval construct a rectangle that extends from the x-‐axis to any point on the curve that is above the subinterval.

•  For each n, the total area of the rectangles can be viewed as an approxima>on to the exact area under the curve over the interval [a, b]. As n increases these approxima>ons will get beBer and beBer and will approach the exact area as a limit.

Area as a Limit; Sigma Nota.on

•  The nota>on we will use is called sigma nota>on or summa>on nota>on because it uses the uppercase Greek leBer (sigma) to denote various kinds of sums. To illustrate how this nota>on works, consider the sum in which each term is of the form k2, where k is one of the integers from 1-‐5. In Sigma nota>on this can be wriBen as

which is read “the summa>on of k2 from 1 to 5”.

22222 54321 ++++

Sigma Nota.on

•  In general, we can look at sigma nota>on like this:

Example 1

•  Let’s look at some examples of sigma nota>on.

=+∑=

=+−∑=

)12()1(k

Other Sums

•  Here are two other ideas we need to know 1.  If the upper and lower limits of summa>on are the

same, we evaluate that number once in the func>on.

2.  If the func>on we are evalua>ng is a constant, we add that number to itself n >mes, or Cn.

33 2kk

==++++=5

10)5(2222222i

Changing the limits of Summa.on

•  A sum can be wriBen in more than one way using Sigma Nota>on with different limits of summa>on. For example, these three are all the same.

10864225

++++=∑=ii

Example 2

•  Express the following in sigma nota>on so that the lower limit of summa>on is 0 rather than 3.

Example 2

•  Express the following in sigma nota>on so that the lower limit of summa>on is 0 rather than 3.

•  We need to subtract 3 from the upper and lower limits of summa>on. If you subtract 3 from the limits, you must add 3 to k in the func>on. Always do the opposite.

Changing Limits

•  Be careful… you add/subtract from k, not the exponent. For example, change the lower limit of summa>on from 3 to 1.

∑∑∑=

− ==5

1)2(27

12 333k

Proper.es of Sums

•  Theorem 6.4.1

a) b) c)

∑∑==

kk acca

∑∑∑===

kk baba

∑∑∑===

−=−n

kk baba

Summa.on Formulas

•  Theorem 6.4.2 a) b) c)

2)1(...21

+=+++=∑

6)12)(1(...21 222

2 ++=+++=∑

nnnnkn

2)1(...21∑

=⎥⎦

⎤⎢⎣

⎡ +=+++=

Example 3

•  Evaluate ∑=

Example 3

∑ ∑ ∑∑= = ==

+=+=+30

)1(k k kk

kkkkkk

Example 3

∑ ∑ ∑∑= = ==

+=+=+30

)1(k k kk

kkkkkk

99202)31(30

6)61)(31(30

Defini.on of Area

•  We will now turn to the problem of giving a precise defini>on of what is meant by “area under a curve.” Specifically, suppose that the func>on f is con>nuous and nonnega>ve on the interval [a,b], and let R denote the region bounded below by the x-‐axis, bounded on the sides by the ver>cal lines x = a and x = b, and bounded above by the curve y = f(x).

Defini.on of Area

•  Divide the interval [a, b] into n equal subintervals by inser/ng n – 1 equally spaced points between a and b and denote those points by 121 ,..., −nxxx

Defini.on of Area

•  Each of these subintervals has width (b – a)/n, which is customarily denoted by

nabx −

Defini.on of Area

•  Over each interval construct a rectangle whose height is the value of f at an arbitrarily selected point in the subinterval. Thus, if denote the points selected in the subintervals, then the rectangles will have heights and areas

*1 ,..., nxxx

)(),...(),( **2

*1 nxfxfxf

xxfxxfxxf n ΔΔΔ )(,...)(,)( **2

Defini.on of Area

•  This can be expressed more compactly in sigma nota>on as: (k = # of rectangle)

•  We will repeat the process using more and more subdivisions, and define the area of R to be the “limit” of the areas of the approxima>ng regions Rn as n increases without bound. That is, we define the area A as

kk Δ≈∑

kknΔ≈ ∑

=+∞→

)(lim1

Defini.on of Area

•  Defini>on 6.4.3 (Area under a curve) •  If the func>on f is con>nuous on [a, b] and if f(x) > 0 for all x in [a, b], then the area under the curve y = f(x) over the interval [a, b] is defined by

kknΔ= ∑

=+∞→

)(lim1

Points

•  The values of can be chosen arbitrarily, so it is conceivable that different choices of these values might produce different values of A. Were this to happen, then the defini>on of area would not be acceptable. This does not happen. We will get the same area regardless of which points we choose.

*1 ,..., nxxx

Points

•  The three ways we will look at this is: 1.  The lec endpoint of each subinterval 2.  The right endpoint of each subinterval 3.  The midpoint of each subinterval

Points

•  To be more specific, suppose that the interval [a, b] is divided into n equal parts of length by the points and let x0 = a and xn = b. Then

for k = 0, 1, 2,…,n

nabx /)( −=Δ*1

*1 ,..., −nxxxxkaxk Δ+=

Points

•  We will look at each point as:

•  Lec endpoint •  Right endpoint •  Midpoint

xkaxx kk Δ−+== − )1(1*

xkaxx kk Δ+== −1*

xkaxx kk Δ−+== − )2/1(1*

Example 5

•  Use the defini>on of area using the right endpoint of each subinterval to find the area between the graph of f(x) = x2 and the interval [0, 1].

