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Welcome to Math 103:207Integral Calculus with Applications to the Life
Sciences
Cole Zmurchokhttp://www.math.ubc.ca/~zmurchok
January 6, 2015
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January 6, 2015
Introductions & Information
Integral Calculus & the Life Sciences
Areas and sums
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Introductions & Information
http://www.math.ubc.ca/~zmurchok
Course Information:
I Schedule & Important Dates
I Notes
I WebWork & Old-School Homework (OSH)
I Pizza
I Resources
Office Hours: Wednesday 9:30 to 10:30 and Thursday 9:00 to10:00 in LSK 300BExpectations:
I Phones & laptops
I Homework professionally completed
I Lecture preparation: read & work
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Integral Calculus & cats
Cats tend to land on their feetwhen falling (0.3 seconds to flipover).
Question
What is the minimum heightfrom which a cat can fall toensure it lands on its feet?
Annals of Improbable ResearchDiamond, J. (1988) Why cats have nine lives, Nature 332.Thanks to Joseph M. Mahaffy for the falling cat idea.
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Boardwork: Falling Cat
Newton’s law of motion provides a model:
ma = −mg
I m is the mass of the cat
I a is the acceleration of cat
I −mg is the force of gravity (g = 980.7 cm/sec2)
Let h(t) be the height of the cat at time t. Initially the cat is atrest at height h0:
I h(0) = h0
I v(0) = h′(0) = 0 (initial velocity)
h′′ = −g.
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Boardwork: Falling Cat
Main theme of this course is “integral calculus”. The integral isthe inverse operator of the derivative:∫
f(x)dxddx−−→ f(x)
ddx−−→ f ′(x)
Integrating h′′ = −g, we find
v(t) = h′(t) = −gt,
which satisfies v(0) = 0. Integrating h′ = −gt, we find theheight of the cat at time t:
h(t) =−gt2
2+ h0,
which satisfies h(0) = h0.
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Boardwork: Falling Cat
The height of the cat at time 0.3 seconds is h(0.3):
h(0.3) =−g(0.3)2
2+ h0 = −44.1315 + h0,
so the cat must be above 44.1315 cm to have sufficient time toflip over before hitting the ground.
Note
This is consistent with the Annals of Improbable Research data,which suggests that a cat dropped upside down from a height of2 to 6 feet (30.84 cm to 182.88 cm) will always land on its feet.
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Boardwork: Areas and sums
Self-study
Review how to find the area, volume, perimeters of basic shapes(Section 1.1 to 1.3)
Integral calculus originated as a method to calculate the areaand volume of everyday objects (size of plot of land, volume ofbarrel, etc.).The fundamental idea is to cut up the geometric shape intosmaller pieces, and approximate those smaller pieces by regularshapes that can be easily quantified.
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Areas and sums
Example Area under y = x2 from 0 to 1.
Dissect the area under the curve into N pieces, each with width1N . The kth piece can be approximated by a rectangle with
dimensions 1N by
(kN
)2.
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Areas and sums
Thus, the area, A, is approximately
A ≈ 1
N
(1
N
)2
+1
N
(2
N
)2
+ · · ·+ 1
N
(N − 1
N
)2
+1
N
(N
N
)2
=1
N3
(12 + 22 + 32 + · · ·+ (N − 1)2 + N2
)=
1
N3
N(N + 1)(2N + 1)
6
=1
6
(1 +
1
N
)(2 +
1
N
).
As N →∞, A→ 13 . This is the key idea behind integration,
which we will make precise next week.
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Sums
Summation notation: a.k.a. sigma notation
ak + ak+1 + · · ·+ an =
n∑j=k
aj
Here, the sum is indexed by the auxiliary variable j, which wecould choose to be anything, e.g. ♣ or ∆. k signifies the termwhich starts off the series, and n signifies when the series ends.
Example Write the sum 3 + 6 + 11 + 18 in sigma notation.
3 + 6 + 11 + 18 =4∑
i=1
(i2 + 2)
works. Can you find another expression that works?
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SumsExample
I Find the value of 3 + 6 + 11 + 18 =∑4
i=1(i2 + 2).
3 + 6 + 11 + 18 = 38
or
4∑i=1
(i2 + 2) =
4∑i=1
i2 +
4∑i=1
2 =4(4 + 1)(2 · 4 + 1)
6+ 4 · 2 = 38.
Important Formula
n∑i=1
i =n(n + 1)
2,
n∑i=1
i2 =n(n + 1)(2n + 1)
6,
n∑i=1
i3 =
(n(n + 1)
2
)2
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Gauss’ formula
Theorem (Gauss’s formula)
n∑i=1
i =n(n + 1)
2
Proof.
2
n∑i=1
i = 1 + 2 + 3 + · · ·n− 2 + n− 1 + n
+ n + n− 1 + n− 2 + · · ·+ 3 + 2 + 1
= n + 1 + n + 1 + n + 1 + · · ·+ n + 1 + n + 1 + n + 1
= n(n + 1)
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SumsExample
n∑i=1
i3 + i2 + i + 1 =
n∑i=1
i3 +
n∑i=1
i2 +
n∑i=1
i +
n∑i=1
1
=
(n(n + 1)
2
)2
+n(n + 1)(2n + 1)
6
+n(n + 1)
2+ n
Example
5∑k=2
(k − 1)3
Use a substitution (change summation index), i.e., let j = k− 1:
5∑k=2
(k − 1)3 =4∑
j=1
j3 =
(4(4 + 1)
2
)2
= 100
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Sums
Example
n∑i=1
(i− 1)2 =
n∑i=1
i2 − 2i + 1 =
n∑i=1
i2 −n∑
i=1
2i +
n∑i=1
1
=n(n + 1)(2n + 1)
6− 2
n∑i=1
i + n
=n(n + 1)(2n + 1)
6− n(n + 1) + n
Alternately:
n∑i=1
(i− 1)2 =
n−1∑j=0
j2 = 0 +
n−1∑j=1
j2
=(n− 1)(n− 1 + 1)(2(n− 1) + 1)
6=
n(n− 1)(2n− 1)
6which is the same!
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Sums
Example Sums can have infinitely many terms, yet can still bere-indexed
1− 2 + 3− 4 + 5− 6 + · · · =∞∑j=1
(−1)j+1j =
∞∑j=0
(−1)j(j + 1)
Example Telescoping sum
n∑k=1
1
k(k + 1)=
n∑k=1
(1
k− 1
k + 1
)= 1− 1
2+
1
2− 1
3+
1
3− 1
4+
1
4− 1
5+ · · ·+ 1
n− 1
n + 1
= 1− 1
n + 1
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Sums
Example If∑n
k=1 ak = A,∑n
k=1 bk = B, and c, d ∈ R, find∑nk=1 cak + dbk.
n∑k=1
cak + dbk =
n∑k=1
cak +
n∑k=1
dbk = c
n∑k=1
ak + d
n∑k=1
bk = cA+ dB
This property is called linearity.
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Lecture 1 Ending
Recap:
I Areas and sums
I Prelude to integration
I Gauss’ formula
I Working with summation notation
Questions?
For next class
Try to do the following problems, we will start Thursday’s classwith them.
I Find the sum of 1 + 12 + 1
4 + 18 + · · ·+ (12)n.
I Find the sum of 1 + 2 + 22 + 23 + · · ·+ 2n.
I Review formulas for areas & volumes.
I Read Chapter 1.