Lesson 26: The Fundamental Theorem of Calculus (handout)

. V63.0121.001: Calculus I
. Sec on 5.4: The Fundamental Theorem
. May 2, 2011
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Notes
Sec on 5.4
The Fundamental Theorem of
Calculus
V63.0121.001: Calculus I
Professor Ma hew Leingang
New York University

May 2, 2011

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Notes
Announcements

Today: 5.4
Wednesday 5/4: 5.5
Monday 5/9: Review and
Movie Day!
Thursday 5/12: Final
Exam, 2:00–3:50pm

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Notes
Objectives
State and explain the
Fundemental Theorems
of Calculus
Use the first fundamental
theorem of calculus to
find deriva ves of
func ons defined as
integrals.
Compute the average
value of an integrable
func on over a closed
interval.
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. 1
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. V63.0121.001: Calculus I
. May 2, 2011

Notes
Outline
Recall: The Evalua on Theorem a/k/a 2nd FTC
The First Fundamental Theorem of Calculus
Area as a Func on
Statement and proof of 1FTC
Biographies
Differen a on of func ons defined by integrals
“Contrived” examples
Erf
Other applica ons

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Notes
The definite integral as a limit
Defini on
If f is a func on 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 n→∞
i=1

b−a
where ∆x = , and for each i, xi = a + i∆x, and ci is a point in
n
[xi−1 , xi ].

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Notes
The definite integral as a limit

Theorem
If f is con nuous on [a, b] or if f has only finitely many jump
discon nui es, then f is integrable on [a, b]; that is, the definite
∫ b
integral f(x) dx exists and is the same for any choice of ci .
a

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. 2
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. V63.0121.001: Calculus I
. May 2, 2011

Notes
Big time Theorem

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

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Notes
The Integral as Net Change

Corollary
If v(t) represents the velocity of a par cle moving rec linearly, then
∫ t1
v(t) dt = s(t1 ) − s(t0 ).
t0

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Notes

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

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. 3
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. V63.0121.001: Calculus I
. May 2, 2011

Notes

Corollary
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

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Notes
My ﬁrst table of integrals
.
∫ ∫ ∫
[f(x) + g(x)] dx = f(x) dx + g(x) dx
∫ ∫ ∫
xn+1
xn dx = + C (n ̸= −1) cf(x) dx = c f(x) dx
∫ n+1 ∫
1
ex dx = ex + C dx = ln |x| + C
∫ ∫ x
ax
sin x dx = − cos x + C ax 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

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Notes
Outline
Area as a Func on
Biographies
Erf
Other applica ons

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. 4
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. V63.0121.001: Calculus I
. May 2, 2011

Notes
Area as a Function
Example
∫ x
Let f(t) = t3 and define g(x) = f(t) dt. Find g(x) and g′ (x).
0

Solu on
x
Dividing the interval [0, x] into n pieces gives ∆t = and
n
ix
ti = 0 + i∆t = .
n
x x3 x (2x)3 x (nx)3
. Rn = · 3 + · 3 + ··· + · 3
x n n n n n n
0 x4 ( 3 ) x4 [ ]2
. = 4 1 + 2 + 3 + · · · + n = 4 1 n(n + 1)
3 3 3
n n 2
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Notes
Area as a Function
Example
∫ x
Let f(t) = t3 and define g(x) = f(t) dt. Find g(x) and g′ (x).
0

Solu on
x
Dividing the interval [0, x] into n pieces gives ∆t = and
n
ix
ti = 0 + i∆t = .
n
x4 n2 (n + 1)2
. Rn =
x 4n4
0
. x4
So g(x) = lim Rn = and g′ (x) = x3 .
x→∞ 4 .

Notes
The area function in general
Let f be a func on which is integrable (i.e., con nuous or with
finitely many jump discon nui es) 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.
Ques on: What does f tell you about g?
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. 5
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. V63.0121.001: Calculus I
. May 2, 2011

Notes
Envisioning the area function
Example

y
Suppose f(t) is the func on
graphed to the right. Let
∫ x
g(x) = f(t) dt. What can
0
you say about g? .
x
2 4 6 8 10f

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Notes
Features of g from f
y Interval sign monotonicity monotonicity concavity
of f of g of f of g
g [0, 2] + ↗ ↗ ⌣
.
fx
2 4 6 810 [2, 4.5] + ↗ ↘ ⌢
[4.5, 6] − ↘ ↘ ⌢
[6, 8] − ↘ ↗ ⌣
[8, 10] − ↘ → none

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

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Notes
Another Big Time Theorem
Theorem (The First Fundamental Theorem of Calculus)
Let f be an integrable func on on [a, b] and deﬁne
∫ x
g(x) = f(t) dt.
a

If f is con nuous at x in (a, b), then g is diﬀeren able at x and

g′ (x) = f(x).

