Download 1 Introduction and Definitions 2 Example: The Area of a Circle

Lecture 1: A Review of Calculus I
the Fundamental Theorem of Calculus
Rahul Krishna
[Preliminaries and Policies]
[Hand out Syllabus]
1
Introduction and De…nitions
Calculus revolves around two fundamental questions: …rst, how to calculate the tangent line to a smooth
curve, and second, how to compute the area under a smooth curve. These lead, respectively, to the notions
of derivatives and integrals.
Let’s quickly review integral calculus quickly. I’m going to base the review on a single and very old
example:
2
Example: The Area of a Circle
Let D be a disc of radius R; and denote the area of D by A: Recall the ancient formula:
A = R2
Why is this true? Well, if we approximate the disc by circumscribed [polygon on inside] n-gons An [draw
Picture of circle with inscribed triangle and square] [draw picture of one swoop of angle 2 =n], we see that
the polygon is made of n triangles of area R2 cos n sin n : Taking n ! 1; we …nd the area of the n-gons An
tends towards
lim Area (An ) = lim nR2 cos
n!1
n!1
= R2
lim cos
n!1
n
sin
n
n
lim
n!1
sin =n
=n
= R2 (1) ( ) = R2
since limx!0 sinx x = 1:
If we approximate the disc by inscribed [polygon on the outside] n-gons Bn ; we see that the polygon is
sin
made of n triangles of area R2 cos n : Then the area of the big n-gons Bn tends to
n
lim Area (Bn ) = lim nR2
n!1
n!1
= R2
lim
1
sin
cos n
n
n!1
1
cos n
lim
n!1
sin =n
=n
= R2 (1) ( ) = r2
so we have Area(An )
Area(D)
Area(Bn ) and limn!1 Area(An ) = limn!1 Area(Bn ) = R2 : By
2
squeeze, Area(D) = R :
What can we learn from this example, besides that the area of a circle is R2 ? Well, there are a few
ideas present.
If we want to compute areas, we need to approximate areas we can’t calculate (of continuous objects
like discs) by pieces whose areas we can compute. Above, we used triangles. (Funny point: what objects
do we know the areas of? Rectangles. Triangles, which are half-rectangles. Anything else? So the idea
is to reduce hard problems to lots of easy problems by approximation.) [Discrete/continuous table.]
1
The importance of drawing pictures. Enough said.
If we want to formally de…ne areas of smooth objects, we should have a lower bound and an upper
bound which "sandwich" the actual value. Above, we had the n-gons An for our lower bound, and Bn
for our upper bound.
Di¤erent ways of approaching a problem can be enlightening.
Limiting processes are at the heart of approximating continuous by discrete. [Draw Table of Continuous
v. Discrete]
The general method for computing areas of continuous objects is horrendous, even for this simple
example. Is there a way to cheat? (There is, and it’s called the Fundamental Theorem of Calculus.)
Later we will generalize this calculation to areas in polar coordinates.
3
3.0.1
De…nitions
Limits [Can cut this.]
Let me write down the formal de…nition of a limit of a sequence now and discuss if for a minute.
De…nition 1 (The limit of a sequence.) Let x1 ; x2 ; ::: be an in…nite sequence of real numbers. We say that
x1 ; x2 ; ::: converges to a limit x if, for all " > 0; there exists an N = N (") so that for all n N;
jx
xn j < "
We write xn ! x or limn!1 xn = x:
1
Notation 1 If I slip up, it’s likely that I’ll denote x1 ; x2 ; ::: as fxn gn=1 ; or, even shorter, as fxn g. Keep
in mind that this is just notation, so you shouldn’t be scared of it. However, I’ll try to avoid building up an
excessive amount of notation since that can get confusing.
This seems like an awfully di¢ cult de…nition to parse, so let me explain it. [Draw picture of a limit.]
The de…nition says that xn ! x if, for any strictly positive (think small!) tolerance " > 0; there is a point,
depending on the tolerance, after which every element of the sequence lies within that tolerance.
3.0.2
Back to Area I: Integrals
Here’s a related problem to calculating the area of a circle: calculating the (signed) area under a curve.
Using what we know from the proof above, we might want to approximate an area A under a curve with
a Riemann Sum. [Draw Picture.] For simplicity, let f (x) be a positive function on an interval [a; b]: We
want a lower bound and an upper bound on the area. If we take a partition a = x0 < x1 < ::: < xn = b; and
set xk = xk xk 1 ; then we have
n
X
f xmin
k
xk
Area (A)
k=1
n
X
f (xmax
) xk
k
k=1
where xk 1
xmin
xk satis…es f xmin
= minxk 1 x xk f (x) and where xk 1
xmax
xk satis…es
k
k
k
max
max
min
f (xk ) = maxxk 1 x xk f (x) : [Continue picture, labelling xk and xk in each partition interval.] This
is good, but it’s too dependent on the partition we take.
