Lesson 26: The Fundamental Theorem of Calculus (Section 021 handout)

V63.0121.021, Calculus I Section 5.4 : The Fundamental Theorem December 9, 2010

Section 5.4
Notes
The Fundamental Theorem of Calculus

V63.0121.021, Calculus I

New York University

December 9, 2010

Announcements

Today: Section 5.4
”Thursday,” December 14: Section 5.5
”Monday,” December 15: (WWH 109, 12:30–1:45pm) Review and
Movie Day!
Monday, December 20, 12:00–1:50pm: Final Exam (location still
TBD)

Announcements
Notes

Today: Section 5.4
”Thursday,” December 14:
Section 5.5
”Monday,” December 15:
(WWH 109, 12:30–1:45pm)
Review and Movie Day!
Monday, December 20,
12:00–1:50pm: Final Exam
(location still TBD)

V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 2 / 32

Objectives
Notes

State and explain the
Fundemental Theorems of
Calculus
Use the first fundamental
theorem of calculus to find
derivatives of functions
defined as integrals.
Compute the average value
of an integrable function
over a closed interval.


1


Outline
Notes

Recall: The Evaluation Theorem a/k/a 2nd FTC

The First Fundamental Theorem of Calculus
Area as a Function
Statement and proof of 1FTC
Biographies

Differentiation of functions defined by integrals
“Contrived” examples
Erf
Other applications


The definite integral as a limit
Notes

Definition
If f is a function defined on [a, b], the definite integral of f from a to b
is the number
b n
f (x) dx = lim f (ci ) ∆x
a ∆x→0
i=1


Big time Theorem
Notes

Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a, b] and f = F for another function F , then
b
f (x) dx = F (b) − F (a).
a


2


The Integral as Total Change
Notes

Another way to state this theorem is:
b
F (x) dx = F (b) − F (a),
a

or the integral of a derivative along an interval is the total change between
the sides of that interval. This has many ramiﬁcations:

Theorem
If v (t) represents the velocity of a particle moving rectilinearly, then
t1
v (t) dt = s(t1 ) − s(t0 ).
t0


Notes

b
F (x) dx = F (b) − F (a),
a


Theorem
If MC (x) represents the marginal cost of making x units of a product, then
x
C (x) = C (0) + MC (q) dq.
0


Notes

b
F (x) dx = F (b) − F (a),
a


Theorem
If ρ(x) represents the density of a thin rod at a distance of x from its end,
then the mass of the rod up to x is
x
m(x) = ρ(s) ds.
0


3


My ﬁrst table of integrals
Notes

[f (x) + g (x)] dx = f (x) dx + g (x) dx

x n+1
x n dx = + C (n = −1) cf (x) dx = c f (x) dx
n+1
1
e x dx = e x + C dx = ln |x| + C
x
x ax
sin x dx = − cos x + C a dx = +C
ln a

cos x dx = sin x + C csc2 x dx = − cot x + C

sec2 x dx = tan x + C csc x cot x dx = − csc x + C
1
sec x tan x dx = sec x + C √ dx = arcsin x + C
1 − x2
1
dx = arctan x + C
1 + x2


Outline
Notes


Area as a Function
Biographies

Erf
Other applications


Area as a Function
Notes
Example
x
Let f (t) = t 3 and deﬁne g (x) = f (t) dt. Find g (x) and g (x).
0

Solution

Dividing the interval [0, x] into n pieces
x ix
gives ∆t = and ti = 0 + i∆t = . So
n n
x x 3 x (2x)3 x (nx)3
Rn = · + · + ··· + ·
n n3 n n3 n n3
x4 3 3 3 3
= 4 1 + 2 + 3 + ··· + n
n
x x4 1 2
0 = 4 2 n(n + 1)
n


4


Area as a Function
Notes
Example
x
Let f (t) = t 3 and define g (x) = f (t) dt. Find g (x) and g (x).
0

Solution

Dividing the interval [0, x] into n pieces
x ix
gives ∆t = and ti = 0 + i∆t = . So
n n
x 4 n2 (n + 1)2
Rn =
4n4
x4
So g (x) = lim Rn = and g (x) = x 3 .
x→∞ 4
0 x


The area function in general
Notes

Let f be a function which is integrable (i.e., continuous or with finitely
many jump discontinuities) on [a, b]. Define
x
g (x) = f (t) dt.
a

The variable is x; t is a “dummy” variable that’s integrated over.
Picture changing x and taking more of less of the region under the
curve.
Question: What does f tell you about g ?


Envisioning the area function
Notes

Example
Suppose f (t) is the function graphed below:
y

x
2 4 6 8 10f

x
Let g (x) = f (t) dt. What can you say about g ?
0


5


features of g from f
Notes

y
Interval sign monotonicity monotonicity concavity
of f of g of f of g
g
[0, 2] +
x
2 4 6 8 10f [2, 4.5] +
[4.5, 6] −
[6, 8] −
[8, 10] − → none

We see that g is behaving a lot like an antiderivative of f .


Another Big Time Theorem
Notes

Theorem (The First Fundamental Theorem of Calculus)
Let f be an integrable function on [a, b] and deﬁne
x
g (x) = f (t) dt.
a

If f is continuous at x in (a, b), then g is diﬀerentiable at x and

g (x) = f (x).


