53
Area/Sigma Nota.on Objec.ve: To define area for plane regions with curvilinear boundaries. To use Sigma Nota.on to find areas.
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:
The Rectangle Method
• 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.
The Rectangle Method
• 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 ++++
∑=
5
1
2
kk
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.
=∑=
8
4
3
kk
=∑=
5
1
2k
k
=+∑=
5
0
)12(k
k
=+−∑=
5
0
)12()1(k
k 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.
∑=
=2
2
33 2kk
∑=
==++++=5
1
10)5(2222222i
∑=
=n
kCnC
1
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
1
++++=∑=ii
∑=
+4
022
jj
∑=
−7
3
42k
k
Example 2
• Express the following in sigma nota>on so that the lower limit of summa>on is 0 rather than 3.
=∑=
−7
3
25k
k
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.
∑=
−7
3
25k
k
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
3227
23
1)2(27
3
12 333k
k
k
k
k
k
Proper.es of Sums
• Theorem 6.4.1
a) b) c)
∑∑==
=n
kk
n
kk acca
11
∑∑∑===
+=+n
kk
n
kkk
n
kk baba
111
)(
∑∑∑===
−=−n
kk
n
kkk
n
kk baba
111
)(
Summa.on Formulas
• Theorem 6.4.2 a) b) c)
2)1(...21
1
+=+++=∑
=
nnnkn
k
6)12)(1(...21 222
1
2 ++=+++=∑
=
nnnnkn
k
2
1
3333
2)1(...21∑
=⎥⎦
⎤⎢⎣
⎡ +=+++=
n
k
nnnk
Example 3
• Evaluate ∑=
+30
1
)1(k
kk
Example 3
• Evaluate ∑=
+30
1
)1(k
kk
∑ ∑ ∑∑= = ==
+=+=+30
1
30
1
30
1
2230
1
)1(k k kk
kkkkkk
Example 3
• Evaluate ∑=
+30
1
)1(k
kk
∑ ∑ ∑∑= = ==
+=+=+30
1
30
1
30
1
2230
1
)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 −
=Δ
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
**2
*1 ,..., nxxx
)(),...(),( **2
*1 nxfxfxf
xxfxxfxxf n ΔΔΔ )(,...)(,)( **2
*1
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
xxfAn
kk Δ≈∑
=
)(1
*
xxfAn
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
xxfAn
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.
**2
*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
*2
*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
• 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].
• We will use the same problem solving process for each point we choose.
1. Find xΔ
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].
• We will use the same problem solving process for each point we choose.
1. Find xΔ nnx 101
=−
=Δ
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].
• We will use the same problem solving process for each point we choose.
1. Find 2. Find
xΔ nnx 101
=−
=Δ
*kx
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].
• We will use the same problem solving process for each point we choose.
1. Find 2. Find
xΔ nnx 101
=−
=Δ
*kx
nk
nkxk =+=10*
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].
• We will use the same problem solving process for each point we choose.
1. Find 2. Find
xΔ nnx 101
=−
=Δ
*kx
nk
nkxk =+=10*
∑∑==
+∞→=⎟
⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=Δn
k
n
kkn n
knn
kxxf1
3
22
1
* 1)(lim
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].
• We will use the same problem solving process for each point we choose.
1. Find 2. Find
xΔ nnx 101
=−
=Δ
*kx
nk
nkxk =+=10*
⎥⎦
⎤⎢⎣
⎡ ++==⎟
⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=Δ ∑∑==
+∞→ 6)12)(1(11)(lim 3
13
22
1
* nnnnn
knn
kxxfn
k
n
kkn
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].
• We will use the same problem solving process for each point we choose.
1. Find 2. Find
xΔ nnx 101
=−
=Δ
*kx
nk
nkxk =+=10*
31
62
6)12)(1(1lim 3
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.
∑=
+∞→=
n
kn na
1
111lim) ∑=
+∞→=
n
knk
nb
12 2
11lim)
∑=
+∞→=
n
knk
nc
1
23 3
11lim) ∑=
+∞→=
n
knk
nd
1
34 4
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
• 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].
1. Find xΔ nnx 303
=−
=Δ
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].
1. Find 2. Find
xΔ nnx 303
=−
=Δ
*kx
nnk
nkxk 2
333)2/1(0* −=−+=
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].
1. Find 2. Find
xΔ nnx 303
=−
=Δ
*kx
nnk
nkxk 2
333)2/1(0* −=−+=
nnnkxxf
n
kkn
32339)(lim
2
1
*
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛ −−=Δ∑=
+∞→
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].
1. Find 2. Find
xΔ nnx 303
=−
=Δ
*kx
nnk
nkxk 2
333)2/1(0* −=−+=
nnnk
nk
nnnkxxf
n
kkn
349
218993
2339)(lim 222
22
1
*⎟⎟⎠
⎞⎜⎜⎝
⎛−+−=⎟
⎟⎠
⎞⎜⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛ −−=Δ∑=
+∞→
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].
1. Find 2. Find
xΔ nnx 303
=−
=Δ
*kx
nnk
nkxk 2
333)2/1(0* −=−+=
333
2
222
2
427
25427273
49
21899lim
nnk
nk
nnnnk
nk
n−+−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−+−
+∞→
180032727 =−+−=
Example 6
• Let’s quickly look at this as a right endpoint and compare the two answers.
nnnkxxf
n
kkn
32339)(lim
2
1
*
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛ −−=Δ∑=
+∞→
nnkxxf
n
kkn
339)(lim2
1
*
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛−=Δ∑=
+∞→
Example 6
• Let’s quickly look at this as a right endpoint and compare the two answers.
3
22 2727339limnk
nnnk
n−=⎟
⎟⎠
⎞⎜⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛−+∞→
333
2
222
2
427
25427273
49
21899lim
nnk
nk
nnnnk
nk
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
xxfAn
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
kn
kxk222)1(0* −=−+=
nnnk
nnnkxxf
n
kkn
2442122)(lim 221
* −−=⎥⎦
⎤⎢⎣
⎡−⎟⎠
⎞⎜⎝
⎛ −=Δ∑=
+∞→
022 =−
Example 7 • Again, let’s compare this to the right endpoint. • Using the right endpoint, n
kxk20* +=
nnnk
nnnkxxf
n
kkn
2442122)(lim 221
* −−=⎥⎦
⎤⎢⎣
⎡−⎟⎠
⎞⎜⎝
⎛ −=Δ∑=
+∞→
022 =−
nnk
nnkxxf
n
kkn
24212)(lim 21
* −=⎥⎦
⎤⎢⎣
⎡−⎟⎠
⎞⎜⎝
⎛=Δ∑=
+∞→
Homework
• Page 384 • 3-‐15 odd • 27, 39-‐51 odd