2
Example 1 f (x) = x2 on the interval [0; 1]: First we take the partition x0 = 0; x1 = 41 ; x2 = 12 ; x3 = 34 ;
x4 = 1: With this partition, we have, since xmin
= xk 1 that
k
4
X
f xmin
k
xk = f (0)
k=1
1
4
1
+f
4
1
1
+
4
16
14 1
7
=
=
16 4
32
= (0)
1
+f
4
1
2
1
+f
4
1
+
4
4
16
1
+
4
1
2
1
+f
4
3
4
3
4
9
16
1
4
1
4
while since xmax
= xk ;
k
4
X
f (xmax
) xk = f
k
k=1
1
+f
4
1
4
1 1
+
16 4
15
30 1
=
=
16 4
32
4
16
=
1
+
4
9
16
1
1
+ f (1)
4
4
1
+
4
16
16
1
4
Now, suppose we take the partition x00 = 1; x01 = 15 ; x02 = 25 ; x03 = 53 ; x04 = 45 ; x05 = 1: We get
5
X
min
f (x0k )
x0k = f (0)
k=1
1
+f
5
1
1
+
5
25
30 1
6
=
=
25 5
25
= (0)
1
5
1
+f
5
1
+
5
2
5
1
+f
5
3
5
1
+f
5
4
5
16
25
1
5
4
25
1
+
5
9
25
1
+
5
1
+f
5
3
5
1
+f
5
4
5
1
5
while
5
X
max
f (x0k )
x0k = f
k=1
1
5
1
+f
5
2
5
1
1
+ f (1)
5
5
1 1
4 1
9 1
16 1
25 1
+
+
+
+
25 5
25 5
25 5
25 5
25 5
55 1
11
=
=
25 5
25
So the approximate areas depend quite a lot on the partitions we take to calculate them. The way to
resolve this is in the following:
Rb
De…nition 2 Let f (x) be a function on an interval [a; b]: We de…ne the lower integral
f (x) dx to be
a
the maximum of the lower Riemann sum approximations
=
n
X
f xmin
k
xk
k=1
where the maximum is taken over all possible partitions a = x0 < x1 < ::: < xn = b: We similarly de…ne the
Rb
upper integral a f (x) to be the minimum of the upper approximations
n
X
f (xmax
) xk
k
k=1
over all partitions of [a; b]:
3
De…nition 3 A function f (x) is integrable on [a; b] if its lower integral is equal to its upper integral. We
Rb
set its de…nite integral a f (x) dx to be this value.
This is the just the squeeze theorem! We’ve bounded the area below by a sequence of lower Riemann
sums, and above by a sequence of upper Riemann sums, and we say the area is well de…ned (the functions
is integrable) if the two squeeze together in the limit.
Note that this de…nition, while rigorous, is incredibly di¢ cult to apply. So the question remains: how
can we cheat?
Before we go on, let me note something we are not going to prove, but will use:
Theorem 1 Continuous functions are integrable.
I might write up a proof of this and post it if there is any interest in seeing it done, since it is incredibly
important. But for now, we’ll skip it.
3.0.3
A Litle Review: Derivatives
Recall that the derivative of a function f at a point x0 is de…ned by (assuming the limit exists–if it does, we
say f is di¤erentiable at x0 )
f 0 (x0 ) =
f (x0 + h)
df
(x0 ) := lim
h!0
dx
h
f (x0 )
[Draw picture.] Note that this exactly gives us the slope of the tangent line to the graph y = f (x) at the
point (x0 ; f (x0 )) since we approximate the tangent line by secant lines, which tend towards the tangent line.
[What is the equation of the tangent line to y = f (x) at (x0 ; f (x0 ))? Answer: y f (x0 ) = f 0 (x0 ) (x x0 ) :]
d
By varying x0 ; we get a function dx
f (x) := f 0 (x) which we call the derivative of f: This function is
the instantaneous rate of change of f at x:
Remark 1 This leads to the notion of best linear approximation, which is the most transparent way to
translate the de…nition of a derivative to higher dimensions. Other ways seem slightly unnatural. DO NOT
WORRY ABOUT THIS NOW.
d n
x . Well, if we remember anything from Calculus I, we know
Example 2 (Power functions.) Compute dx
n 1
the answer should be nx
: Let’s look at the argument why again:
n
d n
(x + h)
xn
x = lim
h!0
dx
h
1
n n
n n
= lim
x +
x
h!0 h
0
1
since
n
0
= 1 and
n
1
1
h+
n n
x
2
h + ::: +
n n
h
n
n n
h
n
= nxn
2 2
xn
= n; we have
d n
x = lim nxn
n!1
dx
1
+
n n
x
2
2
h + ::: +
1
1
The derivative has some properties, which I will list here (You should know all of them already! Can you
prove them?):
1.