Proving the Fundamental Theorem
Notes
Proof.
Let h > 0 be given so that x + h < b. We have
x+h
g (x + h) − g (x) 1
= f (t) dt.
h h x

Let Mh be the maximum value of f on [x, x + h], and let mh the minimum
value of f on [x, x + h]. From §5.2 we have
x+h
mh · h ≤ f (t) dt ≤ Mh · h
x

So
g (x + h) − g (x)
mh ≤ ≤ Mh .
h
As h → 0, both mh and Mh tend to f (x).


6


Meet the Mathematician: James Gregory
Notes

Scottish, 1638-1675
Astronomer and Geometer
Conceived transcendental
numbers and found evidence
that π was transcendental
Proved a geometric version
of 1FTC as a lemma but
didn’t take it further


Meet the Mathematician: Isaac Barrow
Notes

English, 1630-1677
Professor of Greek, theology,
and mathematics at
Cambridge
Had a famous student


Meet the Mathematician: Isaac Newton
Notes

English, 1643–1727
Professor at Cambridge
(England)
Philosophiae Naturalis
Principia Mathematica
published 1687


7


Meet the Mathematician: Gottfried Leibniz
Notes

German, 1646–1716
Eminent philosopher as well
as mathematician
Contemporarily disgraced by
the calculus priority dispute


Diﬀerentiation and Integration as reverse processes
Notes
Putting together 1FTC and 2FTC, we get a beautiful relationship between
the two fundamental concepts in calculus.
Theorem (The Fundamental Theorem(s) of Calculus)

I. If f is a continuous function, then
x
d
f (t) dt = f (x)
dx a

So the derivative of the integral is the original function.
II. If f is a diﬀerentiable function, then
b
f (x) dx = f (b) − f (a).
a

So the integral of the derivative of is (an evaluation of) the original
function.

Outline
Notes


Area as a Function
Biographies

Erf
Other applications


8


Diﬀerentiation of area functions
Notes
Example
3x
Let h(x) = t 3 dt. What is h (x)?
0

Solution (Using 2FTC)
3x
t4 1
h(x) = = (3x)4 = 1
4 · 81x 4 , so h (x) = 81x 3 .
4 0 4

Solution (Using 1FTC)
u
We can think of h as the composition g ◦ k, where g (u) = t 3 dt and
0
k(x) = 3x. Then h (x) = g (u) · k (x), or

h (x) = g (k(x)) · k (x) = (k(x))3 · 3 = (3x)3 · 3 = 81x 3 .


Diﬀerentiation of area functions, in general
Notes

by 1FTC
k(x)
d
f (t) dt = f (k(x))k (x)
dx a
by reversing the order of integration:
b h(x)
d d
f (t) dt = − f (t) dt = −f (h(x))h (x)
dx h(x) dx b

by combining the two above:

k(x) k(x) 0
d d
f (t) dt = f (t) dt + f (t) dt
dx h(x) dx 0 h(x)

= f (k(x))k (x) − f (h(x))h (x)


Another Example
Notes

Example
sin2 x
Let h(x) = (17t 2 + 4t − 4) dt. What is h (x)?
0

Solution


9


A Similar Example
Notes

Example
sin2 x
Let h(x) = (17t 2 + 4t − 4) dt. What is h (x)?
3

Solution


Compare
Notes
Question
Why is
sin2 x sin2 x
d d
(17t 2 + 4t − 4) dt = (17t 2 + 4t − 4) dt?
dx 0 dx 3

Or, why doesn’t the lower limit appear in the derivative?

Answer


The Full Nasty
Notes

Example
ex
Find the derivative of F (x) = sin4 t dt.
x3

Solution

Notice here it’s much easier than ﬁnding an antiderivative for sin4 .


10


Why use 1FTC?
Notes

Question
Why would we use 1FTC to find the derivative of an integral? It seems
like confusion for its own sake.

Answer

Some functions are difficult or impossible to integrate in elementary
terms.
Some functions are naturally defined in terms of other integrals.


Erf
Notes
Here’s a function with a funny name but an important role:
x
2 2
erf(x) = √ e −t dt.
π 0
It turns out erf is the shape of the bell curve. We can’t find erf(x),
2 2
explicitly, but we do know its derivative: erf (x) = √ e −x .
π
Example
d
Find erf(x 2 ).
dx

Solution


Other functions defined by integrals
Notes

The future value of an asset:
∞
FV (t) = π(s)e −rs ds
t

where π(s) is the profitability at time s and r is the discount rate.
The consumer surplus of a good:
q∗
CS(q ∗ ) = (f (q) − p ∗ ) dq
0

where f (q) is the demand function and p ∗ and q ∗ the equilibrium
price and quantity.


11


Surplus by picture
Notes
consumer surplus
price (p)
producer surplus
supply

p∗ equilibrium

market revenue

demand f (q)

q∗ quantity (q)


Summary
Notes

Functions defined as integrals can be differentiated using the first
FTC: x
d
f (t) dt = f (x)
dx a
The two FTCs link the two major processes in calculus: differentiation
and integration
F (x) dx = F (x) + C

Follow the calculus wars on twitter: #calcwars


Notes

12

Lesson 26: The Fundamental Theorem of Calculus (Section 021 handout)

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