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Averages
D. DeTurck
University of Pennsylvania
September 15, 2015
D. DeTurck Math 104 002 2015C: Averages 1 / 18
Totals and averages
We have used integrals to calculate totals — the total areabetween two curves, the total volume of a solid, etc.One more kind of total:
Total mass: First, consider a long thin tube (so thin that it isessentially one-dimensional) lying along the x-axis between x = aand x = b. And suppose that the linear density of the material ofthe tube is δ(x) grams per centimeter. Then the mass of the littlebit of tube between x and x + dx is
dm = δ(x)dx
and so the total mass of the tube is
m =
ˆ b
aδ(x)dx .
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More mass
Next, consider the region in the plane between the two curvesy = f (x) and y = g(x) for a ≤ x ≤ b and suppose there is alamina (i.e., a thin, flat solid) with the shape of the region. Alsosuppose that the areal density of the material in the lamina varieswith x – say the density of the part of the region over x is δ(x)(grams per square centimeter or something like that). Then themass of the slice of region between x and x + dx is
dm = (f (x)− g(x))δ(x) dx
and so the total mass of the lamina is
m =
ˆ b
a(f (x)− g(x))δ(x) dx .
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3D mass
We can do the same with a solid of revolution around the x-axis ifthe density depends only on x . So if δ(x) is the density (in gramsper cubic centimeter this time), then the mass of the disk orwasher between x and x + dx is
dm = πf (x)2δ(x)dx or dm = π(f (x)2 − g(x)2)δ(x)dx .
So the total mass of the solid is
m =
ˆ b
aπf (x)2δ(x)dx or m =
ˆ b
aπ(f (x)2 − g(x)2)δ(x)dx .
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Averages and weighted averages
You probably learned the formula for the average value of thefunction f (x) for a ≤ x ≤ b:
avg =1
b − a
ˆ b
af (x)dx
Thinking about how this formula works will give us a way togeneralize it quite a bit.
To start, we’ll think of an example:
Section 201 202 203 204 205 206
Students 10 20 15 20 15 20
Avg. grade 8 7 8 9 8 6
The average grade is the average for each section. How can wecalculate the overall class average?
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Class average
Section 201 202 203 204 205 206
Students 10 20 15 20 15 20
Avg. grade 8 7 8 9 8 6
The challenge is that different sections have different numbers ofstudents. But we can get the average because, even though wedon’t know the individual students’ grades, we know for instancethat a total of 10 · 8 = 80 points were achieved by all the studentsin section 201, and 20 · 7 = 140 in section 202, etc. Since there are100 students altogether, we see that the overall class average is:
10 · 8 + 20 · 7 + 15 · 8 + 20 · 9 + 15 · 8 + 20 · 6100
=760
100= 7.6
The essential thing is that we knew the average in each part(section) of the class.
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More generally
We can calculate a weighted average of a set of quantities if wecan divide them up into groups (slices, disks, washers, shells,[infinitesimal] age brackets, etc.) for which
1 We know the average value of the quantity for each group
2 We know the size (area, volume, mass, population,etc) ofeach group
An example: Find the average value of x in the region in the firstquadrant bounded by y = 1− x2
Concentrate on the vertical slice, of width dx overx . Its area is (1− x2)dx and the average value ofx in the slice is simply x . So the total number of“x points” (like “exam points”) in the slice isx(1− x2)dx .
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Getting the average
An example: Find the average value of x in the region in the firstquadrant bounded by y = 1− x2
To get the average, we need to add up all the “x points” in all theslices for 0 ≤ x ≤ 1, and divide by the total area:
avg x =
ˆ 1
0x(1− x2)dx
ˆ 1
0(1− x2)dx
=1423
=3
8
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The average y coordinate
For the same example: Find the average value of y in the regionin the first quadrant bounded by y = 1− x2
To find the average y coordinate, we could use horizontal slicesand proceed the same way as before. But if you think about it, youknow the average y value of one of the vertical slices — it shouldbe the average of the y value at the top of the slice (namely1− x2) and the y value at the bottom (namely 0), which is12(1− x2). So we can proceed as before:
avg y =
ˆ 1
0
1
2(1− x2)(1− x2)dx
ˆ 1
0(1− x2)dx
=41523
=2
5.
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For the record
Just to make sure, we’ll use horizontal slices to find the averagevalue of y in the region:
Since y = 1− x2, we have x =√
1− y on the curve, so the area ofa horizontal slice is
√1− y dy . So the total number of “y points”
in a slice is y√
1− y dy . And so:
avg y =
ˆ 1
0y√
1− y dy
ˆ 1
0
√1− y dy
=41523
=2
5.
(To do both integrals, use the u-substitution u = 1− y).
The point (avg x , avg y) is called the centroid of the region.
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Taking variable density into account
In some problems, the density of the material depends on avariable in a predictable way, and we can take it into account. Itsimpliy provides an additional weighting factor.
The preceding example with variable density: Find thex-coordinate of the center of mass of the region in the firstquadrant bounded by y = 1− x2 if the density is given byδ(x) = 1 + x . (So the region is more “crowded” as you move tothe right)
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Solving the problem
The preceding example with variable density: Find thex-coordinate of the center of mass of the region in the firstquadrant bounded by y = 1− x2 if the density is given byδ(x) = 1 + x .
First we need the total mass:
mass =
ˆ 1
0(1− x2)δ(x) dx =
ˆ 1
0(1− x2)(1 + x) dx
=
ˆ 1
01 + x − x2 − x3 dx =
11
12.
Then we add up all the “weighted x points” and divide by thetotal mass:
avg x =11112
ˆ 1
0x(1−x2)(1+x) dx =
12
11
ˆ 1
0x+x2−x3−x4 dx =
23
55
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The y -coordinate of the center of mass
Example: Now find the y -coordinate of the center of mass of theregion in the first quadrant bounded by y = 1− x2 if the density isgiven by δ(x) = 1 + x .
We can do this using vertical slices as we did before, since thecenter of mass of a vertical slice is still the average of the top andbottom y -coordinates (since the density is constant in the ydirection. Therefore:
avg y =1
mass
ˆ 1
0(avg y in slice)(height of slice)δ(x) dx
=12
11
ˆ 1
0
1
2(1− x2)(1− x2)(1 + x)dx
=12
11· 7
20=
21
55
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Maxwell-Boltzmann
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Maxwell-Boltzmann
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Maxwell-Boltzmann
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Maxwell-Boltzmann
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Maxwell-Boltzmann
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