d
dx f
2.
d
dx
(f (x) + g (x)) =
3.
d
dx
d
(cf (x)) = c dx
f (x)
(x) = 0 if and only if f (x) = c is a constant function.
d
dx f
(x) +
d
dx g (x)
4
4. (Product Rule)
5. (Quotient Rule)
the lo]
6. (Chain Rule)
d
dx
d
dx
d
dx f
d
dx f
(f (x) g (x)) =
(f (x) =g (x)) =
(x) g (x) + f (x)
d
d
g(x) dx
f (x) f (x) dx
g(x)
g(x)2
d
dx g (x)
[Lo-di-hi-minus-hi-di-lo draw the line square
(g (x)) = f 0 (g (x)) g 0 (x)
Remark 2 The …rst few lectures after this will be "running these rules backwards." This will be clear soon
enough.
Example 3 Let’s di¤ erentiate f (x) = 2 cos x +
f 0 (x) =
2 sin x
=
2 sin x
1
x
ln sin x: This gives
1
ln sin x +
x2
1
ln sin x +
x2
1
1
(cos x)
x
sin x
1
cot x
x
Some additional stu¤:
Remark 3 The derivative is used to maximize and minimize. Points where f 0 (x) = 0 are critical points,
and are often local maxima and minima (can also be in‡ection points). I assume everyone remembers this.
Remark 4 As a rate of change, the derivative of position is velocity, the derivative of velocity is acceleration,
etc.
4
Back to Area II: The Fundamental Theorem of Calculus
Suppose f is a continous function on an interval [a; b] : Then it is integrable by the unproved theorem above,
so the function de…ned by
A (x) = area under f from a to x
Z x
=
f (t) dt
a
is well de…ned.
[Draw picture with f; A (x)] Note that
A (x + h)
A (x) = hf (x )
where f (x ) is the mean value of f on the interval [x; x + h] and x
explicit, and aim only for inequalities, like we did above:
f xmin
h
where xmin
is chosen so f xmin
= minx
h
h
have
y x+h
f (x )
x
x + h: But we could be less
f (xmax
)
h
f (y) and xmax
has f (xmax
) = maxx
h
h
y x+h
f (y) : Hence we
A (x + h) A (x)
f (xmax
)
h
h
max
Now, as h ! 0; xmin
! x which implies, if f is continuous, that f xmin
; f (xmax
) ! f (x) : Therefore,
h ; xh
h
h
if f is continuous, we …nd that A (x) is di¤erentiable and
f xmin
h
A (x + h)
h!0
h
f (x) = lim
A (x)
=
This is the …rst version of the Fundamental Theorem of Calculus:
5
d
A (x)
dx
Theorem 2 (Fundamental Theorem of Calculus, RVersion I) Let f be a continuous function on an interval
x
[a; b]: Then for all x 2 [a; b] ; the function A (x) = a f (t) dt is di¤ erentiable at x; and
d
dx
Example 4 Let g (x) =
R px
0
Z
x
f (t) dt = f (x)
a
d
dx g (x)
sin tdt: Then
p 1
= sin x 2p
x
There is another (more useful?) form.
Theorem 3 (Fundamental Theorem of Calculus, Version II) Let f be a function on an interval [a; b] and
let F (x) be a di¤ erentiable function such that F 0 (x) = f (x) : Then
Z
b
f (x) dx = F (b)
F (a)
a
Rx
Proof. By the …rst form of the Fundamental Theorem, we know that g (x) = a f (t) dt is an antiderivative
d
g (x) = f (x) : Any other antiderivative must di¤er from g (x) by a constant c:
for f; that is, it satis…es dx
Rb
Thus F (x) = g (x) + c: Now we want to compute g (b) = a f (x) dx; so, after noting that g (a) = 0; we have
Z
b
f (x) dx = g (b)
a
= g (b) g (a)
= (F (b) c) (F (a)
c) = F (b)
F (a)
This says something about net change. Basically, the fundamental theorem says the net change is the
sum of the in…ntesimal changes. [Refer to discrete/continuous table.]
Example 5 Suppose that a particle moves along a line with velocity v (t) = sin t + 1t m=s from time t = 1
to t = 2 : Let’s calculate the total displacement between these two times. We …nd an antiderivative (only
good up to constant!): x (t) = cos t + ln t: Then the displacement is
x (2 )
x (1) = cos 2 + ln 2
= cos 1 1 + ln 2
( cos 1 + ln 1)
meters.
4.1
Old Example Revisited:
The power of calculus: again, let D be a disc of radius R: Then
Area (D) =
Z
R
0
= r2
Isn’t that easier!
Remark: Shells versus discs?
6
2 rdr