Example 5

•  We will use the same problem solving process for each point we choose.

1.  Find xΔ

Example 5

1.  Find xΔ nnx 101

Example 5

1.  Find 2.  Find

xΔ nnx 101

Example 5

xΔ nnx 101

nkxk =+=10*

Example 5

xΔ nnx 101

nkxk =+=10*

∑∑==

+∞→=⎟

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛=Δn

* 1)(lim

Example 5

xΔ nnx 101

nkxk =+=10*

⎥⎦

⎤⎢⎣

⎡ ++==⎟

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛=Δ ∑∑==

+∞→ 6)12)(1(11)(lim 3

* nnnnn

Example 5

xΔ nnx 101

nkxk =+=10*

6)12)(1(1lim 3

3 ==⎥⎦

⎤⎢⎣

⎡ +++∞→ n

nnnnnn

Theorem 6.4.4

•  Here are a few limits that you may or may not use. They can make the end of the problems easier.

+∞→=

111lim) ∑=

+∞→=

11lim)

+∞→=

11lim) ∑=

+∞→=

11lim)

Example 6

•  Use the defini>on of area using the midpoint to find the area under the curve f(x) = 9 – x2 over the interval [0, 3].

Example 6

1. Find xΔ nnx 303

Example 6

1. Find 2. Find

xΔ nnx 303

nkxk 2

333)2/1(0* −=−+=

Example 6

1. Find 2. Find

xΔ nnx 303

nkxk 2

333)2/1(0* −=−+=

nnnkxxf

32339)(lim

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ −−=Δ∑=

+∞→

Example 6

1. Find 2. Find

xΔ nnx 303

nkxk 2

333)2/1(0* −=−+=

nnnkxxf

218993

2339)(lim 222

*⎟⎟⎠

⎞⎜⎜⎝

⎛−+−=⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ −−=Δ∑=

+∞→

Example 6

1. Find 2. Find

xΔ nnx 303

nkxk 2

333)2/1(0* −=−+=

25427273

21899lim

n−+−=⎟⎟

⎞⎜⎜⎝

⎛−+−

+∞→

180032727 =−+−=

Example 6

•  Let’s quickly look at this as a right endpoint and compare the two answers.

nnnkxxf

32339)(lim

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ −−=Δ∑=

+∞→

nnkxxf

339)(lim2

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛−=Δ∑=

+∞→

Example 6

•  Let’s quickly look at this as a right endpoint and compare the two answers.

22 2727339limnk

n−=⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛−+∞→

25427273

21899lim

n−+−=⎟⎟

⎞⎜⎜⎝

⎛−+−

+∞→

Net Signed Area

•  In our defini>on of area, we assumed that f was con>nuous and nonnega>ve over the interval [a, b].

•  If f is both nega>ve and posi>ve over the interval, our defini>on no longer represents the area between the curve y = f(x) and the interval [a, b]; rather, it represents a difference of areas-‐ the area above the x-‐axis minus the area below the x-‐axis. We call this the net signed area.

Defini.on 6.4.5

•  Net Signed Area-‐ If the func>on f is con>nuous on [a, b] then the net signed area A between y = f(x) and the interval [a, b] is defined by

kknΔ= ∑

=+∞→

)(lim1

Example 7

•  Use our defini>on of area with the lec endpoint to find the net signed area between the graph of

y = x – 1 and the interval [0, 2]

Example 7

•  Use our defini>on of area with the lec endpoint to find the net signed area between the graph of

y = x – 1 and the interval [0, 2] 1. Find 2. Find

nnx 202

=ΔxΔ

*kx nn

kxk222)1(0* −=−+=

nnnkxxf

2442122)(lim 221

* −−=⎥⎦

⎤⎢⎣

⎡−⎟⎠

⎞⎜⎝

⎛ −=Δ∑=

+∞→

022 =−

Example 7 •  Again, let’s compare this to the right endpoint. •  Using the right endpoint, n

kxk20* +=

nnnkxxf

2442122)(lim 221

* −−=⎥⎦

⎤⎢⎣

⎡−⎟⎠

⎞⎜⎝

⎛ −=Δ∑=

+∞→

022 =−

nnkxxf

24212)(lim 21

* −=⎥⎦

⎤⎢⎣

⎡−⎟⎠

⎞⎜⎝

⎛=Δ∑=

+∞→

Homework

•  Page 384 •  3-‐15 odd •  27, 39-‐51 odd

Area/SigmaNotaon* - bcsoh.org 6.4.pdf · The*AreaProblem* •...

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Area/SigmaNotaon* - bcsoh.org 6.4.pdf · TheAreaProblem •...