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. 6
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. V63.0121.001: Calculus I
. May 2, 2011

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

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

Notes
Proving the Fundamental Theorem
Proof.
From §5.2 we have
∫ x+h
mh · h ≤ f(t) dt ≤ Mh · h
x

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

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Meet the Mathematician: James Notes
Gregory
Sco sh, 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

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. 7
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. V63.0121.001: Calculus I
. May 2, 2011

Meet the Mathematician: Isaac Notes
Barrow

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

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Meet the Mathematician: Isaac Notes
Newton

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

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Meet the Mathematician: Gottfried Notes
Leibniz

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

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. 8
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. V63.0121.001: Calculus I
. May 2, 2011

Differentiation and Integration as Notes
reverse processes
Pu ng together 1FTC and 2FTC, we get a beau ful rela onship
between the two fundamental concepts in calculus.
Theorem (The Fundamental Theorem(s) of Calculus)

I. If f is a con nuous func on, then
∫
d x
f(t) dt = f(x)
dx a
So the deriva ve of the integral is the original func on.

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Differentiation and Integration as Notes
reverse processes
Pu ng together 1FTC and 2FTC, we get a beau ful rela onship
between the two fundamental concepts in calculus.
Theorem (The Fundamental Theorem(s) of Calculus)

II. If f is a differen able func on, then
∫ b
f′ (x) dx = f(b) − f(a).
a

So the integral of the deriva ve of is (an evalua on of) the
original func on.
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Notes
Outline
Area as a Func on
Biographies
Erf
Other applica ons

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. 9
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. V63.0121.001: Calculus I
. May 2, 2011

Notes
Differentiation of area functions
Example
∫ 3x
Let h(x) = t3 dt. What is h′ (x)?
0

Solu on (Using 2FTC)
3x
t4 1
h(x) = = (3x)4 = 1
4 · 81x4 , so h′ (x) = 81x3 .
4 0 4

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Notes
Differentiation of area functions
Example
∫ 3x
Let h(x) = t3 dt. What is h′ (x)?
0

Solu on (Using 1FTC)
∫ u
We can think of h as the composi on g ◦ k, where g(u) = t3 dt
0
and 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 = 81x3 .
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Notes
Differentiation of area functions, in general
by 1FTC ∫
d k(x)
f(t) dt = f(k(x))k′ (x)
dx a
by reversing the order of integra on:
∫ ∫
d b d h(x)
f(t) dt = − f(t) dt = −f(h(x))h′ (x)
dx h(x) dx b
by combining the two above:
∫ (∫ ∫ 0 )
d k(x) d k(x)
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)
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. V63.0121.001: Calculus I
. May 2, 2011

Notes
Another Example
Example
∫ sin2 x
Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)?
0

Solu on

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Notes
A Similar Example
Example
∫ sin2 x
Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)?
3

Solu on

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Notes
Compare
Ques on
∫ 2 ∫ 2
d sin x d sin x
Why is (17t2 + 4t − 4) dt = (17t2 + 4t − 4) dt?
dx 0 dx 3
Or, why doesn’t the lower limit appear in the deriva ve?

Answer

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. 11
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. V63.0121.001: Calculus I
. May 2, 2011

Notes
The Full Nasty
Example
∫ ex
Find the deriva ve of F(x) = sin4 t dt.
x3

Solu on

No ce here it’s much easier than finding an an deriva ve for sin4 .
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Notes
Why use 1FTC?
Ques on
Why would we use 1FTC to find the deriva ve of an integral? It
seems like confusion for its own sake.

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

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Notes
Erf
∫ x
2
e−t dt
2
erf(x) = √
π 0

erf measures area the bell curve. erf(x)
We can’t find erf(x), explicitly,
but we do know its deriva ve: .
′ 2 x
erf (x) = √ e−x .
2

π

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. 12
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. V63.0121.001: Calculus I
. May 2, 2011

Notes
Example of erf
Example
d
Find erf(x2 ).
dx

Solu on

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Notes
Other functions deﬁned by integrals
The future value of an asset:
∫ ∞
FV(t) = π(s)e−rs ds
t

where π(s) is the proﬁtability at me 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 func on and p∗ and q∗ the
equilibrium price and quan ty.
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Notes
Surplus by picture
consumer surplus
price (p)
producer surplus
supply

p∗ equilibrium

market revenue

demand f(q)
.
q∗ quan ty (q)
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. 13
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. V63.0121.001: Calculus I
. May 2, 2011

Notes
Summary
Func ons defined as integrals can be differen ated using the
first FTC: ∫
d x
f(t) dt = f(x)
dx a
The two FTCs link the two major processes in calculus:
differen a on and integra on
∫
F′ (x) dx = F(x) + C

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Notes

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Notes

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. 14
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Lesson 26: The Fundamental Theorem of Calculus (handout)

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Lesson 26: The Fundamental Theorem of Calculus (handout)